ELEMENTARY TOPOLOGY I 1. Introduction 3 2. Metric Spaces 4 2.1

ELEMENTARY TOPOLOGY I

ALEX GONZALEZ

1. Introduction3 2. Metric spaces4 2.1. Continuous functions8 2.2. Limits 10 2.3. Open subsets and closed subsets 12 2.4. Continuity, convergence, and open/closed subsets 15 3. Topological spaces 19 3.1. Interior, closure and boundary 21 3.2. Continuous functions 23 3.3. Basis of a topology 25 3.4. The subspace topology 26 3.5. The product topology 28 3.6. The quotient topology 31 3.7. Other constructions 33 4. Connectedness 34 4.1. Path-connected spaces 38 4.2. Connected components 40 5. Compactness 41 5.1. Compact subspaces of Rn 44 6. Countability and separation axioms 47 6.1. The countability axioms 47 6.2. The separation axioms 49 REFERENCES 54 Contents

1 2 ALEX GONZALEZ ELEMENTARY TOPOLOGY I 3

1. Introduction

When we consider properties of a “reasonable” function, probably the first thing that comes to mind is that it exhibits continuity: the behavior of the function at a certain point is similar to the behavior of the function in a small neighborhood of the point. What’s more, the composition of two continuous functions is also continuous. Usually, when we think of a continuous functions, the first examples that come to mind are maps f : R → R: • the identity function, f(x) = x for all x ∈ R; • a constant function f(x) = k; • polynomial functions, for instance f(x) = xn, for some n ∈ N; • the exponential function g(x) = ex; • trigonometric functions, for instance h(x) = cos(x). The set of real numbers R is a natural choice of domain to begin to study more general properties of continuous functions. After all, we are familiar with many of the properties of the real line, and it is (relatively) easy to draw the graphs of functions R → R. So, let’s review the definition of continuity for a function f : R → R: the function f is continuous at the point a ∈ R if lim f(x) = f(a) x→a (in particular we are assuming that this limit exists!). Another way of putting the above definition is the following: for each ε > 0 there exists some δ > 0 such that |f(x) − f(a)| < ε for all x ∈ R such that |x − a| < δ. The function f is then said to be continuous on all R if it is continuous for all a ∈ R. So, basically, the definition of continuity depends only on the absolute value, which is essentially a rule to measure the distance between any two numbers in R. It would seem that continuity relies on the existence of a rule to measure distances, or more precisely, a metric. 4 ALEX GONZALEZ

2. Metric spaces

Let X be a set. Roughly speaking, a metric on the set X is just a rule to measure the distance between any two elements of X.

Deﬁnition 2.1. A metric on the set X is a function d: X × X → [0, ∞) such that the following conditions are satisﬁed for all x, y, z ∈ X:

(M1) Positive property: d(x, y) = 0 if and only if x = y;

(M2) Symmetry property: d(x, y) = d(y, x); and

(M3) Triangle inequality: d(x, y) ≤ d(x, z) + d(z, y).

A note of waning! The same set can be given different ways of measuring distances. Strange as it may seem, the set R2 (the plane) is one of these sets. We will see different metrics for R2 pretty soon. Example 2.2. The set X = R with d(x, y) = |x−y|, the absolute value of the difference x − y, for each x, y ∈ R. Properties (M1) and (M2) are obvious, and d(x, y) = |x − y| = |(x − z) + (z − y)| ≤ |x − z| + |z − y| = d(x, z) + d(z, y).

Let’s see now some examples of metrics on the set X = R2. Example 2.3. The set X = R2 with p 2 2 d (x1, x2), (y1, y2) = (x1 − y1) + (x2 − y2) 2 for each (x1, x2), (y1, y2) ∈ R . Again, properties (M1) and (M2) are easy to check. Property (M3), the triangle inequality, gets its name from this example. Write x = (x1, x2), y = (y1, y2) and z = (z1, z2). Then,

x d(x, y)

d(x, z) y

d(z, y)

Can you think of other examples of metrics on R2? If we compare the metrics in Examples 2.2 and 2.3, we can see that the metric on R2 is some sort of “extension” of the metric on R1 to a higher dimension. But we can try other ways of extending it. ELEMENTARY TOPOLOGY I 5

Example 2.4. The functions d, d0 : R2 × R2 → [0, ∞) deﬁned by d (x1, x2), (y1, y2) = |x1 − y1| + |x2 − y2| 0 d (x1, x2), (y1, y2) = max{|x1 − y1|, |x2 − y2|} are also metrics on R2 (the details will be checked in Examples 2.5 and 2.8 below).

We have just see three diﬀerent metrics on R2. But why stopping at R2? We can extend these metrics to Rn for all n ≥ 1.

Example 2.5. Let X = Rn and let d: Rn × Rn → [0, ∞) be deﬁned by

n X d (x1, x2, . . . , xn), (y1, y2, . . . , yn) = |xi − yi|. i=1

Then, d is a metric on Rn (sometimes known by the name of Manhattan metric). Let’s check the details: conditions (M1) and (M2) follow easily. As for the triangle inequality, we have

Example 2.6. The Euclidean metric on Rn is deﬁned by the formula v u n uX 2 d (x1, x2, . . . , xn), (y1, y2, . . . , yn) = t (xi − yi) , i=1

n for each (x1, x2, . . . , xn), (y1, y2, . . . , yn) ∈ R . As usual (M1) and (M2) are easy to check, but (M3) is not trivial at all.

n Indeed, let x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) and z = (z1, z2, . . . , zn) ∈ R : Let also ri = xi − zi and si = zi − yi, for i = 1, . . . , n. We have to prove that v v v u n u n u n uX 2 uX 2 uX 2 d(x, y) = t (ri + si) ≤ t ri + t si = d(x, z) + d(z, y) (1) i=1 i=1 i=1 6 ALEX GONZALEZ

Notice that both sides of the inequality are positive. Thus, by squaring the above, it is equivalent to prove that v v n n n u n u n X 2 X 2 X 2 uX 2uX 2 (ri + si) ≤ ri + si + 2t ri t si . (2) i=1 i=1 i=1 i=1 i=1 The left part of the inequality expands to n n n n X 2 X 2 X 2 X (ri + si) = ri + si + 2 risi. i=1 i=1 i=1 i=1 Replacing this in (2) and simplifying, we deduce that (1) holds if and only if n 2 n n X X 2 X 2 risi ≤ ri si . i=1 i=1 i=1 The last inequality is the Cauchy-Schwartz inequality, which we prove below as Lemma 2.7.

n Lemma 2.7 (Cauchy-Schwartz inequality). For each (a1, . . . , an), (b1, . . . , bn) ∈ R , n 2 n n X X 2 X 2 aibi ≤ ai bi . i=1 i=1 i=1

Pn Pn 2 Proof. Consider the expression i=1 j=1(aibj − ajbi) . By expanding the brackets, n n n n n n n n X X 2 X 2 X 2 X 2 X 2 X X (aibj −ajbi) = ai bj + aj bi −2 aibi ajbj . i=1 j=1 i=1 j=1 j=1 i=1 i=1 j=1 Now, collect the terms (and reindex the sums) to get n n n n n 1 X X X X X 2 (a b − a b )2 = a2 b2 − a b . 2 i j j i i i i i i=1 j=1 i=1 i=1 i=1 Since the left part of the equality is positive, this proves the statement. Example 2.8. Finally, the box metric on Rn is deﬁned by d (x1, . . . , xn), (y1, . . . , yn) = max{|xi − yi| i = 1, . . . , n} Properties (M1) and (M2) are easily seen to hold. Let’s check property (M3). Let x = n (x1, . . . , xn), y = (y1, . . . , yn) and z = (z1, . . . , zn) ∈ R . Then, for each i = 1, . . . , n,

|xi − yi| = |(xi − zi) + (zi − yi)| ≤ |xi − zi| + |zi − yi| ≤ d(x, z) + d(z, y). Thus, the triangle inequality holds. As an exercise, consider the set R2 with this metric. Fix then a point a ∈ R2 and draw the set 2 Da = {x ∈ R d(a, x) ≤ 1}. ELEMENTARY TOPOLOGY I 7

All these examples should serve as a warning of how flexible the notion of metric is: we should not be surprised to see that statements that go against all intuition are, in fact, true. The following is a good examples of this: it is a process to define a metric on every set. This procedure is not rather descriptive, but it is nonetheless important. Example 2.9. Let X be any set. The discrete metric on X is defined by 1, if x 6= y d(x, y) = 0, if x = y Properties (M1) and (M2) are obvious, so let’s check property (M3). Let x, y, z ∈ X. If x = y, then d(x, y) = 0, and there is nothing to check. Suppose then that x 6= y. Then, either z = x 6= y, or z = y 6= x, or z 6= x, y. Regardless the situation, we have then 1 = d(x, y) and 1 ≤ d(x, z) + d(z, y) ≤ 2, and the triangle inequality holds. One could say that this metric is rather bad: if we think of a metric as a rule to measure the distance between points x, y in the set X, then in this case all points are far (if x 6= y) or really close (if x = y), but we cannot distinguish anything beyond this. However, this metric is not useless: it becomes a nice tool to check if geometric intuition really works and provides counterexamples. Example 2.10. Consider the set of integer numbers Z, and let p be a prime number. The p-adic valuation of n ∈ Z is the maximum non-negative integer νp(n) such that pνp(n) divides n. We can now define the p-adic metric on Z by 0, if a = b; d(a, b) = 1 , otherwise. pνp(a−b) Property (M1) holds by definition of d, and property (M2) holds by definition of the p-adic valuation: νp(a − b) = νp(b − a). Let’s prove the triangle inequality. Let then a, b, c ∈ Z. We can assume that these are all different numbers. In this case, property (M3) looks like 1 1 1 ≤ + , pνp(a−b) pνp(a−c) pνp(c−b) and this is a consequence of the following property of the p-adic valuation:

• νp(a − b) ≥ min{νp(a − c), νp(c − b)}.

Let’s prove the above property. Let then k = νp(a − b), m = νp(a − c), n = νp(c − b)} and l = min{νp(a − c), νp(c − b). By deﬁnition of the p-adic valuation, there exist x, y, z ∈ Z such that p does not divide any of them, and such that pkx = a − b = a − c + c − b = pmy + pnz = pl(pm−ly + pn−lz), and now it is clear that k ≥ l. 8 ALEX GONZALEZ

2.1. Continuous functions. We can introduce now continuity of functions between metric spaces.

Deﬁnition 2.11. Let (X, dX ) and (Y, dY ) be metric spaces, and let f : X → Y be a set map. The map f is continuous at the point a ∈ X (with respect to the metrics dX , dY ) if for each ε > 0 there exists some δ > 0 such that

if dX (a, x) < δ then dY (f(a), f(x)) < ε. If f is continuous at every element a ∈ X, then we say that f is continuous. The map f is a homeomorphism if the inverse map f −1 exists (as a map of sets f −1 : Y → X) and both f and f −1 are continuous. Similarly, two metric spaces (X, dX ) and (Y, dY ) are homeomorphic if there exists a homeomorphism f : X → Y .

Broadly speaking, the notion of homeomorphism is to metric spaces as the notion of isomorphism is to groups or other algebraic structures. Notice that the condition of f −1 existing is already rather restrictive: you know of many continuous maps f : R → R which do not have an inverse function.

Notice that the above deﬁnition depends on the given metrics dX and dY . In order to specify that the map f is continuous with respect to the metrics dX and dY on X and Y respectively, we may also write

f :(X, dX ) −→ (Y, dY ). This is actually rather important since the map f : X → Y can be continuous with a choice of metrics and discontinuous with another choice of metrics. Let’s see this with an example. Example 2.12. Let X = R = Y , and let f : R → R be the map 0, if x ∈ f(x) = Q 1, if x∈ / Q Let then dEuc and ddis be the Euclidean and the discrete metrics on R. We have then

• f :(X, dEuc) → (Y, dEuc) is not continuous;

• f :(X, dEuc) → (Y, ddis) is not continuous;

• f :(X, ddis) → (Y, dEuc) is continuous; and

• f :(X, ddis) → (Y, ddis) is continuous. Of course, this is an extreme example, but it perfectly describes the importance of keeping in mind which are the ambient metrics. As an exercise, ﬁll in the details to show each of the points above. Remark 2.13. The discrete metric also illustrates the following (not intuitive) situation: a bĳective map f : X → Y of sets which is continuous with respect to a choice of metrics dX and dY may not have a continuous inverse! Indeed, the identity map (R, ddis) → (R, dEuc) is continuous, but the inverse is not. ELEMENTARY TOPOLOGY I 9

Example 2.14. The map µ: R2 → R deﬁned by µ(a, b) = a · b (multiplication of real numbers) is a continuous function with respect to Euclidean metrics on R2 and R, which we will denote here by d2 and d1 respectively to avoid confusion. Fix then some (a, b) ∈ R2. To prove that µ is continuous at (a, b) we have to check that for each ε > 0 there exists some δ > 0 such that

if d2((a, b), (u, v)) < δ then d1(µ(a, b), µ(u, v)) = d1(ab, uv) < ε. Let’s ignore ε for a second, and suppose that

p 2 2 d2((a, b), (u, v)) = (a − u) + (b − v) < δ for some δ > 0. Then, we have |a − u|, |b − v| < δ, and

d1(ab, uv) = |ab − uv| = |b(a − u) + a(b − v) + (a − u)(v − b)| ≤ ≤ |b| · |a − u| + |a| · |b − v| + |a − u| · |b − v| < δ2 + δ(|a| + |b|).

Let’s go back now to the question of the continuity of µ at (a, b) ∈ R2. Let then ε > 0, and consider the quadratic equation on δ: δ2 + (|a| + |b|)δ − ε = 0. √ −(|a|+|b|)+ (|a|+|b|)2+4ε This equation has always a positive solutions, namely δ = 2 , and the above discussion implies that, for this particular choice of δ, if d2((a, b), (u, v)) < δ then d1(ab, uv) < ε as required. Exercise 2.15. Let (X, d) be a metric space and let z ∈ X. We can then deﬁne a map

δz X / [0, ∞)

x / δz(x) = d(z, x)

Prove that the map δz is continuous with respect to the metric d on X and the Euclidean metric on [0, ∞) ⊆ R. Example 2.16. Let d be the p-adic metric on the set of integers Z, and ﬁx some α ∈ Z. Then, the maps f g (Z, d) / (Z, d)(Z, d) / (Z, d) k / α · k k / k + α are continuous.

Let’s prove ﬁrst that f is continuous. For every a, b ∈ Z, we have 1 1 d(f(a), f(b)) = d(α · a, α · b) = = = pνp(α·a−α·b) pνp(α(a−b)) 1 1 = = · d(a, b). pνp(α) · pνp(a−b) pνp(α) 10 ALEX GONZALEZ

Since 1 ≤ 1, it follows that, for each a ∈ and each ε > 0, pνp(α) Z if d(a, b) < ε then d(f(a), f(b)) ≤ d(a, b) < ε and hence f is a continuous map. In particular notice that if p does not divide α then f is an isometry. Let’s check now that the map g is continuous. In this case, for each a, b ∈ Z we have 1 1 d(f(a), f(b)) = d(a + α, b + α) = = = d(a, b), pνp(a+b−(b+α)) pνp(a−b) so in fact this map is always an isometry (in particular, it is continuous).

Once we start working with functions, it is natural to consider compositions of functions.

Proposition 2.17. Let f :(X, dX ) → (Y, dY ) and g :(Y, dY ) → (Z, dZ ) be continuous maps of metric spaces. Then, the composition map, (g ◦ f):(X, dX ) → (Z, dZ ), is a continuous map of metric spaces.

Proof. Set for simplicity h = (g ◦ f): X → Z, and ﬁx an element a ∈ X. We have to check that, for each ε > 0 there is some δ > 0 such that, if dX (a, x) < δ, then dZ (h(a), h(x)) < ε. Let b = f(a) ∈ Y . Since g is continuous, there exists some γ > 0 such that, if dY (b, y) < γ, then dZ (g(b), g(y)) = dZ (h(a), g(y)) < ε.

Similarly, since f is continuous, there exists some δ > 0 such that, if dX (a, x) < δ, then dY (f(a), f(x)) = dY (b, f(x)) < γ. Combining the above, we get the following:

dX (a, x) < δ =⇒ dY (f(a), f(x)) = dY (b, f(x)) < γ =⇒

=⇒ dZ (g(b), g(f(x))) = dZ (h(a), h(x)) < ε which proves that the composition is continuous at a ∈ X.

2.2. Limits. Convergence of sequences is another feature that depends on the existence of a metric.

Deﬁnition 2.18. Let (X, d) be a metric space and let {xn}n∈N be a sequence of elements of X. Then, the the sequence {xn}n∈N converges to the element x ∈ X if, for every ε > 0 there exists some N ∈ N such that

d(x, xn) < ε for all n ≥ N.

When this is the case, the element x is called the limit of the sequence {xn}n∈N, and we write x = limn→∞ xn. ELEMENTARY TOPOLOGY I 11

Theorem 2.19. Let (X, d) be a metric space. If the sequence {xn}n∈N in X has a limit, then it is unique.

Proof. Suppose the sequence {xn}n∈N converges to the points x, y ∈ X, and suppose x 6= y. This means that, for each ε > 0 there exist Nx,Ny ∈ N such that

d(x, xn) < ε, for all n ≥ Nx and d(y, xn) < ε, for all n ≥ Ny. d(x,y) In particular, let ε = 2 > 0, and let N = max{Nx,Ny}, where Nx,Ny are as above (with respect to this particular choice of ε). Then, for each n ≥ N, we have d(x, xn), d(y, xn) < ε. On the other hand, by the triangle inequality, d(x, y) d(x, y) ≤ d(x, x ) + d(x , y) < 2 · ε = 2 · = d(x, y), n n 2 which is impossible. Thus, x = y.

Example 2.20. Let X be a set with the discrete metric. Then, a sequence {xn}n∈N is convergent if and only if there exists some N ∈ N such that xn = xN for all n ≥ N. Studying the convergence of sequences in a set with the discrete metric is not really interesting.

Example 2.21. Let X be the set of all continuous functions f : [0, 1] → R, and consider the following two metrics on X: Z 1

dint(f, g) = |f(x) − g(x)|dx d∞(f, g) = max{|f(x) − g(x)| x ∈ [0, 1]}. 0

Now ﬁx some α > 0 and let {fn}n∈N be the sequence where fn is the function with graph

(0, α)

(0, 0) (1, 0) 1 1 1 1 1 ( 2 − n , 0) ( 2 , 0) ( 2 + n , 0)

Then, the sequence {fn}n∈N converges to the function g = 0 (constant on [0, 1] with image 0) with respect to the metric dint. Indeed, for each n Z 1 Z 1 1 1 dint(0, fn) = |0 − fn(x)|dx = fn(x) dx = and lim = 0. 0 0 n n→∞ n

However, the same sequence {fn} does not converge to the constant function g = 0 with respect to the metric d∞, since for any n

d∞(0, fn) = max{|0 − fn(x)| x ∈ [0, 1]} = α > 0. 12 ALEX GONZALEZ

In fact, this sequence does not converge to any continuous function with respect to the metric d∞, but we are not ready yet to prove this.

2.3. Open subsets and closed subsets. The idea of open/closed subset again relies in the existence of a metric.

Deﬁnition 2.22. Let (X, d) be a metric space. For each element a ∈ X and each ε ∈ (0, ∞) ⊆ R, the open ball with center a and radius ε is the subset

Ba,ε = {x ∈ X d(a, x) < ε} ⊆ X.

A subset U ⊆ X is open with respect to the metric d if for each y ∈ U there is some ε(y) ∈ (0, ∞) such that

By,ε(y) ⊆ U.

A subset U ⊆ X is closed with respect to the metric d if the set X \ U is open with respect to the metric d.

The notation ε(y) above is there to stress the idea that the radius of the ball By,ε depends on the element y. Note also that we have to specify with respect to which metric the subset U is open in X. The following example illustrates the reason.

Example 2.23. Let dEuc and ddis be, respectively, the Euclidean and the discrete metrics on R2, and let a ∈ R2 be any point. Then, U = {a} ⊆ R2 is open with respect to ddis but not open with respect to dEuc.

Lemma 2.24. Let (X, d) be a metric space. Then for each x ∈ X and for each ε > 0,

(1) Bx,ε = {y ∈ X d(x, y) < ε} is an open subset of X with respect to d; and

(2) Dx,ε = {y ∈ X d(x, y) ≤ ε} is a closed subset of X with respect to d.

Proof. To prove part (1), we have to show that, for each y ∈ Bx,ε there exists some δ > 0 such that By,δ ⊆ Bx,ε. The following sketch already hints which is the right choice of δ. ELEMENTARY TOPOLOGY I 13

Let then δ = ε − d(x, y) > 0. Then, for each

z ∈ By,δ the triangle inequality implies d(x, z) ≤ d(x, y) + d(y, z) < ε < d(x, y) + δ =

= d(x, y) + ε − d(x, y) = ε y

and hence z ∈ Bx,ε. This proves that δ d(x, y) x

the subset Bx,ε is open.

To prove (2), notice that X \ Dx,ε = {y ∈ X d(x, y) > ε}, and we have to prove that this subset is open. Let then y ∈ X \ Dx,ε, as in the following picture.

Let δ = d(x, y) − ε > 0, and suppose that for

some z ∈ By,δ, we have d(x, z) ≤ ε. Then y ε ≥ d(x, z) ≥ |d(x, y) − d(y, z)| = δ d(x, y) = |ε + δ − d(y, z)| > ε + δ − δ = ε, ε where the second inequality (starting from the left) follows from Exercise 1 in List 1. But this

ε x is impossible, and thus for each z ∈ By,δ we

have d(x, z) > ε. Thus Dx,ε is closed

The following is the most important and basic result about open subsets in a metric space. Theorem 2.25. Let then (X, d) be a metric space, and let O be the collection of all open subsets of X with respect to d. Then, 14 ALEX GONZALEZ

(1) the total subset, X, and the empty set, ∅, are elements of O;

(2) if {Uα}α∈Γ ⊆ O is any family (possibly inﬁnite) of open subsets, then [ Uα ∈ O; α∈Γ

(3) if {U1,...,Un} ⊆ O is any ﬁnite family of open subsets, then n \ Ui ∈ O. i=1

Proof. Property (1) is easy: since X contains all possible open balls, it is open. Also, the empty set ∅ contains no points at all, and thus it also satisﬁes the condition to be an open subset.

Let now U = {Uα}α∈Γ ⊆ O be a family (possibly inﬁnite) of open subsets of X, and let V = ∪ΓUα ⊆ X. For each u ∈ V , there is at least one Uα ∈ U which contains u. Furthermore, since Uα is open, there exists some ε > 0 such that d Bu,ε ⊆ Uα ⊆ V. Hence V is open with respect to d.

Finally, let U = {U1,...,Un} ⊆ O be a ﬁnite family of open subsets. Let also V = n ∩i=1Ui, and let u ∈ V . Since u ∈ Ui for all i, there exist ε1, . . . , εn > 0 such that Bd ⊆ U u,di i i ε = min{ε , . . . , ε } Bd ⊆ Bd i for all . Let then 1 n . Then, u,ε u,εi for all , and thus d Bu,ε ⊆ V and V is open with respect to d.

There is a similar result for closed subsets.

Proposition 2.26. Let (X, d) be a metric space. Then, the following holds:

(1) X and ∅ are closed subsets; Sn (2) if {U1,...,Un} ⊆ O is any ﬁnite family of closed subsets, then the union i=1 Ui is also a closed subset;

(3) if {Uα}α∈Γ ⊆ O is any family (possibly infinite) of closed subsets, then the T intersection α∈Γ Uα is also a closed subset. Example 2.27. Let X be a set with the discrete metric. Then, for each x ∈ X the one-element subset {x} ⊆ X is both open and closed. To check that it is open just ELEMENTARY TOPOLOGY I 15 notice that {x} = Bx,1 (open ball of radius 1). To check that it is closed, notice that the complement [ [ X \{x} = {y} = By,1 y∈X\{x} y∈X\{x} is a (possibly infinite) union of open subsets, and hence is open. Example 2.28. Consider R2 with the Euclidean metric. Then, for each u ∈ R2, the subset {u} ⊆ R2 is closed but not open. Indeed, it is clear that {u} is not open. To check that it is closed, consider the complement U = R2 \{u}. We can describe U as 2 U = {v ∈ R d(u, v) > 0}. d(u,v) Fix some v ∈ U and set ε = 2 > 0. Then, for each w ∈ Bv,ε we have d(u, v) d(u, w) ≥ |d(u, v) − d(v, w)| = d(u, v) − d(v, w) > > 0, 2 and thus w ∈ U (in the above list of (in)equalities we are using some result from the first exercise list). Exercise 2.29. Let (X, d) be a metric space and let x ∈ X be any element. Prove that the subset {x} ⊆ X is closed with respect to d (in other words, regardless of the metric, a point is always a closed subset).

2.4. Continuity, convergence, and open/closed subsets. The concepts that we have introduced in the previous sections are not independent from each other. In this section we will study some of their interactions. Theorem 2.30. Let (X, d) be a metric space and let V ⊆ X be a subset. Then the following statements are equivalent. (1) V is a closed subset of X.

(2) the limit of every convergent sequence {xn}n∈N ⊆ V is an element of V .

Proof. Suppose ﬁrst that (2) is true. We prove by contradiction that V c is open. c c c Assume then that V is not open, and let u ∈ V be an element for which Bu,ε 6⊆ V , for any ε > 0. This means, in particular, that for any n ∈ the ball B 1 contains N u, n (at least) an element xn of V . We can form then the sequence {xn}n∈N ⊆ V , but this sequence clearly converges to the element u∈ / V , hence the contradiction. Suppose now that (1) is true. We have to show that every convergent sequence of elements of V has its limit in V . Let then {xn}n∈N ⊆ V be a convergent sequence, c c and suppose v = lim xn ∈ V . Since V is open by hypothesis, there exists some c ε > 0 such that Bv,ε ⊆ V . On the other hand, since v = lim xn, there exists some 0 0 ε < ε and some N ∈ N such that d(v, xn) < ε for all n ≥ N, and this implies that c xn ∈ Bv,ε0 ⊆ Bv,ε ⊆ V , which contradicts the assumption that xn ∈ V for all n. 16 ALEX GONZALEZ

Let’s analyze the deﬁnition of continuity. Let then f :(X, dX ) → (Y, dY ) be a continuous function, and let a ∈ X. The map f is continuous at a if, for each ε > 0 there is some δ > 0 such that

if dX (a, x) < δ then dY (f(a), f(x)) < ε.

The left part above says that x ∈ Ba,δ, while the right part says that f(x) ∈ Bf(a),ε. This suggests some relation between continuity and open subsets, which we formu- late properly as the following result. You can take it as a set of equivalent deﬁnitions of the notion of continuity.

Theorem 2.31. Let (X, dX ) and (Y, dY ) be metric spaces, and let f : X → Y be a map. Then, the following statements are equivalent:

(1) f is a continuous function;

(2) for each open subset U ⊆ Y , f −1(U) is an open subset of X;

(3) for each closed subset V ⊆ Y , f −1(V ) is a closed subset of X;

(4) for any sequence {xn}n∈N in X, if {xn} converges to z ∈ X, then the sequence {f(xn)}n∈N in Y converges to f(z).

A note of warning: given a map of sets f : X → Y , the notation f −1(U) stands simply for the set of pre-images in X of the elements of U, −1 f (U) = {x ∈ X f(x) ∈ U}. We are not assuming the existence of any inverse function! For instance consider the function f : R → R defined by f(x) = x2. Clearly, this function does not have an inverse, but the pre-image of the subset {1} ∈ R makes perfect sense: −1 f ({1}) = {x ∈ R f(x) = 1} = {−1, 1}. Remark 2.32. With this set of equivalent definitions of continuity, we do not need to define continuity at a point, we can define continuous functions directly: a function between metric spaces f : X → Y is continuous if and only if for each open subset U ⊆ Y the pre-image f −1(U) is an open subset of X.

Be careful with the following: condition (2) above concerns only the pre-image of open subsets of Y , but it says nothing about the image of open subsets of X. The following is, in general, false: if U ⊆ X is an open subset, then f(U) ⊆ Y is an open subset.

Proof of Theorem 2.31. Since this proof is rather long, let’s split it in smaller parts, each proving an equivalence of statements.

• Statements (1) and (2) are equivalent. ELEMENTARY TOPOLOGY I 17

First we prove that (1) implies (2). Suppose then that f is a continuous map, and let U ⊆ Y . Let also a ∈ f −1(U), and let u = f(a). Since U is open, there exists some ε(u) > 0 such that Bu,ε(u) ⊆ U. Also, since f is continuous, there exists some δ(a) > 0 such that

if dX (a, x) < δ(a) then dY (f(a), f(x)) = dY (u, f(x)) < ε(u), −1 which implies that f(Ba,δ(a)) ⊆ U, and hence Ba,δ(a) ⊆ f (U). This proves that −1 f (U) is open with respect to dX . Let’s see now that (2) implies (1). Suppose then that f −1(U) is an open subset of X for each open subset U ⊆ Y , and ﬁx a ∈ X. For each ε > 0, the open ball Bf(a),ε is −1 an open subset of Y , and thus f (Bf(a),ε) is an open subset of X. Furthermore, −1 a ∈ f (Bf(a),ε), and this implies that there exists some δ > 0 such that −1 Ba,δ ⊆ f (Bf(a),ε). The above inclusion is equivalent to say that f is continuous at a ∈ X. • Statements (2) and (3) are equivalent. Let V ⊆ Y be a closed subset, and let U = V c, which is an open subset by hypothesis. By (2), f −1(U) is an open subset of X. Furthermore, we have f −1(U) = f −1(V c) = f −1(Y \ V ) = X \ f −1(V ), and thus f −1(V ) is a closed subset, and this proves that (2) implies (3). Replacing “open” by “closed” proves the converse. • Statements (1) and (4) are equivalent.

First we prove that (1) implies (4). Let then {xn}n∈N be a sequence in X which converges to z ∈ X. Since f is continuous (at z), for each ε > 0 there exists some δ > 0 such that if dX (z, x) < δ then dY (f(z), f(x)) < ε. Also, since the sequence is convergent, we know that there exists some N ∈ N such that dX (z, xn) < δ for all n ≥ N. Thus, for all n ≥ N, we have

dY (f(z), f(xn)) < ε, which implies (4). Finally, we have to check that (4) implies (1). Suppose otherwise that (1) is not true on a ﬁxed element a ∈ X: there exists some ε > 0 such that for each δ > 0 there is some x ∈ X such that

dX (a, x) < δ but dY (f(a), f(x)) ≥ ε. By ﬁxing such ε, we can then deﬁne the sets 1 An = {x ∈ X dX (a, x) < and dY (f(a), f(x)) ≥ ε}, n 18 ALEX GONZALEZ which is non-empty for all n ∈ N. Next choose for each n an element xn ∈ An: the sequence {xn}n∈N clearly converges to a ∈ X, while the sequence {f(xn)}n∈N does not converge to f(a), contradicting (4).

The above result implies a crucial consequence: it is not the metric(s) what matters, but the collection(s) of open/closed subsets! ELEMENTARY TOPOLOGY I 19

3. Topological spaces

We ﬁnished the previous chapter realizing that, when we are talking about continuity of functions, metrics are not so relevant as open subsets. We thus shift the focus to collections of open subsets instead of metrics.

Deﬁnition 3.1. Let X be a set. A topology on X is a collection T of subsets of X satisfying the following conditions

(T1) the total set, X, and the empty set, ∅, are elements of T;

(T2) if {Uγ}γ∈Γ is a (possibly inﬁnite) family of elements of T, then [ Uγ ∈ T; γ∈Γ

(T3) if {U1,...,Un} ⊆ T is a ﬁnite family of elements of T, then n \ Ui ∈ T. i=1

A topological space is a pair (X, T) where T is a topology on X. The elements of T are called open subsets, while V ⊆ X is a closed subset if V c ∈ T.

Example 3.2. Let X be any set, and let T1 be the collection of all the subsets of X, and let T2 = {∅,X}. Then, both T1 and T2 are topologies on X. More precisely,

(1) T1 is called the discrete topology on X, while

(2) T2 is called the indiscrete topology on X.

The topology T1 is induced by the discrete metric, whereas the T2 is as far from the discrete topology as it can be, hence the name.

Example 3.3. Let (X, d) be a metric space, and let T be the collection of open subsets of X with respect to the metric d. Then, T is a topology on X by Theorem 2.25, and (X, T) is a topological space. In other words, every metric on X deﬁnes a topology on X.

Notice that different metrics on the same set X could give rise to the same topology. For example, let X = Rn. The Manhattan metric (Example 2.5), the Euclidean metric (Example 2.6) and the box metric (Example 2.8) all define the same topology on X. Another example of this situation is the following: all metrics on a given finite set define the same topology, the discrete metric.

Example 3.4. Let X = {a, b, c} and let T = {∅, {a}, {a, b},X}. Then, T is a topology on X which is not induced by any metric. 20 ALEX GONZALEZ

a b c

Notice the following: there are no open subsets B,C ∈ T such that b ∈ B, c∈ / B and c ∈ C, b∈ / C. In other words, the elements b and c are “at distance zero” in this topology (of course the last statement does not make any sense). How many topologies can you deﬁne on the set X = {a, b}? Example 3.5. Let T be the collection of subsets of R which, in addition to ∅ and R, contains all the subsets of the form (x, ∞), for x ∈ R. Clearly, this is a topology on R, with less open subsets than the Euclidean topology, since for example the subset (0, 1) is not open in the topology T. This is another example of a topology which is not induced by a metric on R. Indeed, suppose that there exists a metric d on R which induces the above topology. Since open balls are in particular open subsets, for each x ∈ R and each ε > 0, we have

Bx,ε = R or Bx,ε = (y, ∞) for some y ∈ R. Furthermore, if Bx,ε = R for all x ∈ R and all ε > 0, then d cannot induce the topology T on R (why?). Thus, there exists at least one x ∈ R and some ε > 0 such that

Bx,ε = (y, ∞) for some y ∈ R. This way, for each n ∈ N, we have

Bx, ε = (yn+1, ∞) ⊆ Bx, ε = (yn, ∞) 2n+1 2n for some yn, yn+1 ∈ R. This means that, for any z > x and all n ∈ N, ε d(x, z) < , 2n which is impossible since z 6= x.

Let’s see another example of all the unexpected situations that can happen when dealing with topological spaces. Example 3.6. This example is usually called the line with two origins. Let X = R∪{z} (here z denotes simply an extra element), and let

(1) T1 = {U ⊆ R U is open with the Euclidean metric};

(2) T2 = {∅} ∪ {W = V ∪ {z} V ∈ T1 is such that {0} ∈/ V }. ELEMENTARY TOPOLOGY I 21

Finally, let T = T1 ∪ T2. In other words, an element of T is a union U ∪ W , with U ∈ T1 and W ∈ T2. This is a topology in X (check the details), and it has the following property: if A, B ∈ T are such that 0 ∈ A, z∈ / A and z ∈ B, 0 ∈/ B, then A ∩ B 6= ∅. In other words, it is impossible to separate 0 from z!

Example 3.7. This example is called the finite complement topology. Let X be a set, and let T be the collection of subsets U ⊆ X for which U c = X \ U is either empty, finite, or is all of X. Clearly Xc = ∅ and ∅c = X, and this takes care of condition (T1). Let {Uα}α∈Γ be a family of elements of T (we may assume that Uα 6= ∅,X without loss of generality). Then, [ \ X \ Uα = (X \ Uα) α∈Γ α∈Γ is a finite set, since X \ Uα is finite for all α, and condition (T2) follows. Conditions (T3) follows similarly.

3.1. Interior, closure and boundary. This section reviews some basic concepts re- lated to open/closed subgroups.

Deﬁnition 3.8. Let (X, T) be a topological space, and let A ⊆ X.

• The interior of A, A◦, is the greatest open subset of X contained in A.

• The closure of A, A, is the smallest closed subset of X which contains A.

• The boundary of A, ∂(A), is the intersection A ∩ X \ A.

Lemma 3.9. Let (X, T) be a topological space and let A ⊆ X. Then,

(1) the interior of A is the union of all the open subsets which are contained in A: [ A◦ = U; U∈T,U⊆A

(2) the closure of A is the intersection of all the closed subsets which contain A: \ A = V. V c∈T,A⊆V

Proof. We prove part (i) and leave part (ii) as an exercise. Let W = ∪U∈T,U⊆AU. Since T is a topology, W ∈ T and W ⊆ A. This means that W ⊆ A◦. On the other hand, ◦ ◦ ◦ A ∈ T and A ⊆ A. By deﬁnition, A ⊆ W . Lemma 3.10. Let (X, T) be a topological space. Then the following holds.

(1) A subset A ⊆ X is open if and only if A◦ = A.

(2) For every A ⊆ X, if B = A◦ then B◦ = B.

(3) If A◦ ⊆ B ⊆ A then A◦ = B◦. 22 ALEX GONZALEZ

(4) Let {Uα}α∈Γ be a collection (possibly inﬁnite) of subsets of X, and let U = S T α∈Γ Uα and V = α∈Γ Uα; then [ ◦ ◦ ◦ \ ◦ (Uα) ⊆ U V ⊆ (Uα) . α∈Γ α∈Γ

Tn (5) Let {U1,...,Un} be a ﬁnite collection of subsets of X and let V = i=1 Ui. Then, n ◦ \ ◦ V = (Ui) . i=1

Proof. Properties (1) and (2) are obvious by definition. Property (3) is proved as follows. If A◦ ⊆ B, then by definition A◦ ⊆ B◦ ⊆ B ⊆ A. Also, since B◦ is an open subset of A, it follows that B◦ ⊆ A◦. Property (4) follows from the fact that, for each γ ∈ Γ, we have ∩ΓUα ⊆ Uγ ⊆ ∪ΓUα. Finally, let’s prove property (5). The intersection n ◦ n I = ∩i=1(Ui) is an open subset of ∩i=1Ui, and hence I ⊆ V . This combined with property (4) finishes the proof.

A similar list of properties exists for the closure of a subset. We state it below without proof (the proof is similar to the result above, so it is left as an exercise).

Lemma 3.11. Let (X, T) be a topological space. Then the following holds.

(1) A subset A ⊆ X is closed if and only if A = A. (2) For every A ⊆ X, if B = A then B = B. (3) If A ⊆ B ⊆ A then A = B;

(4) Let {Uα}α∈Γ be a collection (possibly inﬁnite) of subsets of X, and let U = S T α∈Γ Uα and V = α∈Γ Uα; then [ \ Uα ⊆ U V ⊆ Uα. α∈Γ α∈Γ

Sn (5) Let {U1,...,Un} be a ﬁnite collection of subsets of X and let V = i=1 Ui. Then, n [ V = Ui. i=1

Finally, let’s see some interesting properties of the interaction of interior, closure and boundary.

Lemma 3.12. Let (X, T) be a topological space. Then, for any subsets A, B ⊆ X, (1) the closure of the complement of A is the complement of the interior of A: Ac = X \ A = X \ (A◦) = A◦c; ELEMENTARY TOPOLOGY I 23

(2) the interior of the complement of B is the complement of the closure of B: (Bc)◦ = (X \ B)◦ = X \ B = Bc;

(3) ∂A ⊆ A; and

(4) ∂B◦ ∩ B◦ = ∅.

c Proof. To prove (1), notice that A◦ = A \ ∂(A) = A ∩ ∂(A) . Thus, A◦c = X \ (A◦) = X \ A ∩ ∂(A)c = (X \ A) ∪ X \ ∂(A)c = c = Ac ∪ ∂(A)c = Ac ∪ ∂(A) = Ac ∪ ∂(Ac) = Ac

To prove (2) just replace A by Bc everywhere above. To check property (3), let B = A. Then ∂B = b ∩ X \ B = B ∩ X \ B ⊆ B. To prove (4), notice that Y ∩ Y c = ∅ for any Y ⊆ X. Thus, we just need to check that c ∂B◦ ⊆ B◦ . Indeed, ∂B◦ = ∂(B◦)c = ∂Bc ⊆ Bc = (B◦)c, where (from left to right) the ﬁrst equality holds by deﬁnition, the second equality holds by (2), the inequality holds by (3) and the last equality holds by (1). Exercise 3.13. Let (X, T) be a topological space. Prove that, for any subset A ⊆ X,

◦ def ◦ ∂(A) = A \ A = {x ∈ A x∈ / A }. Deduce that, if A is an open and closed subset of X, then ∂(A) = ∅.

3.2. Continuous functions. Continuity of functions is also deﬁned in terms of open subsets.

Deﬁnition 3.14. Let (X, TX ) and (Y, TY ) be topological spaces. A map f : X → Y is a continuous map if, for all U ∈ TY , −1 f (U) ∈ TX . The map f : X → Y is a homeomorphism if f is bĳective (as a map between sets) and both f and f −1 are continuous. The map f : X → Y is an imbedding if the restriction to f : X → f(X) is a homeomorphism (in particular, f is injective as a map of sets).

Proposition 3.15. Let f :(X, TX ) → (Y, TY ) and g :(Y, TY ) → (Z, TZ ) be continuous maps. Then, the composition g ◦ f :(X, TX ) → (Z, TZ ) is continuous.

−1 −1 −1 Proof. For any W ∈ TZ , we have U = (g ◦ f) (W ) = f (g (W )) ∈ TX because −1 −1 V = g (W ) ∈ TY (g is continuous) and U = f (V ) ∈ TX (f is continuous). 24 ALEX GONZALEZ

Exercise 3.16. Let X = {1, 2, 3, 4} and let Y = {1, 2,..., 8}. Consider also the collections TX = {∅, {1}, {4}, {1, 3}, {2, 4}, {1, 2, 4}, {1, 3, 4},X} TY = {∅, {1}, {1, 2}, {1, 2, 3},...,X} Finally, let f : X → Y be the map deﬁned by 2n, if n is even f(n) = n + 1, if n is odd

Prove that TX and TY are topologies for X and Y respectively, and show that f is a continuous function.

Theorem 3.17. Let (X, TX ) and (Y, TY ) be topological spaces, and let f : X → Y be a map. Then the following are equivalent.

(1) f is a continuous map.

(2) For every subset A ⊆ X, we have f(A) ⊆ f(A).

(3) For every closed subset V ⊆ Y , the set f −1(V ) is closed in X.

(4) For each x ∈ X and each V ∈ TY containing f(x), there is some U ∈ TX containing x and such that f(U) ⊆ V .

Proof. We prove that (1) implies (2), (2) implies (3), and (3) implies (1). Afterwards we show that (4) is equivalent to (1).

Assume ﬁrst that (1) holds, and let A ⊆ X be a subset. For each x ∈ A we have to check that f(x) ∈ f(A). Fix such x and let V ∈ TY be any open subset containing f(x). Since f is continuous, the subset U = f −1(V ) is an open subset that contains the element x. Note that U ∩A 6= ∅, hence there exists y ∈ A∩U, and f(y) ∈ V ∩f(A). Since every open subset containing f(x) intersects f(A) nontrivially, this means that f(A) ⊆ f(A).

Assume now that (2) holds. Let B ⊆ Y be a closed subset, and let A = f −1(B). We have to show that A = A. We have f(A) ⊆ B, and thus if x ∈ A then f(x) ∈ f(A) ⊆ f(A) ⊆ B = B. This implies that x ∈ A = f −1(B), and thus A = A.

Suppose next that (3) holds, and let V ⊆ Y be an open subset. Then W = V c = Y \V is closed, and f −1(W ) = f −1(Y \ V ) = X \ f −1(V ) By hypothesis f −1(W ) is closed, and thus f −1(V ) is open.

Next we prove that (1) implies (4). Le x ∈ X and let V ∈ TY containing f(x). Then the set U = f −1(V ) is an open subset containing x. Conversely, assume that (4) holds. −1 Let V ∈ TY and let x ∈ f (V ). Then f(x) ∈ V , and by hypothesis there is some ELEMENTARY TOPOLOGY I 25

−1 Ux ∈ TX containing x and such that f(Ux) ⊆ V . Thus Ux ⊆ f (V ). It follows that −1 S f (V ) = x∈f −1(V ) Ux, which is an element of TX .

3.3. Basis of a topology. Very often we will not be able to specify all the open subsets of a given topology, but rather we will specify a subcollection that generates the whole topology.

Deﬁnition 3.18. Let (X, T) be a topological space. A basis for T is a collection B ⊆ T such that each open set is a union of sets in B.

Example 3.19. Consider R with the topology induced by the Euclidean metric, and let

B1 = {(x0, x1) ⊆ R x0, x1 ∈ Q} B2{(x0, x1) ⊆ R x0, x1 ∈ R \ Q} Both collections above are bases for the same topology, their intersection is empty, and while B1 has countably many elements, B2 does not!

An alternative characterization of the basis of a topology is as follows. Lemma 3.20. Let (X, T) be a topological space, and let B be a subset of T. Then, B is a basis for T if and only if it satisﬁes the following properties.

(B1) For each U ∈ T and each x ∈ U there exists some V ∈ B such that x ∈ V ⊆ U.

(B2) Let V1,V2 ∈ B and let x ∈ V1 ∩ V2. Then there exists V ∈ B such that x ∈ V ⊆ V1 ∩ V2.

Proof. Suppose ﬁrst that B is a basis for T. In order to check condition (B1), let U ∈ T and x ∈ U. By deﬁnition of basis, there exists some collection {Vγ}γ∈Γ ⊆ B such that [ U = Vγ. γ∈Γ

In particular, there is some Vγ such that x ∈ Vγ ⊆ U.

Let now V1,V2 ∈ B and let x ∈ V1 ∩ V2. Since V1,V2 ∈ T and T is a topology, it follows that V1 ∩V2 ∈ T. Thus, by property (B1), there exists V ∈ B such that x ∈ V ⊆ V1 ∩V2, and property (B2) holds. Suppose now that B satisﬁes properties (B1) and (B2) above. We have to check that each element U ∈ T is a union of elements of B. Fix an open subset U ∈ T. For each x ∈ U there exists some Vx ∈ B satisfying condition (B1), and we can write [ U = Vx. x∈U Hence B is a basis for T.

We can use Lemma 3.20 to formalize this idea of a topology generated by a basis. 26 ALEX GONZALEZ

Deﬁnition 3.21. Let X be a set. A basis for a topology on X is any collection B of subsets of X satisfying the following properties.

(B1’) For each x ∈ X there exists some V ∈ B such that x ∈ V .

(B2) Let V1,V2 ∈ B and let x ∈ V1 ∩ V2. Then there exists V ∈ B such that x ∈ V ⊆ V1 ∩ V2.

Given a basis B for a topology on X, we can deﬁne the following collection of subsets:

TB = {U ⊆ X for each x ∈ U there exists V ∈ B such that x ∈ V ⊆ U}.

Proposition 3.22. The collection of subsets TB is a topology on X with basis B.

We call TB the topology generated by B. Again, keep in mind that the same topology may be generated by more than one basis.

Proof. First, let’s prove that TB is a topology on X.

(T1) Both ∅ and X satisfy the condition to be elements of TB.

(T2) Let {Uγ}γ∈Γ ⊆ TB. We have to check that U = ∪ΓUγ ∈ TB. Each element x ∈ U is contained in Uγ, for some γ ∈ Γ. Now Uγ ∈ TB, and this means that there exists some V ∈ B such that

x ∈ V ⊆ Uγ ⊆ U.

Thus, U ∈ TB.

(T3) Let {U1,...,Un} ⊆ TB. We have to show that U = ∩iUi ∈ TB. Let then x ∈ U. This means that x ∈ Ui for all i, and thus there exist V1,...,Vn ∈ B such that

x ∈ Vi ⊆ Ui for each i = 1, . . . , n. Since B satisﬁes condition (B2), it follows that there exists some V ∈ B such that

x ∈ V ⊆ Ui for each i = 1, . . . , n.

Thus, x ∈ V ⊆ U and U ∈ TB.

The fact that B is a basis for TB is a consequence of Lemma 3.20.

3.4. The subspace topology. Let (X, T) be a topological space, and let A ⊆ X be a subset. We can then deﬁne

TA = {U ∩ A U ∈ T}.

Lemma 3.23. The collection TA is a topology on A.

This is known as the subspace topology for obvious reasons. ELEMENTARY TOPOLOGY I 27

Proof. (T1) ∅ = ∅ ∩ A, A = X ∩ A ∈ TA.

(T2) Let {Uγ}γ∈Γ ⊆ TA. By deﬁnition of TA, for each γ ∈ Γ there exists some Wγ ∈ T S such that Uγ = A ∩ Wγ. Since T is a topology, we have W = γ∈Γ Wγ ∈ T, and hence [ [ [ U = Uγ = (Wγ ∩ A) = Wγ ∩ A = W ∩ A ∈ TA. γ∈Γ γ∈Γ γ∈Γ

(T3) Let {U1,...,Un} ⊆ TA. Again, for each i there exists some Wi ∈ T such that Ui = Wi ∩ A. Since T is a topology, we have n n n \ \ \ U = Ui = (Wi ∩ A) = Wi ∩ A ∈ T. i=1 i=1 i=1 Lemma 3.24. Let (X, T) be a topological space and let A ⊆ X, with the subspace topology. Then, V ⊆ A is open if and only if there is some U ∈ T such that V = U ∩ A.

Proof. The proof is immediate. Lemma 3.25. Let (X, T) be a topological space, let A ⊆ X be a subspace. Then, the inclusion map incl: A → X is continuous.

Proof. Let U ∈ T. We have to check that f −1(U) is an open subset of A. By deﬁnition −1 of the inclusion map, we have f (U) = U ∩ A, which is indeed open.

The subspace topology can be characterized by a universal property. Theorem 3.26. Let (X, T) be a topological space, let A ⊆ X be a subspace, and let incl: A → X be the inclusion map. The subspace topology TA on A is the only topology satisfying the following property.

(S) A map f :(Z, TZ ) → (A, TA) is continuous if and only if the composition map incl ◦f :(Z, TZ ) → (X, T) is continuous.

Proof. Suppose ﬁrst that f is a continuous map. The inclusion map incl: A → X is continuous by Lemma 3.25, and it follows that incl ◦f is a continuous map by Proposition 3.15. Conversely, suppose that incl ◦f is continuous. We have to check −1 that f is continuous: for each V ∈ TA, f (V ) ∈ TZ . Let U ∈ TX be such that V = U ∩ A. Then, (incl ◦f)−1(U) = f −1(U ∩ A) = f −1(V ) is open, and f is continuous.

Theorem 3.27. Let (X, TX ) and (Y, TY ) be topological spaces, and suppose that there exist open (closed) subsets A, B ⊆ X such that X = A S B. Consider A and B as topological subspaces of X. Then, the following holds. 28 ALEX GONZALEZ

(1) If h: X → Y is a continuous function, then f = h|A and g = h|B are continuous functions. (2) If f : A → Y and g : B → Y are continuous functions such that f(x) = g(x) for all x ∈ A ∩ B, then there exists a (unique) continuous function h: X → Y such that f = h|A and g = h|B.

Proof. Part (1) is an immediate consequence of Proposition 3.15 and Lemma 3.25: the map f (respectively g) is the composition of the inclusion map incl: A → X (respectively incl: B → X) and h.

To prove (2), let V ∈ TY . Then, h−1(V ) = f −1(V ) ∪ g−1(V ). −1 −1 Since f and g are continuous maps, f (V ) ∈ TA and g ∈ TB. Since A and B are open subsets of X, it follows that f −1(V ) and g−1(V ) are also open subsets of X, and −1 thus h (V ) ∈ TX .

3.5. The product topology. Let now (X, TX ) and (Y, TY ) be topological spaces, let X × Y be the product set, and consider the collection of subsets

B = {U × V U ∈ TX and V ∈ TY }. Lemma 3.28. The collection B is a basis for a topology on X × Y .

Proof. We have to check the conditions in Deﬁnition 3.21. Condition (B1’) states that for each (x, y) ∈ X × Y there must exist some W ∈ B such that (x, y) ∈ W . Let U ∈ TX and V ∈ TY be such that x ∈ U and y ∈ V . Then W = U × V ∈ B satisﬁes condition (B1’). Condition (B2) states that, given U1 × V1,U2 × V2 ∈ B and (x, y) ∈ (U1 × V1) ∩ (U2 × V2), there exists some U × V ∈ B such that

(x, y) ∈ U × V ⊆ (U1 × V1) ∩ (U2 × V2).

Choose U = U1 ∩U2 ∈ TX , V = V1 ∩V2. Then U ×V clearly satisﬁes the condition.

Let TX×Y be the topology generated by the above basis BX×Y , and let πX : X ×Y → X and πY : X × Y → Y be the obvious projection functions. The product topology is determined by the following universal property.

Theorem 3.29. The collection TX×Y is the only topology on X × Y satisfying the following condition.

(P) A map f :(Z, TZ ) → (X × Y, TX×Y ) is continuous if and only if the compositions πX ◦ f :(Z, TZ ) → (X, TX ) and πY ◦ f :(Z, TZ ) → (Y, TY ) are continuous.

Proof. The collection B is contained in TX×Y , and clearly satisﬁes the conditions (B1) and (B2) in Lemma 3.20, and thus TX×Y is the topology generated by B. We have to check then that it satisﬁes property (P), and that it is unique satisfying this condition. The rest of the proof is divided into several smaller parts. ELEMENTARY TOPOLOGY I 29

1. The maps πX :(X × Y, TX×Y ) → (X, TX ) and πY :(X × Y, TX×Y ) → (Y, TY ) are continuous.

−1 For instance, let’s check this for πX : we have to show that πX (U) ∈ TX×Y for all −1 U ∈ TX , and this is immediate because πX (U) = U × Y .

2. If f :(Z, TZ ) → (X × Y, TX×Y ) is continuous then so are the compositions πX ◦ f and πY ◦ f. This is an immediate consequence of Proposition 3.15: the composition of continuous maps is continuous.

3. If the compositions πX ◦ f : Z → X and πY ◦ f : Z → Y are continuous then so is the map f : Z → X × Y .

−1 We have to check that, for each W ∈ TX×Y , f (W ) ∈ TZ . Let’s assume ﬁrst that W = U × V , for some U ∈ TX and V ∈ TY . Then we have f −1(W ) = f −1(U × V ) = f −1((U × Y ) ∩ (X × V )) = f −1(U × Y ) ∩ f −1(X × V ) = −1 −1 −1 −1 −1 −1 = f (πX (U)) ∩ f (πY (V )) = (πX ◦ f) (U) ∩ (πY ◦ f) (V ) ∈ TZ .

This means that the preimage of any element of the basis BX×Y is in TZ . Now, each W ∈ TX×Y is a union of elements of BX×Y , and this shows that f is continuous.

4. The topology TX×Y is unique satisfying property (P). Let T be another topology on X × Y satisfying property (P), and consider the identity map Id: (X × Y, T) → (X × Y, T). This is clearly continuous, and hence πX = πX ◦ Id and πY = πY ◦ Id are continuous too. This means that, for each U ∈ TX and V ∈ TY , we have −1 −1 πX (U) = U × Y ∈ T πY (V ) = X × V ∈ T and hence U × V = (U × Y ) ∩ (X × V ) ∈ T. In other words, T contains the basis BX×Y , and thus TX×Y ⊆ T. On the other hand, we know from part 1. that the maps

πX :(X × Y, TX×Y ) → (X, TX ) πY :(X × Y, TX×Y ) → (Y, TY ) are continuous. Notice that the compositions

Id πX πX :(X × Y, TX×Y ) / (X × Y, T) / (X, TX )

Id πY πY :(X × Y, TX×Y ) / (X × Y, T) / (Y, TY ) are continuous maps, and thus by property (P) the map Id: (X × Y, TX×Y ) → (X × Y, T) is continuous, and this means that T ⊆ TX×Y .

WARNING!. The deﬁnition of the product topology TX×Y on X ×Y does NOT say that all the open subsets in TX×Y are of the form U × V for U ∈ TX and V ∈ TY ! 30 ALEX GONZALEZ

For instance, let TR be the Euclidean topology (induced by the Euclidean metric) on R, and let TR×R be the product topology on R × R. The subsets (0, 2), (1, 3) ⊆ R are open in this topology, and we may consider the subset

W = ([0, 2] × [0, 2]) ∪ ([1, 3] × [1, 3]) ⊆ R × R

which is not of the form U × V for any U, V ∈ TR!

Exercise 3.30. Let f :(X, TX ) → (Y, TY ) be a continuous map, and let

Gf = {(x, f(x)) ∈ X × Y x ∈ X} ⊆ X × Y be the graph of the function, with the subspace topology induced from the product topology on X × Y . Show that Gf is homeomorphic to X.

The previous results can be stated actually for infinite cartesian products. S Definition 3.31. Let {Xα}α∈Γ be a collection of sets, and let X = α∈Γ Xα. The cartesian product of {Xα}α∈Γ is Y Xα = {set functions ω :Γ → X such that ω(α) ∈ Xα ∀α ∈ Γ}. α∈Γ Q The α-th coordinate of a function ω ∈ α∈Γ Xα is the element ω(α) ∈ Xα. The α-th projection map is the function Y πα : Xα → Xα α∈Γ defined by πα(ω) = ω(α).

Let {(Xα, Tα)α∈Γ be a collection of topological spaces. For each α ∈ Γ, let −1 Sα = {πα (U) | U ∈ Tα}. S Let S = α Sα, and let B be the collection of all ﬁnite intersections of elements of S. Q Proposition 3.32. The collection B is a basis for a topology on Xα. The topology α∈Γ Q TB generated by B is called the product topology and is the only topology on α∈Γ Xα satisfying the following condition. Q (P) A map f :(Z, TZ ) → ( α∈Γ Xα, TB) is continuous if and only if the composition πα ◦ f :(Z, TZ ) → (Xα, Tα) is continuous for all α ∈ Γ. ELEMENTARY TOPOLOGY I 31

Example 3.33. There are other, “natural” topologies that we can consider on the Q product Xα, but only one satisfies the universal property above. The box topol- Q α∈Γ ogy on α∈Γ Xα is the topology generated by the basis Y B = { Uα | Uα ∈ Tα}. α∈Γ If the index set Γ is finite (i.e., if we are only considering a finite cartesian product) then the box product is the same as the product topology, but this is not the case when the index set Γ is infinite. Indeed, an open subset in the product topology has the form Y Uα α∈Γ where Uα = Xα for all but finitely many values of α. It is clear then that the box topology has more open subsets than the product topology.

3.6. The quotient topology. Both the subspace topology and the product topology are natural generalizations of constructions available for metric spaces. In this section we study a new construction which is not available for metric spaces. The idea behind the quotient topology is to create new spaces by “pasting” together other spaces. For example, we can construct a cylinder out of a stripe (a subset of R2), by gluing together two non-adjacent sides, as described in the following sketch. y y

Stripe (0, 1) Cylinder (2, 1) (0, 1) (2, 1) quotient a a

(0, 0) (2, 0) x (0, 0) (2, 0) x

In this case, it is easy to provide the cylinder with a metric (just consider the cylinder as a subset of R3 with the Euclidean metric), but we can use this example to analyze a more general construction. As a set, the cylinder is the quotient set of the stripe by the following equivalence relation: (x, y) ∼ (x, y) if x 6= 0, 1 (0, y) ∼ (1, y). The question is then which topology do we put on this quotient set. Notice that the quotient set of the stripe by this equivalence relation is not a subset of R3 (although it can be identiﬁed with a subset of R3). As happened with the product of topological spaces, not any topology is acceptable. 32 ALEX GONZALEZ

Let (X, T) be a topological space, and let ∼ be an equivalence relation on the set X. You may think of the elements of X in the same equivalence class as the elements that we are going to glue together, just as we did with the stripe above. Let then X be the set of equivalence classes, and let π : X −→ X be the map that sends each x ∈ X to its equivalence class. Notice that the map π is surjective (as a map of sets). Deﬁne also −1 T = {U ⊆ X π (U) ∈ T}.

Lemma 3.34. The collection T is a topology on X.

Proof. Since ∅ = π−1(∅) ∈ T and X = π−1(X), condition (T1) to be a topology is clearly seen to hold. To check condition (T2), let {Uγ}γ∈Γ ⊆ T. We have to check that the union ∪ΓUγ ∈ T. Notice that there is an equality −1 [ [ −1 f Uγ = f (Uγ). γ∈Γ γ∈Γ −1 −1 Since f (Uγ) ∈ T by hypothesis, and T is a topology, we have ∪Γf (Uγ) ∈ T, and thus ∪ΓUγ ∈ T. Finally, to check condition (T3), let {U1,...,Un} ⊆ T. Again, there is an equality n n −1 \ \ −1 f Ui = f (Ui), i=1 i=1 and condition (T3) follows immediately.

For obvious reasons, the topological space (X, T) is called the quotient space of (X, T) by the equivalence relation ∼, and the map π : X → X is called the quotient map. Theorem 3.35. Let (X, T) be a topological space, and let ∼ be an equivalence relation on X. Let also (X, T) be the quotient space by ∼. Then, T is the unique topology on X satisfying the following conditions.

(Q1) The map π :(X, T) → (X, T) is continuous.

(Q2) A map f :(X, T) → (Y, TY ) is continuous if and only if the composition f ◦ π is continuous.

Proof. First we have to check that the topology T satisfies conditions (Q1) and (Q2). The map π : X → X is continuous since, by definition, π−1(U) ∈ T for all U ∈ T. Let f :(X, T) → (Y, TY ). If f is continuous, then clearly f ◦ π is continuous because it is a composition of continuous functions. Suppose that f ◦ π :(X, T) → (Y, TY ) −1 is continuous. We have to show that π (V ) ∈ T for all V ∈ TY . Since f ◦ π is continuous, we have (f ◦ π)−1(V ) = f −1(π−1(V )) ∈ T, and hence by definition of T it follows that π−1(V ) ∈ T. ELEMENTARY TOPOLOGY I 33

0 Now, suppose that T is another topology on X which also satisfies properties (Q1) 0 and (Q2) with respect to the map π0 :(X, T) → (X, T). We have to prove that T = T. 0 0 Consider first the identity map Id: (X, T) → (X, T ). Since T satisfies condition (Q1), the map π0 = Id ◦ π is continuous and thus Id is continuous by property (Q2). This 0 0 implies that T ⊆ T. Similarly, we can consider the map Id0 :(X, T ) → (X, T). Again, 0 0 since π = Id◦π is continuous, it follows that Id is continuous, and hence T ⊆ T.

3.7. Other constructions. Combining the subspace topology, the product topology and the quotient topology we can create many new topologies (or topological spaces). Here are some of the most relevant constructions in topology.

Example 3.36. The mapping cylinder. Let f :(X, TX ) → (Y, TY ) be a map of topological spaces. To construct the mapping cylinder of f, let a Mf = [0, 1] × X Y / ∼, where ∼ is the equivalence relation generated by (1, x) ∼ f(x) for all x ∈ X. The topology of Mf is a combination of the product topology and the quotient topology.

Example 3.37. The mapping cone. Let f :(X, TX ) → (Y, TY ) be a map of topological spaces. The mapping cone is a slight variation of the mapping cylinder: a Cf = [0, 1] × X Y / ≡, where ≡ is the equivalence relation generated by (1, x) ≡ f(x) and (0, x) ≡ (0, z) for all x, z ∈ X. Again, the topology of Cf is a combination of the product and the quotient topologies. Example 3.38. The suspension of a topological space. The suspension of (X, T) is SX = ([0, 1] × X)/ ∼, where ∼ is the equivalence relation generated by (0, x) ∼ (0, z) and (1, x) ∼ (1, z) for all x, z ∈ X. The topology on SX is again a combination of the product and the quotient topologies.

Example 3.39. The join of two spaces. The join of (X, TX ) and (Y, TY ) is X ∗ Y = ([0, 1] × X × Y )/ ≡, where ≡ is the equivalence relation generated by (0, x, y) ≡ (0, x, u) and (1, x, y) ≡ (1, z, y) for all x, z ∈ X and all y, u ∈ Y . 34 ALEX GONZALEZ

4. Connectedness

Now that we understand the basics of topological spaces, we can start comparing different topological spaces, trying to find properties that will distinguish one space from another. This chapter deals with the first of such properties: connectedness. Essentially, a topological space is connected if it cannot be formed out of smaller, disjoint pieces.

Deﬁnition 4.1. A topological space (X, TX ) is connected if it satisﬁes the following. (C) If a subset A ⊆ X is both open and closed then either A = X or A = ∅.

Let’s see some examples of connected and non-connected spaces.

Example 4.2. Let X be a set with the discrete topology. Then, X is connected if and only if X is the empty set or a singleton (if X contains more than one element, then each sub-singleton of X is open and closed).

Example 4.3. Every set with the indiscrete topology is connected.

Example 4.4. Consider R with the topology corresponding to the Euclidean metric, and let Q be the subspace of rational numbers. Then, Q is not connected. Indeed, the subset (−∞, π) ∩ Q is open and is also the complement of an open subset, hence closed.

Example 4.5. Consider the set with the topology associated to the p-adic metric. Z This space is not connected, since the subset A = {n · p n ∈ Z} ⊆ Z of integers divisible by p is both open and closed. Indeed, we have

p−1 [ A = B0,1 A = Z \ Bn,1 . n=1 The equality on the left means that A is open, while the equality on the right says that A is the complement of an open subset, hence closed.

As we have seen already several times in this course, it is always useful to have alternative statements for the same deﬁnition, so let’s see some alternative ways of deﬁning connectedness.

Theorem 4.6. Let (X, TX ) be a topological space. Then, the following are equivalent.

(1) (X, TX ) is connected. (2) There does not exist open subsets U, V ⊆ X such that U ∪V = X and U ∩V = ∅.

(3) There does not exist any surjective continuous map f :(X, TX ) → ({0, 1}, Tdisc). ELEMENTARY TOPOLOGY I 35

Proof. Let’s check first that (1) is equivalent to (3). Suppose first that there is a −1 surjective, continuous map f :(X, TX ) → ({0, 1}, Tdisc), and let A = f ({0}) and B = f −1({1}). Since f is surjective, both A and B are non-empty subsets of X. Furthermore, since {0} (respectively {1}) is and open an closed subset of {0, 1} and f is continuous, it follows that A (respectively B) is an open and closed subset of X. Finally, since both A and B are non-empty, it follows that none of them is either the empty set or X, and thus X is not connected. Suppose now that X is not connected, and let U $ X be a non-empty, open and closed subset. Let also V = U c, and notice that it is again a non-empty, open and closed subset of X. We can then define a map f :(X, TX ) → ({0, 1}, Tdisc) by setting f(U) = {0} and f(V ) = {1}, and it is clear that f is surjective and continuous. Finally let’s prove that (2) is equivalent to (3). If there exists a map f as in (3), then U = f −1({0}) and V = f −1({1}) are open subsets of X which clearly satisfy the conditions of (2). Conversely, if U, V ⊆ X satisfy the conditions of (2) then we just have to define f by f(x) = 0 if x ∈ U and f(x) = 1 if x ∈ V .

Theorem 4.7. Let f :(X, TX ) → (Y, TY ) be a continuous map of topological spaces, and suppose X is connected. Then, f(X) ⊆ Y is a connected subspace of Y .

Proof. Exercise.

Corollary 4.8. Let h:(X, TX ) → (Y, TY ) be a homeomorphism of topological spaces. Then, X is connected if and only if Y is connected.

Theorem 4.9. The set R with the topology associated to the Euclidean metric is connected.

Proof. Suppose R = A∪B, where A, B are non-empty open subsets such that A∩B = ∅. Then, A and B are both closed (they are the complement of each other), and thus ∂(A) = A ∩ R \ A = A ∩ B = A ∩ B = ∅, and similarly for B. On the other hand, we will now show that if U $ R is non-empty, then ∂(U) 6= ∅, hence contradicting the above. This part of the proof uses the Euclidean metric on R. Let then a ∈ R \ U. If U ∩ (a, ∞) = ∅ = U ∩ (−∞, a), then U = {a}, and hence U = ∂(U) 6= ∅. Suppose then that U ∩ (a, ∞) 6= ∅ (the other possible case is similar), and let b = inf{x ∈ U ∩ (a, ∞)}. Then we have

inf{d(b, u) u ∈ U} = 0 inf{d(b, v) v ∈ R \ U} = 0, and this implies that b ∈ ∂(U).

Now it is easy to deduce the following. 36 ALEX GONZALEZ

Example 4.10. For all a < b ∈ R, the open interval (a, b) ⊆ R is connected. Indeed, −π π (a, b) is clearly homeomorphic to ( 2 , 2 ) (give the explicit homeomorphism!), and the map −π π tan: , −→ 2 2 R is a homeomorphism.

Proposition 4.11. Let (X, TX ) be a topological space and let U, K ⊆ X be subsets. Suppose in addition that U is connected and U ⊆ K ⊆ U. Then K is connected.

Proof. Let f : K → {0, 1} be a continuous function (with respect to the subspace topology on K and the discrete topology on {0, 1}). Since U is connected and the restriction of f to U is a continuous function, then either f(U) = {0} or f(U) = {1}. Suppose for simplicity that f(U) = {0}.

For each k ∈ K, we have k ∈ U. Let {xn} ⊆ U be a sequence converging to k. Since f is continuous, it follows that the sequence {f(xn)} converges to f(k). On the other hand, f(xn) = 0 for all n, and thus f(k) = 0. This proves that f(K) = 0, and hence K is connected by Theorem 4.6. Theorem 4.12. Consider R with the topology associated to the Euclidean metric, and let a < b ∈ R. Then, [a, b] ⊆ R is connected.

Proof. This follows immediately from Proposition 4.11 since [a, b] is the closure of (a, b), and the latter is connected.

Lemma 4.13. Let (X, TX ) be a topological space, and let {Aγ}γ∈Γ be a collection of connected subsets of X such that

Aγ ∩ Aω 6= ∅ for all γ, ω ∈ Γ. Then, A = ∪ΓAγ is a connected subset of X.

Proof. Let f : A → {0, 1} be a continuous function. Since Aγ is connected for all γ ∈ Γ, we have

f(Aγ) = εγ, where εγ is either 0 or 1. However, for any two γ, ω ∈ Γ, we have Aγ ∩ Aω 6= ∅, and this implies that εγ = εω. Hence, the map f has to be constant on A: it cannot be surjective, and thus A is connected.

Theorem 4.14. Let (X, TX ) and (Y, TY ) be topological spaces, and let (X × Y, T) be the product topological space. Then, X × Y is connected if and only if X and Y are connected.

Proof. If X × Y is connected, then both X and Y are connected, since the projection functions πX : X × Y → X and πY : X × Y → Y are continuous maps. ELEMENTARY TOPOLOGY I 37

Suppose now that X and Y are connected. Fix then some element a ∈ X and let Γ = Y (as sets). Consider also the sets • B = {a} × Y ⊆ X × Y ; and

• Cy = X × {y} ⊆ X × Y , where y ∈ Y .

Note that B is homeomorphic to Y , while Cy is homeomorphic to X for all y ∈ Y . Thus, all the above subsets are connected subspaces of X × Y .

For each y ∈ Y , we have Cy ∩ B = {(a, y)}= 6 ∅, and hence the subset Uy = B ∪ Cy ⊆ X × Y is connected (for all y ∈ Y ) by Lemma 4.13. Now, consider the collection 0 {Uy}y∈Y . For any y, y ∈ Y , we have Uy ∩ Uy0 6= ∅, and hence [ Uy = X × Y y∈Y is connected by Lemma 4.13. Corollary 4.15. A ﬁnite cartesian product of connected topological spaces is connected. In particular, the space Rn (with the topology associated to the Euclidean metric) is connected. Example 4.16. Let X = RN be a countable cartesian product of copies of R. Let also T× and T be the product topology and the box topology respectively (see Example 3.33). Then, (X, T) is not connected. To see this, let • A ⊆ X be the set of all bounded sequences of real numbers (i.e., if a ∈ A then there exists some N ∈ R such that |an| ≤ N for all n). • B ⊆ X be the set of all unbounded sequences of numbers. These two subsets are clearly disjoint, and X = A S B. We claim that they are both open. Let a ∈ X be a point, and consider the open subset

U = (a1 − 1, a1 + 1) × (a2 − 1, a2 + 1) × ... If a ∈ A (if a is bounded), then all the points in U are bounded sequences, thus U ⊆ A. On the other hand, if a ∈ B (if a is unbounded), then every point of U is an unbounded sequence, so U ⊆ B. This way, we can describe A and B as unions of open subsets in the product topology, and thus they are both open.

We have just checked that (X, T) is not connected, and now we prove that (X, T×) is connected. For each n ∈ N, let Xn ⊆ X be the subset of sequences a = (a1, a2,..., ) such that ai = 0 for all i ≥ n + 1. With the subspace topology, Xn is homeomorphic n to R (with the Euclidean topology), and thus Xn is connected for all n ∈ N. Let X = S X ⊆ X. Since (0, 0,...) is a common element of X for all n, it ∞ n∈N n n follows that X∞ is connected by Lemma 4.13.

To ﬁnish the proof, we show that X∞ = X (this step fails if we replace the product topology by the box topology). Let a ∈ X, and let U = Q U be an element of the n∈N n 38 ALEX GONZALEZ basis for the product topology that contains a. By deﬁnition there is some N ∈ N such that Ui = R for all i ≥ N + 1, and

x = (a1, . . . , aN , 0, 0,...) ∈ U ∩ X∞.

This shows that indeed X is the closure of X∞. By Proposition 4.11, X is connected. Lemma 4.17. Let f :(X, d ) → (Y, d ) be a continuous function, and suppose that X X Y is connected. Then, the graph of f, Gf = {(x, y) ∈ X × Y y = f(x)}, is a connected subspace of X × Y .

Proof. Prove that the space Gf is homeomorphic to X (exercise). Then the statement follows. Example 4.18. This example is known as the topologist’s sine curve. Let 2 2 2 Ω = {(x, sin(1/x)) ∈ R x > 0} ∪ {(0, y) ∈ R y ∈ [−1, 1]} ⊆ R . The following is a sketch of Ω.

To show that this space is connected, let 2 Γ = {(x, sin(1/x)) ∈ R x > 0}. First note that the subset Γ is connected because it is the graph of the continuous function f : (0, ∞) → R deﬁned by f(x) = sin(1/x), and (0, ∞) is connected. The space Ω is the closure of Γ, i.e. Ω = Γ (prove it!), and by Proposition 4.11 Ω is connected. We will come back to this example in the next section.

4.1. Path-connected spaces. Intuition may tell us that a space is connected if we can “link together” any two elements by a path. Although this idea is not exactly right, it is the basis of an important concept that we analyze in this section.

Deﬁnition 4.19. Let (X, TX ) be a topological space, and let x, y ∈ X.A path in X from x to y is a continuous map ω : [0, 1] → X with ω(0) = x and ω(1) = y. ELEMENTARY TOPOLOGY I 39

A topological space (X, TX ) is path-connected if, for any two elements x, y ∈ X there is a path from x to y.

Theorem 4.20. Let (X, TX ) be a path-connected space. Then (X, TX ) is connected.

Proof. Suppose that X is not connected, and let f : X → {0, 1} be a continuous, surjective map. Let also x ∈ f −1({0}) and y ∈ f −1({1}). Since X is path-connected, there is some path ω : [0, 1] → X with ω(0) = x and ω(1) = y. However, the composition f ◦ ω : [0, 1] → {0, 1} is both continuous (it is a composition of continuous functions) and is surjective, and this is impossible since [0, 1] is connected by Theo- rem 4.12. Example 4.21. For all n ≥ 1, Rn is path-connected. Example 4.22. A circle in R2 is the image of a path ω : [0, 1] → R which satisﬁes the extra condition ω(0) = ω(1). Thus, a circle is connected.

Exercise 4.23. Let (X, TX ) be a path-connected space, and let x, y, z ∈ X. Let also ω, γ : [0, 1] → X be paths from x to y and from y to z respectively. How can you deﬁne a path from x to z using ω and γ? Corollary 4.24. A cylinder is not homeomorphic to a Moebius band.

Proof. Let M be a Moebius band and C be a cylinder, both seen as subsets of R3. Let also ∂(M) and ∂(C) be the boundaries of M and C are subsets of R3. Clearly, ∂(C) is not connected, while ∂(M) is path-connected (hence connected). Thus, no map h: C → M can be a homeomorphism.

It turns out that not all connected spaces are path-connected, and we have already met the first example of this behavior recently. Example 4.25. Let Ω ⊆ R2 be the subset in Example 4.18, the topologist’s sine curve, and recall that Ω is connected. Let’s prove now that Ω is not path-connected. Let then 2 2 A = {(0, y) ∈ R y ∈ [−1, 1]} Γ = {(x, sin(1/x)) ∈ R x > 0}. If we choose elements a ∈ A and b ∈ Γ, then there is no path in Ω linking a to b! To prove this, suppose otherwise and let ω : [0, 1] → Ω be a path with ω(0) = (0, 0) and 1 ω(1) = ( π , 0). Let also F : [0, 1] → R be the continuous map ω π1 F : [0, 1] −→ Ω −→ R, 1 where π1(x, y) = x. Since F (0) = 0 and F (1) = π , we can use the Intermediate Value 2 Theorem to find some t1 ∈ (0, 1) such that F (t1) = 3π . Using again the Intermediate 2 Value Theorem we can then find some t2 ∈ (0, t1) such that F (t2) = 5π . Iterating the process we can find a sequence {tn}n≥1, such that 2 t ∈ (0, t ) and F (t ) = n+1 n n (2n + 1)π 40 ALEX GONZALEZ

Since the sequence {tn} ⊆ [0, 1] is monotonic decreasing and bounded below, it is convergent: there is some t ∈ [0, 1] such that lim tn = t. Finally, since ω is continuous 2 n the sequence {ω(tn)} converges to ω(t). On the other hand, ω(tn) = ( (2n+1)π , (−1) ), and this sequence is not convergent, hence a contradiction.

4.2. Connected components. Connectivity constitutes then the ﬁrst of a series of properties that we can use to characterize and distinguish topological spaces. Let’s elaborate this idea a little bit more: a connected space is one which is not made of smaller, disjoint pieces, but what happens in general? That is, given a topological space (X, TX ), how many pieces are needed to “construct” X?

Deﬁnition 4.26. Let (X, TX ) be a topological space. A connected component of X is a connected subset U ⊆ X such that • if U $ K ⊆ X, then K is not connected.

The collection of connected components of X is denoted by π0(X).

Alternatively, the set π0(X) can be constructed as follows. Deﬁne a relation on the set X: x ∼ y if there is some connected subset U ⊆ X such that x, y ∈ U. This is clearly an equivalence relation, and then we can deﬁne π0(X) as the set of equivalence classes.

Proposition 4.27. Every continuous map f :(X, TX ) → (Y, TY ) induces a map of sets

f∗ : π0(X) −→ π0(Y ).

Proof. Let U ⊆ X be a connected component of X, and let x ∈ U. Let also V ⊆ Y be a connected component which contains f(x) (each y ∈ Y has to be contained in such a subset). We can deﬁne then f∗(U) = V . To check that this map is well-deﬁned, let z ∈ U be any other element, and let w = f(z) ∈ Y . We have to prove that f(z) ∈ V . Suppose otherwise: f(z) ∈/ V . Then V $ W = V ∪ {f(z)}, and thus W is not connected. But this is impossible, since f(U) has to be connected.

Corollary 4.28. Let f :(X, TX ) → (Y, TY ) be a homeomorphism. Then f deﬁnes a bĳection of sets f∗ : π0(X) → π0(Y ).

In other words, if X and Y are homeomorphic, they have to be made out of the same number of pieces. ELEMENTARY TOPOLOGY I 41

5. Compactness

Compactness can be thought of as a generalization of finiteness. If X is a finite set (with the discrete topology) then studying (continuous) maps out of X is reasonably easy, we just have to check what happens with the function on every element of X. But we want to consider other maps, not only maps out of finite sets! The idea is that maps out of compact spaces behave nicely, as maps out of finite spaces (of course this is rather imprecise).

Deﬁnition 5.1. Let (X, TX ) be a topological space and let A ⊆ X be any subset. An open cover of A in X is a collection {Uγ}γ∈Γ of open subsets of X such that [ A ⊆ Uγ. γ∈Γ

A subcover of {Uγ}γ∈Γ is a subcollection {Vω}ω∈Ω ⊆ {Uγ}γ∈Γ such that A ⊆ ∪ΩVω.

Example 5.2. Consider the family of intervals Un = (n, n + 2) ⊆ R (Euclidean topology), for n ∈ Z. The collection {Un}n∈Z is an open cover of R. This open cover does not contain any proper subcover: if we remove Un for some n ∈ Z then the union of the remaining subsets Um, m 6= n, does not cover R.

Deﬁnition 5.3. Let (X, TX ) be a topological space. A subset A ⊆ X is compact if every open cover of A contains a ﬁnite subcover.

Remark 5.4. Let (X, TX ) be a topological space and A ⊆ X. Let also {Uγ}γ∈Γ be an open cover of A in X. A finite subcover is a finite subcollection {Vω}ω∈Ω ⊆ {Uγ}γ∈Γ which is also an open cover of A. This is already a hint of the idea that compactness generalizes finiteness. WARNING!. As a note of caution, be extremely careful with the above definition. Specially with the bit every open cover... It is not enough to check that an open cover contains a finite subcover, we have to check ALL possible open covers! Indeed, notice that, for any topological space (X, TX ), the collection {X} is a finite open cover of X, but this does not make X compact!

The best way to understand this notion is by looking at some examples. Let’s start with example of spaces which are not compact. Example 5.5. Consider R with the Euclidean topology. Then, the open cover in Example 5.2 does not contain any ﬁnite subcover, and hence R is not compact. Example 5.6. Consider R with the Euclidean topology, and let (0, 1) ⊆ R be the open interval. Then (0, 1) is not compact. Indeed, consider the open cover ( ) 1 U = , 1 n n n∈N It is easy to see that this open cover does not contain any ﬁnite subcover of (0, 1). 42 ALEX GONZALEZ

Example 5.7. Let X = {0} ∪ {1/n | n ∈ N}, with the subspace topology of a subset of R. Then X is a compact. Indeed, let {Uγ}γ∈Γ be an open cover of X, and let α ∈ Γ be such that 0 ∈ Uα. Then Uα must contain all but ﬁnitely many elements of X. For each x ∈ X \ Uα, choose Ux ∈ {Uγ}γ∈Γ containing x. This forms a ﬁnite subcover.

Theorem 5.8. Let (X, TX ) be a compact space. If A ⊆ X is closed then A is compact.

Proof. Let A ⊆ X be a closed subset, and let {Uγ}γ∈Γ be an open cover (recall that the c S subsets Uγ are open in X). Then the collection {A } {Uγ}γ∈Γ is an open cover of X, and there exists some ﬁnite subcover, {W1,...,Wn}, since X is compact. But clearly c this subcover also covers A, and upon removing A from {W1,...,Wn} if necessary this constitutes a ﬁnite subcover of the original cover {Uγ}γ∈Γ. Thus A is compact.

Theorem 5.9. Let f :(X, TX ) → (Y, TY ) be a continuous map of topological spaces. If A ⊆ X is compact then so is f(A) ⊆ Y .

Proof. Let {Uγ}γ∈Γ be an open cover of f(A) in Y . Since f is a continuous map, the −1 collection {f (Uγ)}γ∈Γ is then an open cover of A. Now, since A is compact we know −1 −1 that there is some ﬁnite subcover {f (Uγ1 ), . . . , f (Uγr )}, and thus {Uγ1 ,...,Uγr } is a ﬁnite subcover of f(A).

Corollary 5.10. Let f :(X, TX ) → (Y, TY ) be a homeomorphism. Then X is compact if and only if Y is compact.

Theorem 5.11. Let (X, TX ) and (Y, TY ) be topological spaces, and let A ⊆ X, B ⊆ Y be compact subsets. Then, A × B is a compact subset of X × Y .

Proof. Before we prove the theorem, we show the following.

Step 1. Let a ∈ A and let N ⊆ X × Y be an open subset such that {a} × B ⊆ N. Then there exists some U ⊆ B such that {a} × B ⊆ U × B ⊆ N.

You can think of U × B as a tube around the ray {a} × B. To show the above point, let {Wγ}γ∈Γ be an open cover of {a} × B, where, for each γ, the subset Wγ is an element of the basis for the product topology such that Wγ ⊆ N. In other words, for each γ ∈ Γ,

Wγ = Uγ × Vγ ⊆ N,

where Uγ ∈ TX and Vγ ∈ TY . ∼ Since {a} × B = B and B is compact, there exists a ﬁnite subcover {W1,...,Wn}, convering {a} × B. The following picture illustrates the situation. ELEMENTARY TOPOLOGY I 43

W3 B

W5 W6

Recall that Wk = Uk × Vk for k = 1, . . . , n. Set

U = U1 ∩ ... ∩ Un ∈ TX . Then a ∈ U. Furthermore, if (x, y) ∈ U × B, then there exists some k such that (x, y) ∈ Uk × Vk. Indeed, since {W1,...,Wn} covers {a} × B, there is some k such that (a, y) ∈ Wk = Uk × Vk, and then (x, y) ∈ Wk since U ⊆ Uk. Since Wγ ⊆ N for all γ ∈ Γ, it follows that U × B ⊆ N.

Step 2. Proof of the theorem.

Let {Wγ}γ∈Γ be an open cover of A × B. Given a ∈ A, the ray {a} × B is compact and

may be covered by ﬁnitely many elements Wa,1,...,Wa,ma of the original cover. The

union Na = Wa,1 ∪ ... ∪ Wa,ma is an open susbet containing {a} × B, and by Step 1 we know that there exist a tube Ua × B such that

{a} × B ⊆ Ua × B ⊆ Na.

Thus, for each a ∈ A we may choose some Ua ∈ TX containing the element a such that the tube Ua × B can be covered by ﬁnitely many elements of {Wγ}γ∈Γ. Note that {Ua}a∈A forms an open cover of A, and thus there exist a1, . . . , an ∈ A such that

{Ua1 ,...,Uan } is a ﬁnite subcover, since A is compact.

Note that A × B ⊆ Ua1 × B ∪ ... ∪ Uan × B. Thus, the collection {W ...,W ,W ,...,W ,...,W ,...,W } a1,1 a1,ma1 a2,1 a2,ma2 an,1 an,man is a ﬁnite subcover of A × B. 44 ALEX GONZALEZ

5.1. Compact subspaces of Rn. In this section we study compact subspaces of a very particular type. Be extremely careful with this section because most of the results here apply only to Rn with the Euclidean metric or, in some cases, to metric spaces. We start with a deﬁnition.

Deﬁnition 5.12. Let (X, dX ) be a metric space. A subset A ⊆ X is bouned if there exist x ∈ X and ε > 0 such that A ⊆ Bx,ε.

Note that the above deﬁnition does not make much sense for topological spaces! Let us now characterize compact subsets of R. Theorem 5.13 (Heine-Borel Theorem for R). Consider R with the Euclidean metric, and let A ⊆ R. Then, A is compact if and only it is closed and bounded.

Proof. Suppose ﬁrst that A is compact. To see that it is bounded, consider the following open cover of A:

{Un = B0,n n ∈ N}. Since it is an open cover of A and A is compact, there exists some ﬁnite subcover, namely {Un1 ,...,Unr }. Let N = max{n1, . . . , nr}. Clearly, Un1 ,...,Unr ⊆ UN , and hence A ⊆ UN = B1,N , i.e. A is bounded. Let’s see also that A is closed. Suppose otherwise, and let x ∈ A \ A. For each n ∈ N, let c Un = B 1 = \ B 1 . x, n R x, n By deﬁnition, each Un is the complement of a closed subset of R and thus is open. Furthermore, we have [ [ \ Un = \ B 1 = \ B 1 = \{x} ⊇ A. R x, n R x, n R n∈N n∈N n∈N

In other words, {Un}n∈N is an open cover of A in R, and since A is compact, it contains a ﬁnite subcover, {Un1 ,...,Uns }. Furthermore, by deﬁnition of the subsets

Un, we have Un1 ,...,Uns ⊆ UN , where N = max{n1, . . . , ns}, and also A ⊆ UN . But then we have A ∩ B 1 = ∅, x, N which contradicts the hypothesis that x ∈ A. Hence A = A. Let’s prove now the converse. Assume then that A is closed and bounded, and let {Uγ}γ∈Γ be an open cover of A. Since A is closed and bounded, there exist m and M ∈ A such that m ≤ a ≤ M ∀a ∈ A.

Now, for each x ∈ R, deﬁne the following sets:

• Wx = (−∞, x]; and

• Ax = A ∩ Wx. ELEMENTARY TOPOLOGY I 45 Define also B = x ∈ R Ax is covered by a finite subcover of {Uγ} , and note that B is non-empty, since m ∈ B. Indeed, Am = A ∩ Wm = {m} by definition of m, and it is clear that Am is covered by a finite subcover of {Uγ}. Notice also that, to finish the proof, it is enough to show that B is not bounded above: if this is the case, then M ≤ y for some y ∈ B, and then by definition of B we have that A = AM = Ay is covered by a finite subcover of {Uγ}. We will prove that B is not bounded above by contradiction. Suppose then that u = sup B exists: y ≤ u for all y ∈ B. If u∈ / A, then there exists some ε > 0 such that Bu,ε ∩ A = ∅ (this is because A is closed). This implies that

Au−ε = Au+ε. But notice that u−ε < u, and hence u−ε ∈ B. The above implies then that u+ε ∈ B, and this contradicts the hypothesis that u = sup B.

Thus we have to assume that u ∈ A. In this case, since {Uγ} is an open conver of A, there is some Uω ∈ {Uγ}such that u ∈ Uω. Furthermore, since Uω is open, we can ﬁnd some δ > 0 such that [u − δ, u + δ] ⊆ Uω.

Now, since u − δ < u, the set Au−δ is covered by a ﬁnite subcover of {Uγ}, namely

{Uγ1 ,...,Uγm }, and then {Uω,Uγ1 ,...,Uγm } is a ﬁnite subcover of Au+δ. Again, this implies that u + δ ∈ B, which contradicts the hypothesis that u = sup B. Hence, B is not bounded above, and A is compact.

We can use this theorem to deduce that some other subsets of R are compact. The following example is not intuitive at all. Example 5.14. Let K ⊆ R be the Cantor set (see Exercise 6 in List 3): start from 1 2 1 2 I0 = [0, 1] ⊆ R, then form I1 = [0, 3 ] ∪ [ 3 , 1] by removing the middle third ( 3 , 3 ) of I0, and keep iterating the process. We end up with a sequence {In}n∈N of subsets of R, and ∞ \ K = In. n=0 Since each In is closed and K is an (inﬁnite) intersection of closed subsets, we know that K is closed. Also, it is clearly bounded, since K ⊆ I0, and hence by the Heine- Borel it is compact.

Proposition 5.15. Let (X, dX ) be a metric space, and let A ⊆ X be a compact subspace. Then, A is closed and bounded.

Proof. First we check that A is bounded. Fix some x ∈ X and let Un = Bx,n for each n ∈ N. The collection {Un}n∈N is thus an open cover of A in X, and since A is compact there exists some ﬁnite subcover, {Un1 ,...,Unr }. Let then N = max{n1, . . . , nr}: we have A ⊆ UN = Bx,N and hence A is bounded. 46 ALEX GONZALEZ

To see that A is also closed, suppose otherwise and let x ∈ A) \ A. For each n ∈ N deﬁne then c Un = B 1 = X \ B 1 . x, n x, n Then, Un is open for all n and we have [ [ \ Un = X \ B 1 = X \ B 1 = X \{x} ⊆ A. x, n x, n n∈N n∈N n∈N

Thus, {Un}n∈N is an open cover of A, and since A is compact there is some ﬁnite subcover, {Un1 ,...,Uns }. Furthermore, by setting N = max{n1, . . . , ns}, we have A ⊆ UN , which implies A ∩ B 1 = ∅, x, N contradicting the hypothesis that x ∈ A. Thus, A = A. Theorem 5.16 (Heine-Borel Theorem for Rn). Consider Rn with the Euclidean metric, and let A ⊆ Rn be a subset. Then, A is compact if and only if it is closed and bounded.

Proof. By Proposition 5.15, if A is compact then it is closed and bounded. Suppose then that A is closed and bounded. In particular, there is some interval [m, M] ⊆ R such that n A ⊆ I = [m, M] × ... × [m, M] ⊆ R . Now, the subset I ⊆ Rn is compact by Theorem 5.11, since I is the direct product of compact subsets of R, and then by Lemma 5.8 A is compact since it is a closed subset of I.

We can use now the Heine-Borel Theorem to prove that a certain subset of Rn is compact. Example 5.17. Consider R2 with the Euclidean metric. Then, every closed disk and every circle is compact. More generally, in the Euclidean space Rn, the sphere Sn−1, n−1 n 2 2 S = {(x1, . . . , xn) ∈ R x1 + ... + xn = 1} is a compact space. Example 5.18. The cylinder and the Moebius band are both compact subsets of R3. Thus compactness is not a good property to distinguish these two subsets. Fortunately connectivity has done the job for us already. Example 5.19. Consider the set Z with the discrete metric. Then, Z is closed and bounded, but not compact. Indeed, Z is closed as a subset of itself, and it is bounded because Z is contained in the open ball of radius 2 around any element. However, every singleton {n} in is an open subset, and hence the collection {n} is an Z n∈Z open cover which does not contain any ﬁnite subcover. ELEMENTARY TOPOLOGY I 47

6. Countability and separation axioms

In this chapter we analyze certain properties of diﬀerent nature: countability (of bases for a given topology) and separation (of diﬀerent subsets in a given topology). The ultimate goal is to give conditions for a given topological space to be metrizable (i.e., the given topology is actually the topology associated to some metric).

6.1. The countability axioms. Recall the deﬁnition of basis of a topology, see Sec- tion 3.3.

Deﬁnition 6.1. Let (X, TX ) be a topological space. X has a countable basis at x if there is a countable collection B of open subsets satisfying the following: (i) x ∈ U for each U ∈ B; and

(ii) if V ∈ TX is such that x ∈ V , then there exists some U ∈ B such that U ⊆ V . If X has a countable basis at every element x ∈ X, then X satisfies the first countability axiom. We also say that X is first-countable.

Theorem 6.2. Let (X, dX ) be a metric space, and let TX be the topology associated to dX . Then (X, TX ) is ﬁrst-countable.

Proof. For each x ∈ X, the collection B = {Bx, 1/n} satisﬁes conditions (i) and (ii) in the deﬁnition above.

Theorem 6.3. Let (X, TX ) be a ﬁrst-countable space. Then the following holds.

(i) Let A ⊆ X and x ∈ X. Then, x ∈ A if and only if there exists a sequence {xn}n∈N of points of A converging to x.

(ii) Let f :(X, TX ) → (Y, TY ). Then f is continuous if and only if, for every convergent sequence {xn}n∈N in X with limit x, the sequence {f(xn)}n∈N converges to f(x).

Proof. Part (i) is proved by modifying appropriately the proof of Theorem 2.30. The necessary modiﬁcations in the proof of Theorem 2.31 imply part (ii).

Deﬁnition 6.4. A topological space (X, TX ) satisﬁes the second countability axiom if it has a countable basis for TX . We also say that X is second-countable.

Remark 6.5. If (X, TX ) is second-countable then it is ﬁrst-countable, but the converse is not true in general.

Lemma 6.6. Let (X, TX ) be a second-countable space. Then, every discrete subspace A ⊆ X is countable.

Proof. Let B ⊆ TX be a countable basis. Since A is discreet, every singleton {a} ⊆ A is an open subset of A, which means that there exists some U ∈ TX such that U ∩ A = {a}. Hence, for each a ∈ A we may choose a basis element Ua ∈ B such 48 ALEX GONZALEZ that Ua ∩ A = {a}. This produces an injective map A → B, where a 7→ Ua (if a 6= b then Ua ∩ Ub = ∅, so this map is indeed injective), and thus A is countable.

Example 6.7. Let Y = R, and let dY be the Euclidean metric (i.e. absolute value). Let also d: R × R → [0, ∞) be the map deﬁned by d(x, y) = min{|x − y|, 1}.

As an exercise, we can show that this is a metric on Y . Now, consider X = Πγ∈ΓR, an infinite (possibly uncountable) product of copies of R. Define also dX : X×X → [0, ∞) by dX = sup{dY (xγ, yγ) | γ ∈ Γ}. This defines a metric on X, called the uniform metric. Again, details that this is a metric are left as an exercise. Since (X, dX ) is a metric space, it is first-countable by Theorem 6.2. We claim that it is not second-countable. Let A ⊆ X be the subset of elements (xγ)γ∈Γ such that xγ ∈ {0, 1} for each γ ∈ Γ. Then A is a discreet subspace, because dX ((xγ), (yγ)) = 1 if (xγ) 6= (yγ). By Lemma 6.6, if (X, TX ) was second-countable then A should be countable, but A is not countable. Proposition 6.8. The following holds. (i) A subset of a first-countable space is again first-countable. (ii) A countable product of first-countable spaces is again first-countable. (iii) A subset of a second-countable space is again second-countable. (iv) A countable product of second-countable spaces is again second-countable.

Proof. The proof is left as an exercise.

Deﬁnition 6.9. Let (X, TX ) be a topological space. A subset A ⊆ X is dense if A = X.

Theorem 6.10. Let (X, TX ) be second-countable, and let A ⊆ X. Then the following holds. (i) Every open covering of A contains a countable subcover. (ii) There exists a countable subset of X that is dense in X.

Proof. Let B be a countable basis for X, and let BA = {W ∩A | W ∈ B} be a countable basis for the subspace topology on A. Since BA is a countable basis, we may write it as {W1,W2,...}.

Part (i). Let {Uγ}γ∈Γ be an open covering of A. For each n ∈ N, let Uγn ∈ {Uγ} be such that Wn ⊆ Uγn or, if no such Uγn exists, set Uγn = ∅. Let also

Ω = {Uγn | n ∈ N and Uγn 6= ∅} ⊆ {Uγ}γ∈Γ. ELEMENTARY TOPOLOGY I 49

The collection Ω is countable, and also covers A. Indeed, for each a ∈ A there is some Uα ∈ {Uγ}γ∈Γ such that a ∈ Uα, and thus Uα contains some element of BA by deﬁnition of basis.

Part (ii). From each nonempty basis element Wn ∈ BA, choose some xn ∈ Wn, and let D = {xn | Wn ∈ BA , Wn 6= ∅}. Then D is dense in A: for each a ∈ A, every basis element in BA that contains a must intersect D, so x ∈ D.

6.2. The separation axioms. Given a metric space, we have already seen (in the exercise lists) that every two elements can be separated from each other by two disjoint subsets. We have also seen examples of topological spaces where this property does not hold any more (for instance X = {a, b} with the indiscrete topology). In this section we analyze diﬀerent levels of separability in topological spaces and their consequences.

Deﬁnition 6.11. A space (X, TX ) is Hausdorﬀ if the following condition holds.

(H) For each two distinct elements x, y ∈ X, there exist U, V ∈ TX such that x ∈ U, y ∈ V and U ∩ V = ∅.

A space (X, TX ) is regular if every singleton {x} ⊆ X is closed and the following condition holds.

(R) For each x ∈ X and each closed subset B ⊆ X, there exist U, V ∈ TX such that x ∈ U, B ⊆ V and U ∩ V = ∅.

A space (X, TX ) is normal if every singleton {x} ⊆ X is closed and the following condition holds.

(N) For each pair of distinct closed subsets A, B ⊆ X there exist U, V ∈ TX such that A ⊆ U, B ⊆ V and U ∩ V = ∅.

Lemma 6.12. Let (X, TX ) be a Hausdorﬀ space. Then every singleton {x} ⊆ X is a closed subset.

Proof. Fix x ∈ X and let y 6= x be another element of X. By deﬁnition there exist U, V ∈ TX such that x ∈ U, y ∈ V and U ∩ V = ∅. This implies that V ∩ {x} = ∅, and thus y does not belong to {x}. Remark 6.13. In view of the previous result, it is clear that normal implies regular, and regular implies Hausdorﬀ.

Theorem 6.14. Let (X, dX ) be a metric space, and let TX be the topology associated to dX . Then (X, TX ) is normal.

Proof. Clearly, a metric space is Hausdorﬀ, and thus singletons are closed subsets. Let A, B ⊆ X be closed and disjoint subsets. Note that the points in A are not limit points of B, and reciprocally the points in B are not limit points of A. Thus, 50 ALEX GONZALEZ

(1) for each a ∈ A there is some ε(a) such that Ba,ε(a) ∩ B = ∅; and

(2) for each b ∈ B there is some ε(b) such that Bb,ε(b) ∩ A = ∅. S S Let U = a∈A Ba,ε(a)/2 and V = b∈B Bb,ε(b)/2. Then clearly A ⊆ U and B ⊆ V , and we have to show that U ∩ V = ∅. Suppose there exists some x ∈ U ∩ B. Then there exist a ∈ A and b ∈ B such that x ∈ Ba,ε(a)/2 ∩ Bb,ε(b)/2. Suppose for simplicity that ε(a) ≤ ε(b). Then,

dX (a, b) ≤ dX (a, x) + dX (b, x) < ε(a)/2 + ε(b)/2 ≤ ε(b).

Thus a ∈ Bb,ε(b), which contradicts our choice of ε(b). Thus U ∩ V = ∅.

Theorem 6.15. Let (X, TX ) be a Hausdorﬀ space, and let {xn}n∈N be a sequence. If {xn}n∈N is convergent then its limit is unique.

Proof. Suppose that {xn}n∈N converges to x, and let y 6= x be a different element of X. By definition there exist U, V ∈ TX such that x ∈ U, y ∈ V and U ∩ V = ∅. Since the sequence converges to x, the open subset U contains all but finitely many terms of the sequence, which implies that V contains at most finitely many terms of the sequence. Thus {xn}n∈N cannot converge to y. Example 6.16. Let X = R. Let also K = {1/n | n ∈ N}, and let B = {(a, b) | a, b ∈ R} ∪ {(a, b) \ K | a, b ∈ R}.

This forms a basis for a topology on X, namely TX . The space (X, TX ) is easily seen to be Hausdorﬀ. However, it is not regular, as we next show: the set K is a closed subset of X, but it cannot be separated from {0}.

Suppose otherwise, and let U, V ∈ TX be disjoint subsets such that 0 ∈ U and K ⊆ V . Let also W ∈ B be such that 0 ∈ W ⊆ U. Then W must be of the form W = (a, b) \ K (if W = (a, b) then W ∩ K 6= ∅). Choose n ∈ N large enough so that 1/n ∈ (a, b), and let Z ∈ B be such that 1/n ∈ Z ⊆ V . Notice that Z = (c, d) for some c, d ∈ R. Finally, choose z ∈ R so that z < 1/n and z > max{c, 1/(n + 1)}. Then z ∈ U ∩ V , contradicting the hypothesis that they are disjoint.

Example 6.17. Let X = R, and let B be the collection of intervals of the form [a, b), with a < b. Then B forms a basis for a topology on X, namely TB. In this example we show that (X, TB) is a normal space, while the product space X × X is not. To prove that X is normal, ﬁrst note that (a, b) is open in this topology for all a < b. S Indeed, we can write (a, b) = x∈(a,b)[x, b). Thus, it is easy to check that singletons are closed subsets in this topology. To prove normality, let A, B ⊆ X be closed, disjoint subsets. For each a ∈ A let [a, xa) be an open subset not intersecting B, and similarly for each b ∈ B let [b, yb) be an open subset not intersecting A. Then, S S U = a∈A[a, xa) and V = b∈B[b, yb) are disjoint open subsets separating A and B. ELEMENTARY TOPOLOGY I 51

Next we show that X × X is not normal (with the product topology). Notice that X × X is regular, since X is regular. Now, suppose that X × X is normal, and let L = {(x, −x) | x ∈ X}. Then the following holds (it is left as an exercise to prove these properties actually hold):

(1) L is closed in X × X; and

(2) L has the discrete topology as a subspace of X × X.

Let now A ⊆ L be any closed subset. Since L itself is closed, it follows that A is also closed in X × X. Actually, both A and L \ A are closed in X × X, and there are disjoint open subsets UA and VA separating A and L \ A. Let D ⊆ X × X be the subset of points with rational coordinates, which is a dense subset of X × X, and deﬁne a map θ : {subsets of L} → {subsets of D} as follows.   θ(A) = ∅,A = ∅ θ(A) = θ(A) = D,A = D  θ(A) = D ∩ UA, otherwise

Here, UA is the subset of X × X ﬁxed in the previous paragraph. We claim that the map θ is injective. Let A ⊆ L be a proper subset. Then θ(A) = D ∩ UA 6= ∅,D, since D ∩ UA,D ∩ VA 6= ∅. Now, if A, B ⊆ L are proper, distinct subsets, then one of them contains an element that is not contained in the other. For simplicity suppose that (x, −x) ∈ A, (x, −x) ∈/ B. Then (x, −x) ∈ L \ B, and thus

(x, −x) ∈ UA ∩ VB.

Notice that UA ∩ VB is open and nonempty, and thus K = D ∩ (UA ∩ VB) 6= ∅, since D is dense. By deﬁnition, the elements of K are elements of D ∩ UA, but not of D ∩ UB, and thus θ(A) 6= θ(B).

Next we construct an injective map φ: {subsets of D} → L. Since D is countable and L has the cardinality of R, it is enough to construct a map from {subsets of N} into R. For each subset S ⊆ N, deﬁne ∞ k φ(S) = Σk=1ak/10 , where ak = 0 if k ∈ S and ak = 1 if k∈ / S. This is clearly injective. Combining the map φ with the map θ, we obtain an injective map {subsets of L} −→ L, but no such map can exists. Hence a contradiction.

Lemma 6.18. Let (X, TX ) be a topological space whose singletons are closed subsets. Then the following holds.

(i) X is regular if and only if given a point x ∈ X and U ∈ TX containing x, there is a some V ∈ TX such that x ∈ V ⊆ V ⊆ X.

(ii) X is normal if and only if given a closed subset A ⊆ X and U ∈ TX containing A, there is a some V ∈ TX such that A ⊆ V ⊆ V ⊆ X. 52 ALEX GONZALEZ

Proof. We only prove part (i), part (ii) being an easy exercise. Suppose X is regular, c and let x ∈ X, U ∈ TX be given. Let B = U , which is a closed subset not containing x. Since X is regular, there exist disjoint subsets V,W ∈ TX such that x ∈ V and B ⊆ W . We claim that V ∩ B = ∅: if y ∈ B, then W ∈ TX is such that y ∈ W and W ∩ V = ∅. Thus, V ⊆ U as desired. To prove the converse, let x ∈ X and let B ⊆ X be a closed subset not containing x. We have to show that there are disjoint open subsets separating x and B. Let U = c B ∈ TX . Then x ∈ U, and there exists some V ∈ TX be such that x ∈ V ⊆ V ⊆ U. c Let W = V . Then x ∈ V , B ⊆ W and V ∩ W = ∅. Thus X is regular. Theorem 6.19. The following holds. (i) A subset of a Hausdorff space is Hausdorff. (ii) A product of Hausdorff spaces is Hausdorff. (iii) A subset of a regular space is regular. (iv) A product of regular spaces is regular.

Proof. Point (i). Let (X, TX ) be a Hausdorﬀ topological space, let A ⊆ X, and let x, y ∈ A. Since X is Hausdorﬀ, there exist U, V ∈ TX such that x ∈ U, y ∈ V and U ∩ B = ∅. Then U ∩ A, V ∩ A ∈ TA, x ∈ U ∩ A, y ∈ V ∩ A and (U ∩ A) ∩ (V ∩ A) = ∅.

Point (ii). Let {Xγ}γ∈Γ be a family of Hausdorﬀ spaces, let X = Πγ∈ΓXγ be the product space, and let x = (xγ), y = (yγ) be two distinct elements of X. Since x 6= y, there exists some γ ∈ Γ such that xγ 6= yγ. Choose W, Z ∈ TXγ to be disjoint subsets −1 −1 separating xγ and yγ, and let U = πγ (W ) and V = πγ (Z), where πγ : X → Xγ is the projection map. Then U and V are disjoint open subsets of X that separate x and y.

Point (iii). Let (X, TX ) be a regular space, and let A ⊆ X. By (i), A is Hausdorﬀ, and thus singletons are closed subsets. Let also x ∈ A be a point and B ⊆ A be a closed subset (B is closed in A), x∈ / B. Let C ⊆ X be the closure of B in X, so B = C ∩ A. Then x∈ / C, and the regularity of X implies that there exist U, V ∈ TX such that x ∈ U, C ⊆ V and U ∩ V = ∅. Thus U ∩ A and V ∩ A are disjoint open subsets of A separating x and B.

Point (iv). Let {Xγ}γ∈Γ be a family of regular spaces, and let X = Πγ∈ΓXγ be the product space. By (ii), X is Hausdorﬀ, and thus singletons are closed subsets. Let x = (xγ) ∈ X be a point, and let U ⊆ X be an open subset containing x. Choose a basis open subset W = ΠγUγ such that x ∈ W ⊆ U. Since each Xγ is regular, we may choose for each γ an open subset Vγ such that x ∈ Vγ ⊆ Vγ ⊆ Uγ. Furthermore, whenever Uγ = Xγ, we choose Vγ = Xγ. This way, V = ΠγVγ satisﬁes

x ∈ V ⊆ V = ΠγVγ ⊆ ΠγUγ ⊆ U. By Lemma 6.18 (i) it follows that X is regular. ELEMENTARY TOPOLOGY I 53

Theorem 6.20. Let (X, TX ) be a second-countable, regular space. Then X is normal.

Proof. Let B be a countable basis for TX , and let A, B ⊆ X be disjoint closed subsets. Then for each point a ∈ A there is an open subset Ua ∈ TX such that Ua ∩ B = ∅. Since X is regular, we may choose Va ∈ TX such that a ∈ Va ⊆ Va ⊆ Ua, and thus S we may also choose Wa ∈ B such that a ∈ Wa ⊆ Va. This way, W = a∈A Wa is such that A ⊆ W and W ∩ B = ∅. Furthermore, the collection {Wa}a∈A is countable (after removing any possible repetitions), so we may write it as {Wn}n∈N. A similar process produces a collection {Z } such that B ⊆ Z = S Z and Z ∩ A = ∅. However, n n∈N n∈N n W and Z may not be disjoint. For each n ∈ N, define n n 0 [ 0 [ Wn = Wn \ ( Zi) and Zn = Zn \ ( Wi). i=1 i=1 0 0 0 S 0 0 S 0 Notice that Wn and Zn are open for all n. Set W = n Wn and Z = n Zn. Then, S 0 S 0 0 0 A ⊆ n Wn and B ⊆ n Zn, and, we claim that W ∩ Z = ∅. Suppose otherwise and 0 0 0 0 let x ∈ W ∩ Z . Then x ∈ Wj ∩ Zk for some j, k ∈ N. Suppose for simplicity that 0 j ≤ z. By definition of Wj it follows that x ∈ Wj. On the other hand, by definition of 0 Zk we have x∈ / Wj, which is a contradiction.

Theorem 6.21. Let (X, TX ) be a compact Hausdorﬀ space. Then X is normal.

Proof. Let A, B ⊆ X be disjoint closed subsets. Fix a ∈ A. For each b ∈ B there exist disjoint subsets Ua,Vb ∈ TX containing a and b respectively. Note that {Vb}b∈B is an open cover of B, and B is compact because it is a closed subset of a compact space. Thus there exist b1, . . . , bn ∈ B such that {Ub1 ,...,Ubn } is a ﬁnite subcover. Tn Sn Let U = i=1 Uai and V = i=1 Vbi . Then a ∈ U, B ⊆ V and U ∩ V = ∅.

Now, for each a ∈ A, choose disjoint subsets Wa,Za ∈ TX such that a ∈ Wa and B ⊆ Za. The collection {Wa}a∈A is an open cover of A, and A is compact because it is a closed subset of a compact space, and thus there exist a1, . . . , am ∈ A such that

{Wa1 ,...,Wam } is a ﬁnite subcover. Set m m [ \ W = Waj and Z = Zaj . j=1 j=1 Then A ⊆ W , B ⊆ Z, and W ∩ Z = ∅. Thus X is normal.

54 ALEX GONZALEZ

REFERENCES

The main published references for this course are the following. 1. M. O. Searcoid, Metric Spaces, Springer Undergraduate Mathematics Series, ISBN 978-1-84628-369-7. 2. W. A. Sutherland, Introduction to Metric and Topological Spaces, 2nd Edition, Oxford University Press, ISBN 978-0-19-956308-1. Other general, electronic sources are the following. 1. http://www-history.mcs.st-and.ac.uk/~john/MT4522/index.html 2. http://en.wikibooks.org/wiki/Topology/Metric_Spaces References about the p-adic metric. 1. http://scientopia.org/blogs/goodmath/2013/01/13/define-distance-differently- the-p-adic-norm/ 2. http://en.wikipedia.org/wiki/P-adic_number References about the Cantor set. 1. http://planetmath.org/cantorset 2. http://mathworld.wolfram.com/CantorSet.html 3. http://en.wikipedia.org/wiki/Cantor_set References about the Heine-Borel Theorem. 1. http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem References about Hausdorff spaces (this has not been explicitly defined in the course, but the notion has appeared several times in examples). 1. http://en.wikipedia.org/wiki/Hausdorff_space#Examples_and_counterexamples References about the classification of surfaces. 1. http://www.open.edu/openlearn/science-maths-technology/mathematics-and- statistics/mathematics/surfaces/content-section-0 2. http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Huang.pdf 3. http://www.cis.upenn.edu/~jean/surfclass-n.pdf