Math 525 Some Notes on Sec 23, 24 Connectedness Some Introductory Notes on P.147 of Munkres Remind Us of the Importance of Conne

Math 525 Some notes on sec 23, 24

Connectedness

Some introductory notes on p.147 of Munkres remind us of the importance of connectedness and compactness in analysis. These two notions exhibit some behavioral similarities.

For one, they are topological properties which are deﬁned not only for topological spaces, but also for subsets of topological spaces.

Def Let (X, τ) be a topological space, with A ⊆ X. A is disconnected iﬀ there are open sets U, V in X such that:

(1) U ∩ V ∩ A = φ, (2) U ∩ A =6 φ, (3) V ∩ A =6 φ, (4) A ⊆ U ∪ V .

In this case, we say U and V disconnect or separate A.A connected subset is one which is not disconnected.

A space (X, τ) is connected if X is a connected subset. Thus (X, τ) is disconnected iﬀ there are non-empty, disjoint open sets U and V whose union is X. In this case, U = X − V and

V = X − U, so both U and V are clopen (simultaneously open and closed). Thus we have:

Prop A40 A space (X, τ) is connected iﬀ it has no clopen sets other than φ and X.

It is a relatively simple matter to check that a subset A of a space (X, τ) is connected iﬀ the subspace (A, τA) is connected. In any topological space, singleton sets and φ are connected; thus disconnected spaces can have connected subsets. A discrete space and all of its subsets other than φ and singletons are disconnected. An indiscrete space and all of its subsets are connected.

On p. 153, Munkres deﬁnes a linear continuum as a totally ordered set L with more than one element such that (1) L has the least upper bound property, and (2) whenever x < y in L, there is a z ∈ L such that x

Any set A in (R, τu) which is not an interval, ray or R is disconnected, since:

A not an interval, ray or R ⇒ there are x, y ∈ A and z 6∈ A such that x

⇒ (−∞, z) and (z, ∞) disconnect A.

In summary, we have:

Prop A41 A subset of (R, τu) is connected iﬀ it is an interval, ray, or R.

Prop A42 If (X, τ) is connected and µ ≤ τ, then (X, µ) is connected.

Proof : If(X, µ) is disconnected, then X contains a proper, nonempty µ-clopen subset which is also τ-clopen, so (X, τ) is also disconnected. ¥

Examples Consider our seven standard topologies on R:

• τd and τs are both zero-dimensional, and any zero-dimensional space which is not indiscrete must be disconnected;

• (R, τu) is connected by Prop A41, and since τcof, τo and τi are all coarser than τu, it follows by Prop A42 that (R, τcof), (R, τo) and (R, τi) are all connected;

• (R, τcoc) is connected, since a cocountable subset of R can not have a cocountable comple- ment, so there are no τcoc-clopen sets in R except φ and R. cont Prop A43 If f :(X, τ) −→ (Y, µ) and A ⊂ X is connected, then f(A) is connected.

Proof : If ∃ U, V which disconnect f(A), then f −1(U) and f −1(V ) disconnect A. ¥

Since connectedness is preserved under continuous maps, it is preserved under homeomor- phisms, and is therefore a topological property. An important consequence of Prop A43 is the Intermediate Value Theorem, which follows from Prop A43 and Thm 24.1 of Munkres.

Intermediate Value Theorem (IVT) Let (X, τ) be a connected space, and let Y be a totally ordered set with order topology µ. Assume f : (X, τ) → (Y, µ) is continuous. Then if a and b are two points in X and r is a point in Y strictly between f(a) and f(b), there must exist a point c in X such that f(c)= r.

Proof : Assuming all hypotheses, the sets

U = f(X) ∩ (−∞, r) and V = f(X) ∩ (r, ∞) are disjoint, nonempty sets which are open in f(X). If there were no c in X such that f(c) = r, then we’d have f(X) = U ∪ V , so U and V would separate the space f(X). But f(X) must be connected, since it is the continuous image of a connected space. ¥

Note that if (X, τ) in the IVT is a linear continuum and a < b, then f : [a, b] → (Y, µ) is continuous, and [a, b] and f([a, b]) are connected, so the proof proceeds similarly with X replaced by [a, b]. The conclusion would then be that there must exist a c in [a, b] such that f(c)= r. Since c can not be a or b because f(c)= r and f(a) =6 r =6 f(b), we get the more familiar conclusion that c ∈ (a, b).

Corollary (the beginning calculus version of the IVT): If f : [a, b] → R is continuous and c is strictly between f(a) and f(b), there exists y ∈ (a, b) such that f(y)= c.

Prop A44 If A is connected in (X, τ) and A ⊆ B ⊆ A , then B is connected. In particular, the closure of any connected set is connected. Proof : Assume A is connected in (X, τ) and A ⊆ B ⊆ A . This proof is by contradiction, so suppose U and V disconnect B. Then

U ∩ V ∩ B = φ, U ∩ B =6 φ, V ∩ B =6 φ, and B ⊆ U ∪ V .

Since A ⊆ B, it clearly follows from the ﬁrst and fourth of these conditions that U ∩V ∩A = φ and A ⊆ U ∪ V . If we can show that A ∩ U =6 φ and A ∩ V =6 φ, then U and V would separate A, which contradicts the assumption that A is connected. But we know there is an x ∈ U ∩ B, and since B ⊆ A ,

x ∈ U ∩ B ⇒ x ∈ U ∩ A ⇒ (x ∈ U and either x ∈ A or x is a limit point of A).

Either way, it follows that U ∩A =6 φ. Similary, V ∩B =6 φ ⇒ V ∩A =6 φ. This contradiction shows that B is connected. ¥

Prop A45 If {Ai ¯ i ∈ I} is a collection of connected subsets in (X, τ) and there exists ¯ x ∈ \ Ai, then A = [ Ai is connected. i∈I i∈I

Proof : Assume all hypotheses, and suppose U and V disconnect A. Then x (which is in A) must be in either U or V , but not both; say x ∈ U. Then we must have Ai ⊆ U for all i, since each Ai is connected, and

Ai 6⊆ U for some i ⇒ Ai ∩ V =6 φ ⇒ U, V disconnect Ai.

Finally, since Ai ⊆ U for all i, we have A ⊆ U, and so A ∩ V = φ, a contradiction. Thus A can not be disconnected. ¥

Let (X, τ) be a topological space with x ∈ X. Let Cx be the union of all connected subsets of

X containing x. By Prop A45, Cx is connected, and it is clearly the largest connected subset containing x. Cx is called the connected component of x in (X, τ). Let P = {Cx ¯ x ∈ X} ¯ be the collection of all connected components in (X, τ).

Prop A46 For any topological space (X, τ), P (as just deﬁned) is a partition of X induced by the equivalence relation ∼ deﬁned by:

x ∼ y iﬀ there exists a connected subset of X containing both x and y.

Furthermore, each connected component Cx is closed. Proof : To prove That P is a partition of X, simply note that:

(1) since x ∈ Cx for all x ∈ X, x = ∪P;

(2) if z ∈ Cx ∩ Cy, then by Prop A45, Cx ∪ Cy is connected, and Cx = Cy = Cz.

In checking that ∼ is an equivalence relation, reﬂexivity and symmetry should be clear. If x ∼ y and y ∼ z, then there exists connected sets A and B such that x, y ∈ A and y, z ∈ B.

Then, by Prop A45, A ∪ B is connected and contains both x and z, so x ∼ z.

To show that ∼ induces the partition P and vice-versa, note that:

(1) x, y ∈ Cz for some z ⇒ x ∼ y;

(2) x ∼ y ⇒ x, y both in some connected set A ⇒ A ⊆ Cx ∩ Cy ⇒ Cx ∩ Cy =6 φ

⇒ Cx = Cy (since P is a partition).

Finally, for each x ∈ X, clτ Cx is connected by Prop A44, and since Cx is the largest connected set containing x, clτ Cx = Cx. ¥

If (X, τ) is connected, then for all x ∈ X, Cx = X. If X = R × R with the dictionary order topology, then for any a ∈ R, the sets Ua = {(x, y) ¯ x ≤ a} and Va = {(x, y) ¯ x>a} ¯ ¯ disconnect X. Thus for any point z0 = (x0, y0) ∈ X, Cz0 = {(x, y) ¯ x = x0} is the vertical ¯ 2 line in R containing z0, and so X has an uncountable number of connected components.

As another interesting example, consider the set Q of rationals with the topology inherited from (R, τu), which I’ll denote by (Q, τu). Let x, y ∈ A ⊆ Q and let z ∈ R − Q be such that x

(disconnect) A, and hence the only connected subsets of Q are φ and singletons. Stated another way, Cx = {x} for all x ∈ (Q, τu).