Bootcamp

Christoph Thiele Summer 2012

Metric spaces Definition 1 A metric space is a pair (M, d) consisting of a set M with a function + d : M × M → R0 with the following properties that hold for all x, y, z ∈ M 1. Zero diestance for identical element: d(x, x) = 0. 2. Non-zero distance for non-identical elements: If x 6= y, then d(x, y) > 0 3. Symmetry: d(x, y) = d(y, x) 4. Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z) We list some examples: q 1. D with distance d(x, y) = |x − y| = (x − y)2

q 2. R with distance d(x, y) = |x − y| = (x − y)2

n 3. R , i.e. the set of tuples x = (x1, . . . , xn) with xi ∈ R with distance d(x, y) = q Pn 2 i=1(xi − yi)

n 4. R with distance d(x, y) = sup1≤i≤n |xi − yi| q 2 P∞ 2 5. l (N), the set of sequecnes x : N → R with distance d(x, y) = i=1(xi − yi)

∞ 6. l (N), the set of bounded sequences, with distance d(x, y) = supi∈N |xi − yi| 1 P∞ 7. l (N), the set of absolutely summable sequences, with d(x, y) = i=1 |xi − yi| 8. The set BV (D) of functions of bounded variation on D, i.e. the set of pairs (f, g) with f, g right continuous (say) monotone increasing bounded functions modulo the equivalence relation (f, g) ∼ (f 0, g0) if f + g0 = f 0 + g with distance d((f, g), (h, k)) = inf[| sup(f 0+k0)(x))−(f 0+k0)(0)|+| sup(g0+h0)(x)−(g0+h0)(0)|] x x where the infimum is taken over all representatives of ther respective the equiv- alence classes.

1 9. The space C([0, 1]) of continuous functions f : [0, 1] → R with distance

sup |f(x) − g(x)| x∈[0,1]

+ + 10. The space C(R0 ) of continuous functions f : R0 → R with distance

sup |f(x) − g(x)| + x∈R0

+ + 11. The space C0(R0 ) ⊂ C(R0 ) of functions satisfying

lim sup |f(x)| = 0 x→∞ with the same distance.

2 We show the triangle inequality in the case of l (N). Let xi and yi two sequecnes P 2 P 2 with i xi < ∞ and i yi < ∞. s X X 2 X 2 |xiyi| ≤ ( xi )( yi ) i i i

Note that this inequality is homgeneous, if we multiply xi or yi by a positive real numfber, the inequality remains true. hence it suffices to prove this inequality under the assumpiton X 2 X 2 xi = yi = 1 i i But then the inequality becomes

X X 2 X 2 X 2 2 2 |xiyi| ≤ ( xi ) + ( yi ) = (x1 + yi ) i i i i But then the inequality follwos by the comparison principle since for each i

2 2 ±2xiyi ≤ xi + yi since 2 0 ≤ (xi ± yi) We conclude the Cauchy Schwarz inequality s X X 2 X 2 xiyi ≤ ( xi )( yi ) i i i

But then we have

X 2 X 2 X 2 X (xi + yi) = xi + yi + 2 xiyi i i i i

2 s s s X 2 X 2 X 2 X 2 X 2 X 2 2 ≤ xi + yi + 2 ( xi )( yi ) = ( xi + yi ) i i i i i i

Upon setting xi = ai − bi and yi = bi − ci this becomes the triangle inequality for the sequences ai, bi, ci We also show there tirangle inequality for the space BV (D). For simplicity of notation we restrict attention to the space of functions therein which vanish at 0, which can be expressed using pairs of functions vanishing at 0. Then the distance is expressed as

d((f, g), (h, k)) = inf[sup(f 0 + k0)(x) + sup(g0 + h0)(x)] x x where again the infimum is taken over all choices of representatives of the given class. (exercise: do the reducdtion of the general caes to this caes.) Let (fg, ), (h, i), (j, k) be three pairs of functions. Assume that the first two are chosen appropriate representatives so that

d((f, g), (h, i)) +  ≥ sup(f + i)(x) + sup(g + h)(x)] x x Assume that (j, k) is chosen so that for some equivalent pair h0, i0 to h, i

d((j, k), (h, i)) +  ≥ sup(j + i0)(x) + sup(k + h0)(x)] x x Then we have h + i0 = h0 + i and (f, g) is equivalent to (f + i + h0, g + h + i0). Hence

d((f, g), (j, k)) ≤ sup(f + i + h0 + k)(x) + sup(g + h + i0 + j)(x) x x ≤ sup(f + i)(x) + sup(g + h)(x) + sup(h0 + k)(x) + sup(i0 + j)(x) x x x x d((f, g), (h, i)) +  + d((j, k), (h, i)) +  Since  was arbitrary, this proves the triangle inequality. Furthermore, if (M, d) is a metric space, then for any subset M 0 of M we have the metric subspace (M 0, d0) with d0 the restriction of d to M 0. A space satisfying only properties 1, 3, 4 of a metric space can be transformed into a metric space by passing to equivalence classes.

Lemma 1 Assume a space M and a distance function d on M satisfies properties 1, 3, 4 of a metric space. Then the relation defined by x ∼ y if d(x, y) = 0 is an equiv- alence relation on M The distance function d(x, y) depends only on the equivalence classes of x and y. the space M˜ of equivalence classes with the distance function d˜ induced by d is a metric space.

Proof: We first show that ∼ is an equivalence relation. Reflexivity: For all x ∈ M we have d(x, x) = 0 by property 1 and thus x ∼ x. Symmetry: follows from symmetry of d. Transitivity: If x ∼ y and y ∼ z then d(x, z) ≤ d(x, y) + d(y, z) = 0 + 0 = 0. hence x ∼ z.

3 Next we show that the distance function depends only on the equivalence class of x, y. let x ∼ x0 and y ∼ y0. Then d(x, y) ≤ d(x, x0) + d(x0, y0) + d(y0, y) = d(x0, y0) and similarly we have the reverse ineuqlity. thus d(x, y) = d(x0, y0). Finally we show that M˜ is a metric space with metric d˜. properties 1, 3, 4 follow immediately form the corresponding properties of M. Property 2 not that if x and y are not equivalent, then d(x, y) 6= 0 and hence d˜(x, y) 6= 0. 2 The above reasoning suggests that it makes sense to talk about almost equality of points in a metric space when they have small distance, where smallness may be determined by some parameter  that may depend on the context. If two elements have distance at most , then each is in the  ball around the other.

Definition 2 (Open ball, cone) The open r ball about a point x in a metric space (M, d), denoted by Br(x), is the set of all points y such that d(x, y) <  The cone B(x) associated with a point x in a metric space (M, d) is the set of all pairs (y, r) in M × D which satisfy y ∈ Br(x).

The following lemma shows that a cone has aproperty that can be stated without reference to the point x.

Lemma 2 If B(x) is the cone associated with x, then for all points (y, r) in the cone B(x) there exists  > 0 such that

inf sup d(z, y) +  ≤ r δ z:(z,δ)∈B(x) and for all points (y, r) not in the cone we have

inf sup d(z, y) ≥ r δ z:(z,δ)∈B(x)

Proof: First let (y, r) ∈ B(x). Pick  such that d(x, y) +  = r. Then for all (z, δ) ∈ B(x) we have

d(z, y) +  ≤ d(z, x) + d(x, y) +  = d(z, x) + r ≤ δ + r

Hence sup d(z, y) +  ≤ δ + r z:(z,δ)∈B(x) Taking infimum over δ gives

inf sup d(z, y) +  ≤ r δ z:(z,)∈B(x)

Now let (y, r) ∈/ B(x), then

sup d(z, y) ≥ d(x, y) ≥ r z:(z,δ)∈B(x)

4 and taking infimum over δ proves the desired estimate. 2 A cone is the natural pendant for metric spaces of a Dedekind cut for ordered spaces. This suggest we will want to do extensions of metric spaces like D to obtain + metric spaces like R0 . We begin with a few definitions.

Definition 3 A subset M 0 of a metric space M is called dense if for every (x, r) ∈ 0 M × D we have y ∈ M such that y ∈ Br(x).

Definition 4 A metric space is called separable, if it has a countable dense subspace.

Exercise 1 Prove any number of the following:

1. l2(N) is separable.

2. l∞(N) is not separable.

3. BV (D) is not separable.

4. C([0, 1]) is separable.

Definition 5 Let (M, d) be a metric space and M 0 be a dense subspace. A set B in M 0 × D is called a Dedekind cone, if for all r there is a point (y, r) ∈ B and for all (y, r) ∈ B there exists  such that

inf sup d(y, z) +  ≤ r δ z:(z,δ)∈B and for all (y, r) 6∈ D inf sup d(y, z) ≥ r δ z:(z,δ)∈B

Given a cone B we can define for r ∈ D the set (“ball”) Br of all points y with (y, r) ∈ B.

Lemma 3 The set of Dedekind cones forms a metric space with the distance

d(B,B0) = inf sup d(y, y0) δ (y,δ)∈B,(y0,δ)∈B0

Proof: We first observe that d(B,B) = 0. It suffcies to show for each δ

sup d(y, y0) ≤ 2δ (y,δ)∈B,(y0,δ)∈B0

However, we have for each z ∈ B by the triangle inequality and symmetry

sup d(y, y0) ≤ 2 sup d(y, z) (y,δ)∈B,(y0,δ)∈B0 (y,δ)∈B

Now taking sup in z ∈ B and inf in  proves the desired inequality.

5 0 + Next we show that d(B,B ) is finite (it is clearly in R0 ∪ {∞}). Note that

inf sup d(y, y0) δ (y,δ)∈B,(y0,δ)∈B0

≤ sup d(y, y0) (y,1)∈B,(y0,1)∈B0 0 0 No let y0 and y 0 be points in B1, and B 1 which are assumed to exist, and note that the last display is bounded by

0 0 0 sup d(y, y0) + sup d(y 0, y ) + d(y0, y 0) (y,1)∈B (y0,1)∈B0

The first two terms are bounded by 2 by a previous arguemnt, the last term is some finite number. This proves that d(B,B0) is finite. Next we show that if d(B,B0) = 0 then B = B0. By symmetry it suffices to show B ⊂ B0. Thus let (y, r) ∈ B and pick  such that then

inf sup d(y, z) + 3 ≤ r δ z:(z,δ)∈B

Bick δ such that sup d(z, z0) ≤  (z,δ)∈B,(z0,δ)∈B0 Then by the triangle inequality

sup d(y, z) ≤  + sup d(y, z) z:(z,δ)∈B0 z:(z,δ)∈B

Putting inequalities together proves (y, r) ∈ B0. Symmetry of the function d is clear by definiton. It remains to prove the triangle 0 00 0 0 inequality. Let B,B B be three cones. Not that we have for every y ∈ B δ

sup d(y, y00) (y,δ)∈B,(y00,δ)∈B00

sup (d(y, y0) + d(y0, y00)) (y,δ)∈B,(y00,δ)∈B00 ≤ sup d(y, y0) + sup d(y0, y00) (y,δ)∈B,(y00,δ)∈B00 (y,δ)∈B,(y00,δ)∈B00 by the trangle inequality for the supremum. Erasing trivial suprema and taking supremum over y’ estimates this by

≤ sup d(y, y0) + sup d(y0, y00) (y,δ)∈B,(y0,δ)∈B0 (y0,δ)∈B0,(y00,δ)∈B00

Noting that this is monootne in δ and taking infimum in δ proves the tirangle in- equality. 2

6 Lemma 4 Let (M, d) be a metric space and M 0 a dense subspace with respect to which we form Dedekind cones. Let M,˜ d˜ be the metrics space of Dedekind cones in M. Then there is an isometric embedding M → M˜ mapping x to B(x).

Note we have the conse B(x) for all x ∈ M as defined before, and then restriced to M 0, D. We need to show that d(x, y) = d˜(B(x),B(y)). Injectivitiy of this map follows from the identity, while surjectivity may not hold and neither is claimed. But we have inf sup d(y, y0) δ (y,δ)∈B(x),(y0,δ)∈B(x) ≤ inf sup (d(y, x) + d(x, x0) + d(x0, y0)) δ (y,δ)∈B(x),(y0,δ)∈B(x0) ≤ inf(2δ + d(x, x0)) = d(x, x0) δ 2 A twin lemma allows to determine the distance between B(x) and B.

Lemma 5 Let (M, d) be a metric space and M 0 a dense subspace and (M,˜ d˜) be the respective space of Dedekind cones . Let B be a Dedekind cone of M. Then we have 0 ˜ x ∈ Br(x) for some x ∈ M if and only if d(B(x),B) < r Proof: Let  > 0 such that

inf sup d(x, y0) +  ≤ r δ (y0,δ)∈B

Then inf sup d(y, y0) +  δ (y,δ)∈B(x),(y0,δ)∈B inf sup d(y, x) + d(x, y0) +  δ (y,δ)∈B(x),(y0,δ)∈B ≤ inf δ + inf sup d(x, y0) +  ≤ r δ δ (y0,δ)∈B The converse is proved similarly. 2 Recall that a Dedekind cut of D is determined by relatively few points in the cut, all it takes is an appropriate monotone sequence. This motivates the following variant fo Dedekind cones

Lemma 6 A metric space is dense in the space of Dedekind cones.

Definition 6 A metric space is called complete, if every Dedekind cone is of the form B(x) for some x ∈ M

Lemma 7 Given a metric space M and a dense subspace M 0, then the space of dedekind conses of M is a complete metric space.

7 Alternative approach to completion using sequences

Definition 7 A sequence xn of elements in some metric space is said to converge to an element x if lim sup d(xn, x) = 0 n→∞ Exercise 2 This definition is equivalent to

∀ > 0 : ∃n ∈ N : ∀m ∈ N, m ≥ n : d(xm, x) ≤  In particular, from set theoretic point of view, it may be interesting whether there is a small subset of M such that every eleemnt in M can be approximated by eleemnts in the smaller set.

Definition 8 A subset M 0 ⊂ M is called dense in M, if for everey x ∈ M there exists a sequence y : N → M 0 which converges to x.

Exercise 3 Prove that this definition is equivalent to the previous definiiton.

Definition 9 An (infinite) metric space is called separable, if there is a countable dense set subset.

In order to converge, the elements of a sequence not only have to get closer to x, but they also come closer to each other, as consequecne of the triangle inequality. If d(xn, x) ≤  and d(xm, x) ≤  then

d(xn, xm) ≤ d(xn, x) + d(x, xm) = d(xn, x) + d(xm, x) ≤ 2 Elements coming close to each other can be expressed as

Definition 10 A sequence xn is Cauchy, if

inf inf sup sup d(xn+k, xm+l) = 0 n m k l Lemma 8 If a sequence converges, then it is Cauchy.

Proof: inf inf sup sup d(xn+k, xm+l) n m k l

≤ inf inf sup sup d(xn+k, x) + d(xm+l, x) n m k l

≤ inf sup d(xn+k, x) + inf sup d(xm+l, x) = 0 n k m l Here we have used in the second line the triangle inequality for the supremum, and in the last inequality we have used that for monotone sequences the infimum commutes with finite sums. 2 The converes is not always true, as the example of the space D and a monotone increasing sequence of dyadic numbers whos esupremum is a non-dyadic real number shows.

8 Definition 11 A metrci space M is called complete, if every Cauchy sequences in the space converges to some element in the space. While not every metric space needs to complete, one can to some extend rectify this by embedding the space into a bigger metric space which is complete. Define a distance on Cauchy sequences

inf inf sup sup d(xn+k, ym+l) = 0 n m k l Lemma 9 The disitance on Cauchy sequences satisfies properties 1, 3, 4 of a metric space. Proof: Property one follows by definition of a Cauchy sequence. Property two follows since d is symmetric and we can commute supremum with supremum and infimum with infimum:

inf inf sup sup d(xn+k, ym+l) = 0 n m k l

= inf inf sup sup d(ym+l, xn+1) = 0 m n l k Property four can be seen similar as above:

inf inf sup sup d(xn+k, zm+l) n m k l

≤ inf inf sup sup(d(xn+k, ym+1) + d(ym+l, yn+k) + d(yn+k, zm+l) n m k l Using the triangle inequality as before, we can estimate this by thre etemrs, of which the middle term vanishes since y is Cauchy. 2 Property two does not hold in general, two different Cauchy sequences may con- verge to the same limit. However, we can form equivalence classes of Cauchy sequences as in the above Lemma and thus obtain a metric space M,˜ d˜. We claim that Lemma 10 The space M embeds isometrically into M˜ by the map sending an element x to the constant sequence x. The space M is dense in M˜ . the space M˜ is complete. Proof: For the isometric embedding we need to show that the distance of the constant sequences x and y is equal to d(x, y). This is however trivial. For density, let xn be a Cauchy sequence and let  > 0. We need to find x such ˜ that d(xn, x) ≤ . But there is n, m such that

sup sup d(xn+k, xm,l) ≤  k l by definition of Cauchy sequence. Assume without loss of generality that m ≥ n and pick x = xm then by the triangle inequality

sup sup d(xn+k, x) ≤ 2 k l Thsi proves that x is 2 close to the given Cauchy sequence. For completeness, consider a sequecne of sequences xn(m) whcih is Cauchy. We ˜ claim that the diagonal sequence xn(n) is a limit for this cauchy sequence in M. Proof: exercise

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