Another isometric imbedding problem

[radou]

Homework Statement

I like to deal with equivalence relations, so I decided to skip to this exercise.

Let (X, d) be a metric space. If xn and yn are Cauchy sequences in X, define a relation xn ~ yn if d(xn, yn) --> 0. Denote the equivalence class of x with [x], and the set of all equivalence classes with Y. Define a metric D on Y with D([x], [y]) = lim d(xn, yn), n --> ∞.

The Attempt at a Solution

(a) Show that ~ is an equivalence relation, and show that D is a well-defined metric.

Clearly, for any Cauchy sequence, xn ~ xn, xn ~ yn <==> yn ~ xn and (xn ~ yn & yn ~ zn) ==> xn ~ zn holds (the third arrived from the triangle inequality). It is also easy to check D is a metric, for example the triangle inequality (all limits are as n --> ∞): d(xn, zn) <= d(xn, yn) + d(yn, zn) ==> lim d(xn, zn) <= lim d(xn, yn) + lim d(yn, zn). The other "metric properties" are trivial. The metric D is well-defined, since limits are unique in R.

(b) Define h : X --> Y with h(x) = [(x, x, x, ...)] (i.e. h(x) is the equivalence class of the constant sequence (x, x, x, ...)). Show that h is an isometric imbedding.

Obviously for any x, y in X, d(x, y) = D(h(x), h(y)) = D([(x, x, ...)], [(y, y, ...)]) = d(x, y), so h is an isometry. Let's check for injectivity. Indeed, if x and y are distinct elements of X, y cannot possibly lie in h(x) (the equivalence class of the constant sequence) and vice versa. And further on, h : X --> h(X) is clearly a surjection.

Now, let's check continuity of h.

Let B([x], ε) = {[y] in Y : D([x], [y]) = lim d(xn, yn) < ε (as n --> ∞)} be an open ball around [x] in Y. The inverse image of this open ball under h is defined for constant sequences in Y only. And one can easily see that h^1(B) = {x in X: d(x, xn) < ε, for every n (n indexes the sequence xn)}. Clearly, this is a union of open balls around the elements xn of the sequence xn (again, notation is not quite consistent and clear, but I'll assume it's obvious what I mean), and hence open.

Now, the only thing remaining is to check that the inverse of h is continuous. Take an open ball B(x, ε) in X. We need to check that its image under h (which is the inverse of the inverse of h) is open. Since B(x, ε) = {y in X : d(x, y) < ε)}, and since h is an isomerty, we have d(x, y) = D(h(x), h(y)) < ε, so clearly h(B) = {[y] in Y : D([x], [y]) < ε}, which is an open ball in Y.

I hope these results are correct.

 

[micromass]

All is well, until you get to the continuity of h.

Homework Statement

h^1(B) = {x in X: d(x, xn) < ε, for every n (n indexes the sequence xn)}

I don't really see this. I don't even think it's true...

Now, the only thing remaining is to check that the inverse of h is continuous. Take an open ball B(x, ε) in X. We need to check that its image under h (which is the inverse of the inverse of h) is open. Since B(x, ε) = {y in X : d(x, y) < ε)}, and since h is an isomerty, we have d(x, y) = D(h(x), h(y)) < ε, so clearly h(B) = {[y] in Y : D([x], [y]) < ε}, which is an open ball in Y.

I know what you mean, but still: h(B) is not an open ball in Y, but it's an open ball in h(X).
Compare with this: $$]-1,1[\cap \mathcal{Q}$$ is an open ball in $$\mathbb{Q}$$, but not in $$\mathbb{R}$$.


Overall, I want to make one suggestion. You check continuity with the definition of h. But isn't it easier to show that every isometric imbedding is continuous. If you prove this result in general, then you can avoid the troublesome notation.
Also, I think it's easier to use $$\epsilon,\delta$$ definition in this case. Working with open balls is maybe a bit to complicated...

 

[radou]

I don't really see this. I don't even think it's true...

I know what you mean, but still: h(B) is not an open ball in Y, but it's an open ball in h(X).
Compare with this: $$]-1,1[\cap \mathcal{Q}$$ is an open ball in $$\mathbb{Q}$$, but not in $$\mathbb{R}$$.

Actually, what I meant was h(X). But that doesn't help right now...

Overall, I want to make one suggestion. You check continuity with the definition of h. But isn't it easier to show that every isometric imbedding is continuous. If you prove this result in general, then you can avoid the troublesome notation.
Also, I think it's easier to use $$\epsilon,\delta$$ definition in this case. Working with open balls is maybe a bit to complicated...

A good suggestion indeed, I was over-complicating again!

Btw, an imbedding is continuous by definition, so what I actually need to show is that an isometry is continuous.

Indeed, let f : X --> Y be an isometry. Let's show it's continuous at some arbitraty point c of X. Let ε > 0 be given, and let δ = ε. Then, for all x in X such that dx(x, c) < δ, we have dy(f(x), f(c)) = dx(x, c) < ε = δ ! This indeed trivial!

And again, I don't pick an easier and obvious way to deal with things...

 

[micromass]

That is correct! And maybe a bit simpler than working with the definition of h

 

[radou]

Btw, my suggested proof that the inverse of h is continuous won't work? It's the only thing left to show to prove that h is an imbedding...

 

[micromass]

That part of the proof was alright. But the inverse of an isometry is also an isometry and you have proven that isometries are continuous. This shows immediately that the inverse of h is continuous...

 

[radou]

That part of the proof was alright. But the inverse of an isometry is also an isometry and you have proven that isometries are continuous. This shows immediately that the inverse of h is continuous...

Ah yes, right! OK, moving on...

 

[radou]

(c) Show that h(X) is dense in Y.

We need to show that Cl(h(X)) = Y.

Let's first check inclusion "$$\subseteq$$".

Let [x] be in Cl(h(X)). Then for every ε > 0, B([x], ε)$$\cap$$Y$$\neq \emptyset$$. This means that there is a constant sequence y such that D([x], [y]) = lim d(xn, y) < ε, as n --> ∞ (every element of the sequence yn equals y). Hence, the sequence xn is Cauchy. And so it must belong to Y.

 

[micromass]

That's ok! But now the harder inclusion...

 

[radou]

OK, I may have got an idea.

If [y] is an element of Y, and if we show that there is a sequence in h(X) converging to [y], then [y] is in Cl(h(X)), which is the desired result.

Since yn, is Cauchy, for ε = 1, choose N1 such that for all m, n >= N1 we have d(ym, yn) < ε. Specially, if we fix m = N1, we have, for all n >= N1, d(yN1, yn) < ε. Do this for ε = 1/2, ... , 1/n in general, and for every ε name the positive integer Nn.

I assert that h(yNn) is a sequence in h(X) converging to [y].

Let ε > 0 be given. We wish to find an integer N such that whenever n >= N, we have D(h(yNn), [y]) = lim d (yNn, yn) < ε, as n --> ∞. But, we can find such an integer, as is proposed in the paragraph above. Hence h(yNn) --> [y].

 

[micromass]

That's right!

 

[radou]

That's right!

Thanks!

 

[radou]

(d) Now, this is an intermediate result to prove that Y is complete.

Let A be a dense subset of the metric space (Z, d). If every Cauchy sequence converges to a limit in Z, then Z is complete.

My idea is to take a Cauchy sequence xn in Z, and find a convergent subsequence of xn in A. But I don't really see how. Somehow, I need to use the fact that A is dense in Z, i.e. Cl(A) = Z...

Edit: clearly, for any member xi of the sequence xn there is a sequence xn' of points of A converging to xi. Further on, for any neighborhood of any member of the sequence xn intersects A. These are some apparent facts, but I don't see how to put them together.

 

[micromass]

Well, there are several ways to tackle this problem. But I think you've already showed a way that should work:

clearly, for any member xi of the sequence xn there is a sequence xn' of points of A converging to xi.

Allright, now try to take a diagonal sequence. I think I should be able to do something with that...

 

[radou]

Well, there are several ways to tackle this problem. But I think you've already showed a way that should work:



Allright, now try to take a diagonal sequence. I think I should be able to do something with that...

OK, can you just reveal one hint to me in order to direct my thinking - does this enable us to find a convergent subsequence of the original Cauchy sequence? Or should I apply some other method of showing the sequence converges?

Btw, I tried to consider the set of all members of the original sequence which belong to A. But I don't see a way to prove the set is non-empty and has a countable number of elements... It's stupid, probably.

(Btw, congratulations on your science advisor medal! )

 

[micromass]

Take a sequence $$(x_n)_n$$. The trick is to construct a sequence in A which is very close to $$(x_n)_n$$. We can do this by a diagonal argument. But I think there might be an easier way like this:

Take an arbitrary n, then by density of A, we know that there exists an $$a_n$$ in A, such that $$d(x_n,a_n)<1/n$$. The sequence $$(a_n)_n$$ lies in A an lies very close to $$x_n$$.

In fact, we got that $$d(x_n,a_n)\rightarrow 0$$, so if $$(a_n)_n$$ converges, then $$(x_n)_n$$ converges. So, the only problem is to show that $$(a_n)_n$$ converges (by showing it is Cauchy).

 

[micromass]

(Btw, congratulations on your science advisor medal! )

Thanks a lot I'm honored!

 

[radou]

Take a sequence $$(x_n)_n$$. The trick is to construct a sequence in A which is very close to $$(x_n)_n$$. We can do this by a diagonal argument. But I think there might be an easier way like this:

Take an arbitrary n, then by density of A, we know that there exists an $$a_n$$ in A, such that $$d(x_n,a_n)<1/n$$. The sequence $$(a_n)_n$$ lies in A an lies very close to $$x_n$$.

In fact, we got that $$d(x_n,a_n)\rightarrow 0$$, so if $$(a_n)_n$$ converges, then $$(x_n)_n$$ converges. So, the only problem is to show that $$(a_n)_n$$ converges (by showing it is Cauchy).

Hm, I can't say I understand the alternative approach.

We take an element of the original sequence, say the element is xi. By density, we construct a sequence an such that d(xi, an) --> 0, as n --> ∞. Hence, an converges to xi, right? How does that relate to the convergence of xn? Probably it's some notational misunderstanding here...

 

[micromass]

No, you take an element $$x_n$$. By density, we can find an element of A which is arbitrarily close to $$x_n$$. So, we can find an $$a_n\in A$$ such that $$d(x_n,a_n)<1/n$$. This gives us a sequence $$(a_n)_n$$.


Another way is the following: Consider an element $$x_n$$. We can find a sequence $$x_{n,m}\rightarrow x_n$$. Then consider the diagonal sequence $$(x_{n,n})_n$$.

 

[radou]

OK, I get it, except for one thing: if d(xn, an) --> 0, and if an converges, how does that imply that xn converges? I feel I'm missing an elementary result about sequences...Of course, this is intuitively acceptable and I see it's logical, but still...We mentioned similar things, but in the context of uniformly continuous maps.

 

[micromass]

Yes, we indeed mentioned it form uniform maps. The result was

If $$(x_n)_n$$ and $$(y_n)_n$$ are sequences such that $$d(x_n,y_n)\rightarrow 0$$ and $$x_n\rightarrow x$$, then $$y_n\rightarrow x$$.

The other result we mentioned (but I don't think we need it here) was:

If [tex](x_n)_n[tex] and $$(y_n)_n$$are sequences such that $$x_n\rightarrow x$$ and $$y_n\rightarrow x$$, then $$d(x_n,y_n)\rightarrow 0$$

And another result needed a uniform function:

f is uniform continuous if and only if for every sequences such that $$d(x_n,y_n)\rightarrow 0$$ holds that $$d(f(x_n),f(y_n))\rightarrow 0$$.

But you only need the first result here...

 

[radou]

Yes, we indeed mentioned it form uniform maps. The result was

If $$(x_n)_n$$ and $$(y_n)_n$$ are sequences such that $$d(x_n,y_n)\rightarrow 0$$ and $$x_n\rightarrow x$$, then $$y_n\rightarrow x$$.
[/tex].

Hm, I'll quote what was written in a previous thread:

"2) Let (x_n) and (y_n) be two sequences such that d(x_n,y_n)-->0. If x_n--> x and y_n-->y, then x=y."

So, this is equivalent to:

if xn and yn are sequences such that d(xn, yn) --> 0 then xn --> x implies yn --> x?

 

[micromass]

Yes, both are correct. And their proof is almost the same.

The reason I formulated it differently in the other thread is probably because that form was more convenient in that thread...

 

[radou]

Yes, both are correct. And their proof is almost the same.

The reason I formulated it differently in the other thread is probably because that form was more convenient in that thread...

OK, now I have all the tools I need. (I'll look into the proofs of these statements too, for convenience...)

 

[micromass]

Maybe you will also want to prove the following

If $$d(x_n,y_n)\rightarrow 0$$ and if $$(x_n)_n$$ is Cauchy, then $$(y_n)_n$$ is Cauchy.

 

[radou]

Well, this is indeed simple!

Let ε > 0 be given. Choose N1 such that n >= N1 implies d(xn, an) < ε/3, choose N2 such that n, m >= N2 implies d(xn, xm) < ε/3, and choose N3 such that m >= N3 implies d(xm, am) < ε/3.

Now, by the triangle (generalized) inequality, we have, for n, m >= max {N1, N2, N3}:

d(am, an) <= d(an, xn) + d(xn, xm) + d(xm, am) < ε.

Hence, an is Cauchy and by hypothesis converges in Y.

(Btw, the proof of our statement in post #22 is very easy, only apply the triangle inequality...)

 

[micromass]

Yes, that is good! Now it shouldn't be a problem to prove (e)...

 

[radou]

Since h(X) is proven to be dense in Y, we only need to show that every Cauchy sequence in h(X) converges in Y.

Let [xn] be a Cauchy sequence in h(X). For any ε > 0, there exists N such that n, m >= N implies D([xn], [xm]) = lim d(xn, xm) < ε. Since we're dealing with sequences in h(X), [xn] = [(x, x, ...)] and [yn] = [(y, y, ...)] and hence we have a Cauchy sequence xn in X.

Well, I thought this was obvious and easy at first, but... If I knew that xn converges in X, one could easily see that [xn] converges in Y, too. Hm..

 

[micromass]

Well, let's say that we have the following Cauchy sequence in h(X):

$$(x_1,x_1,x_1,...)$$ is the first element of the sequence
$$(x_2,x_2,x_2,...)$$ is the second element of the sequence
$$(x_3,x_3,x_3,...)$$ is the third element of the sequence

You will want to show that this sequence converges to

$$(x_1,x_2,x_3,...)$$

The above element does indeed lie in Y, since it is a Cauchy sequence...

 

[radou]

OK, I'll think about it tomorrow, since my brain completely stopped working right now. :zzz:

Btw, this section (i.e. the exercises) is very useful, since it filled in a lot of gaps in my knowledge of analysis...

 

[radou]

Well, this is actually pretty easy, and we have already proved it.

As stated, a Cauchy sequence in Y corresponds to a Cauchy sequence xn in X. By (c), h(xn) converges to [(x1, x2, ...)] in Y.

 

[micromass]

Have we already proved that? Well, then it's pretty easy indeed!

 

[radou]

Have we already proved that? Well, then it's pretty easy indeed!

Yes, it's basically in post #10, unless I'm mistaken.

 

[micromass]

Ah yes, it is basically the same thing indeed! That completes your completion exercise

 

[radou]

Ah yes, it is basically the same thing indeed! That completes your completion exercise

Yes... So the purpose of this exercise was to show another way to imbed a metric space into a complete metric space, through a relation defined for Cauchy sequences of the original space.

The other way to imbed some metric space was (Theorem 43.7., and actually I dislike this theorem) through the set of all bounded functions from that space into R...i.e. there is an imbedding of (X, d) into the set of all bounded functions from X to R in the uniform metric.