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Topologically complete space in the product topology

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  • #1
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Homework Statement



One needs to show that a countable product of topologically complete spaces is topologically complete in the product topology.

The Attempt at a Solution



A space X is topologically complete if there exists a metric for the topology of X relative to which X is complete.

So, let X = ∏Xi, where (Xi, di) are complete metric spaces (di denotes the bounded metric deduced from the metric on Xi).

My first guess is the metric that induces the product topology on X, given with:

D(x, y) = sup{di(xi, yi)/i}, where d is the standard bounded metric. Of course, a little variation is that we consider di for each i, i.e. the metric in Xi.

Now, let xn be a Cauchy sequence in X. Given ε > 0, there exists N such that for all m, n >= N, we have D(xn, xm) < ε. In particular, for every m, n, and for every i, we have:

di(xni, xmi)/i <= D(xn, xm) < ε. Let i be fixed. We find that di(xni, xmi) <= i D(xn, ym) < i ε, for all n, m >= N. This implies that the sequence xni is a Cauchy sequence in Xi, hence it converges to xi. I assert that xn converges to the point x of X, where x = (xi) (a bit stupid notation, but I think it's obvious what I mean).

This follows from the fact that for ε > 0, we can find n, m such that D(xn, xm) < ε/2, whenever m, n >= N. By what is written above, for every i, and m, n >= N, we have di(xni, xmi) < i ε/2. If we fix n and i, and let m --> ∞, we have di(xni, xi) <= ε/2. This holds for any i, and for n >= N. Hence D(xn, x) <= ε/2 < ε, whenever n >= N. So, xn --> x.

I think this should work, since I studied some similar proofs in the book and applied them to this problem.
 

Answers and Replies

  • #2
radou
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Btw, I just realized I was being a bit harsh again.

In Munkres, I only know that the metric D induces the product topology on Rω. Actually, I'm not sure if the metric I proposed induces the product topology on an arbitrary countable product of spaces...

Edit: probably not, since there would be a theorem in the book about that... boy, it seems I'll have to think this over.
 
  • #3
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Seems alright!
 
  • #4
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Btw, I just realized I was being a bit harsh again.

In Munkres, I only know that the metric D induces the product topology on Rω. Actually, I'm not sure if the metric I proposed induces the product topology on an arbitrary countable product of spaces...

Edit: probably not, since there would be a theorem in the book about that... boy, it seems I'll have to think this over.
Don't worry about that, the metric D you mentioned does indeed induce the product topology on a countable product. The proof is analogous to the case [tex]\mathbb{R}^\omega[/tex]. But you're right, Munkres should have mentioned it as a separate theorem...
 
  • #5
radou
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Don't worry about that, the metric D you mentioned does indeed induce the product topology on a countable product. The proof is analogous to the case [tex]\mathbb{R}^\omega[/tex]. But you're right, Munkres should have mentioned it as a separate theorem...
OK, thanks!
 
  • #6
radou
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Btw, could we somehow generalize this to an arbitrary product using the uniform metric? I assume this doesn't work?
 
  • #7
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Hmm, very good question. I guess your proof should work for the uniform topology as well (and in that case: for arbitrary factors in your product).
 
  • #8
radou
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Hmm, very good question. I guess your proof should work for the uniform topology as well (and in that case: for arbitrary factors in your product).
Yes, basically, when I look at Theorem 43.5. in Munkres, I don't see why it wouldn't work for an arbitrary product X = ∏Xi, where Xi is complete with respect to the metric di. i.e. X would be complete in the uniform metric based on the metrics di.
 
  • #9
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Ah yes, I forgot about theorem 43.5... Now I'm completely sure that it works for arbitrary products :smile:
 
  • #10
radou
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Ah yes, I forgot about theorem 43.5... Now I'm completely sure that it works for arbitrary products :smile:
OK. Btw, does an isometric embedding preserve completness?

I don't quite get the concept of an exercise which asks to show that an open subspace of a topologically complete space is topologically complete. The hint says to define an imbedding of the open set U into X x R (where X is complete). But I didn't find completeness as a topological invariant on the list of these... Correct me if I'm wrong.
 
  • #11
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Well, if f:X--> Y is a homeomorphism and if X is completely metrizable by a metric d, then Y is also completely metrizable by the metric

[tex]e(f(x),f(y))=d(x,y)[/tex]

This shows that complete metrizability is a topological property.

But beware: completeness is NOT a topological property, indeed, completeness depends crucially on the metric. For example: [tex]\mathbb{R}[/tex] with the usual metric is complete, but if we equip [tex]\mathbb{R}[/tex] with the metric

[tex]d(x,y)=|atan(x)-atan(y)|[/tex]

then this metric induces the same topology on [tex]\mathbb{R}[/tex], but [tex]\mathbb{R}[/tex] is not complete w.r.t. the metric d.

However, the existence of a complete metric is a topological property...
 
  • #12
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The hint in Munkres actually says that

[tex]e(x,y)=d(x,y)+|\phi(x)-\phi(y)|[/tex]

is a complete metric on U... Maybe this will help you...
 
  • #13
radou
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OK, I'll think about all you said.
 

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