Another isometric imbedding problem

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Homework Help Overview

The discussion revolves around a problem in metric spaces, specifically focusing on equivalence relations and isometric embeddings. The original poster presents a framework involving Cauchy sequences and defines a metric on the set of equivalence classes derived from these sequences.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the properties of the defined equivalence relation and the associated metric. There are attempts to demonstrate that the mapping defined by the original poster is an isometric embedding, with specific focus on continuity and the behavior of open balls in the context of the metric space.

Discussion Status

Some participants express skepticism about certain claims regarding continuity and the nature of open balls in the metric space. Suggestions are made to simplify the proof by leveraging general properties of isometric embeddings. There is ongoing exploration of the implications of density and completeness in the context of the problem.

Contextual Notes

Participants note the complexity of notation and definitions, particularly regarding the continuity of the inverse of the mapping. There is also mention of the need to utilize the density of subsets within the metric space to address completeness.

  • #31
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.
 
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  • #32
Have we already proved that? Well, then it's pretty easy indeed!
 
  • #33
micromass said:
Have we already proved that? Well, then it's pretty easy indeed!

Yes, it's basically in post #10, unless I'm mistaken.
 
  • #34
Ah yes, it is basically the same thing indeed! That completes your completion exercise :biggrin:
 
  • #35
micromass said:
Ah yes, it is basically the same thing indeed! That completes your completion exercise :biggrin:

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.
 
  • #36
Yes, now you've shown that every metric space has a completion. The usual way to prove this is by exercise 9. The reason most textbooks prefer exercise 9 is because it can be easily generalized and because the completion is very easy to describe.

One can in fact also show that the completion is unique. This is actually a consequence of exercise 2.

Also note that \mathbb{Q} is an incomplete metric space. It's completion is of course \mathbb{R}. And exercise 9 now gives a very easy idea of how to construct \mathbb{R}! Just take all Cauchy sequences of rational numbers...
 
  • #37
micromass said:
Also note that \mathbb{Q} is an incomplete metric space. It's completion is of course \mathbb{R}. And exercise 9 now gives a very easy idea of how to construct \mathbb{R}! Just take all Cauchy sequences of rational numbers...

Wow, I never thought of it that way! Thanks!

Btw, basically, this doesn't strictly have much to do with topology, right? i.e. it's more about metric spaces...
 
  • #38
Yes, this is more analysis than topology. In fact, entire chapter 7 seems to be more about metric spaces than topology.

If you're going to study functional analysis, then you're going to see much of chapter 7 again. Specifically, the completion is very important in functional analysis! But you're correct, it's not really topology...
 

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