Another isometric imbedding problem

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The discussion revolves around proving properties of an isometric embedding defined from a metric space (X, d) to the set of equivalence classes Y of Cauchy sequences. The participants confirm that the defined relation is an equivalence relation and that the metric D on Y is well-defined. They explore the continuity of the embedding function h and suggest that proving the continuity of isometric embeddings in general could simplify the process. Additionally, they discuss the density of h(X) in Y and the completeness of Y, considering the implications of Cauchy sequences and the density of a subset A in a metric space. The conversation emphasizes the importance of clear notation and logical reasoning in mathematical proofs.
  • #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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