0

The statement of the problem is:

Prove that a metric space (X,d) is separable iff d is topologically equivalent to d' and (X,d') is totally bounded.

(topologically equivalent means they generate the same topology, or equivalently, they have the same open sets)

1) First of all, how can I prove that: if d is topologically equivalent to d' and (X,d') is totally bounded THEN (X,d) separable?

Does it work if I give the same argument as in totally bounded implies separable and I use topologically equivalence, or is the proof different.

2) Second: I dont understand the other implication (if (X,d) is separable then d is topologically equivalent to d' and (X,d') is totally bounded.), so do you know any statement similar to that?

3)Finally, Is it true that if (X,d) is separable then there exists a metric d' such that d is topologically equivalent to d', and (X,d') is totally bounded ??

Raiden
  • 51

1 Answers1

1

Certainly, if two spaces are topologically equivalent (homeomorphic), then one is separable if and only if the other is as well. So, for 1), it suffices to show that $(X,d')$ is separable.

In order to do so, apply the definition of totally bounded to sets of size $1/n$ for positive integers $n$. That is, for each $n$, $X$ has a cover consisting of finitely many $1/n$-balls. Take the union of all the centers of all the balls that result in order to end up with a countable dense subset.

Your statement to 3) is correct, and answers your question about 2).

For your proof of the converse: consider the metric $$ d'(x,y) = \frac 2{\pi}\arctan(d(x,y)) $$ (or, if you prefer, replace $\arctan$ with any increasing, concave non-negative function passing through $(0,0)$). Note that $(X,d')$ is bounded and separable. From there, this proof will do, if you can work with that.

Community
  • 1
Ben Grossmann
  • 225,327
  • In general bounded does not imply totally bounded... How do you know that (X, d') is totally bounded? In fact, how do you know they are topollogically equivalent? – Raiden Dec 02 '14 at 03:01
  • If you accept that $d'$ gives you a metric, showing topological equivalence is easy: all you need to show is that an open ball in one space is also an open ball (or the entire set) in the other. – Ben Grossmann Dec 02 '14 at 03:04
  • Ok, but how do you show that the metric space (X, d') is totally bounded (which is stronger than bounded)? – Raiden Dec 02 '14 at 03:06
  • As for total boundedness, see the link I gave in my answer for a very concise solution. If you are unable to make sense of it (which is understandable), I would suggest posting a new question asking for a clarification or a direct proof that a bounded and separable metric space is totally bounded. – Ben Grossmann Dec 02 '14 at 03:07
  • Is my answer to your 3 original questions clear, though? – Ben Grossmann Dec 02 '14 at 03:12