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I know that this question Separability, total boundness and topological equivalence of metrics has been asked, but the only solution given is not valid.

There is something I already knew: $(Y, d_2)$ is totally bounded, so for each $\varepsilon$ there exists $x_1,\ldots,x_n$ such that $$ Y \subseteq B(x_1,\varepsilon) \cup B(x_2,\varepsilon) \cup \cdots \cup B(x_n,\varepsilon) $$ Now, the hint given is to look at the preimages of those balls ($B(x_1,\varepsilon)$, $B(x_2,\varepsilon)$, $B(x_n,\varepsilon)$) and see what happens...

How do I exactly use this hint? Could you help me?

NOTE that I am trying to prove that $(X,d')$ is totally bounded, NOT $(Y,d_2)$.

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Sigma
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  • The solution given there is correct. Please edit your question to indicate what you think is wrong with it, so that we know what the real problem is. – Brian M. Scott Dec 12 '14 at 00:11
  • What Alex Ravsky proved was that (Y,d2) is totally bounded. What I want to prove is that (X,d') is totally bounded. – Sigma Dec 12 '14 at 00:52
  • Which immediately implies that $\langle X,d'\rangle$ is totally bounded: by definition of $d'$ the two spaces are isometric. – Brian M. Scott Dec 12 '14 at 00:54
  • So you say that (X,d') and (Y, d2) are isometric? And what happens if two spaces are isometric? I was unaware of that concept. – Sigma Dec 12 '14 at 00:58
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    Isometries are to metric spaces what homeomorphisms are to topological spaces: isometric metric spaces are essentially the same space under different names, just as homeomorphic topological spaces are essentially the same space under different names. They are structurally identical as metric spaces. Here this is clear: $d'$ is simply the $d_2$ metric on $Y$ transferred back to $X$ by the bijection $f$. (One does of course have to prove that $d$ and $d'$ are equivalent, by the OP in that question had already done so.) – Brian M. Scott Dec 12 '14 at 01:01
  • Ok, but in the case of metric spaces homeomorphisms don't preserve total boundness... How can I prove (or where can I find a proof) that if (X,d) and (Y,d') are isometric and (X,d) is totally bounded THEN (Y,d') is totally bounded too? – Sigma Dec 12 '14 at 01:20
  • Have you tried? I ask because it’s absolutely immediate. Let $f:X\to Y$ be an isometry. If $F$ is a finite subset of $Y$ such that every point of $Y$ is within $\epsilon$ of some point of $F$ in the $d'$ metric, then every point of $X$ is within $\epsilon$ of some point of $f^{-1}[F]$ in the $d$ metric. – Brian M. Scott Dec 12 '14 at 01:25

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