If two metrics are equivalent and one is totally bounded, is the other totally bounded?

Question

I want to know if the following proof is correct...

If $(X,d)$ is separable then, if $S$ is an open cover of $X$, I can pick a numerable number of open sets in $S$, such that $X$ is included in their union (this is Lindelöf property, equivalent to separability).

But now, if $R$ is an open cover of $X$ in $(X,e)$, it is also an open cover of $X$ in $(X,d)$ (because the sets are the same, and they are also open in $(X,d)$ because of top. equivalence of metrics) an so I can pick a numerable number of open sets in R, such that X is included in their union (in $(X,d)$). Let $R_1, R_2, R_3, \dots, R_n, \dots$ be those sets. Then $\{R_1, R_2, R_3, \dots, R_n, \dots\}$ is a numerable set of open sets in $(X,e)$ such that X is included in their union.

I am trying to find a counterexample of this

if $d$ and $e$ are top. equivalent, and $(X,d)$ is totally bounded then $(X,e)$ is totally bounded.

But thats for metric equivalence. Topological equivalence is when open sets are the same with both metrics. — Igor Tahr, Dec 01 '14 at 22:46
Also related: Prove that a separable metric space is Lindelof without proving it is second countable. It would improve your Question if you give at least a sketch of the counterexample claimed above. — hardmath, Dec 01 '14 at 22:47
Ok, I am writing it, but... Its the prove given above right? — Igor Tahr, Dec 01 '14 at 22:50
What you claim in (A) is false: if $\langle X,d\rangle$ is separable, and $d$ and $e$ are topologically equivalent metrics on $x$, then $\langle X,e\rangle$ is also separable. Separability is a topological property, and if $d$ and $e$ are topologically equivalent metrics, by definition they generate the same topology. — Brian M. Scott, Dec 01 '14 at 22:51
What Brian said: A topological space is separable iff it has a countable dense subset. Of course this is preserved by changing from one metric to another, provided the metrics are topologically equivalent. — hardmath, Dec 01 '14 at 22:55
@JohnK.: If you’ve already shown that the Lindelöf property is equivalent to separability in metric spaces, then yes, your argument is correct. — Brian M. Scott, Dec 01 '14 at 22:57
Note that in your last edit, you changed the property at issue from compactness to total boundedness. — hardmath, Dec 01 '14 at 23:14

score 0 · Answer 1 · answered Apr 16 '15 at 14:08

This is false. $(0,1)$ and $\mathbb{R}$ with their usual metrics are homeomorphic, but $(0,1)$ is totally bounded and $\mathbb{R}$ is not even bounded.

To convert this into an example with two metrics on the same set, take $X = (0,1)$ and define the metric $d$ as usual by $d(x,y) =|x-y|$. Now let $f : (0,1) \to \mathbb{R}$ be your favorite homeomorphism, such as $f(x) = \tan(\pi(x-\frac{1}{2}))$, and set $e(x,y) = |f(x)-f(y)|$.

If two metrics are equivalent and one is totally bounded, is the other totally bounded?

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