Is "total boundedness" a topological property? I have been trying to find a counterexample of ...
If $d$ and $e$ are topologically equivalent metrics (they generate the same open sets), and $(X,d)$ is totally bounded then $(X,e)$ is totally bounded.
Any ideas?
 A: If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1+d(x,y)}$ is another metric giving the same topology but bounded by 1.
(I'll leave it to you to show A. It's a metric, and B. They give the same topology.)

As I was asleep and missed that totally bounded is what you asked about:
Not every metric space has a topologically equivalent metric that's totally bounded, but any subset of e.g. $\mathbb{R}$ does, by a homeomorphism from $\mathbb{R}$ to $(0,1)$, for example $f(t) = \frac{1}{1+e^{-t}}$ gives a homeomorphism $f: \mathbb{R} \rightarrow (0,1)$.
The example another poster gave, that $\{1/n : n \in \mathbb{Z}, n > 0\}$ has the discrete topology but is totally bounded, is a simple example of this phenomenon.
A: Consider $\mathbb N$. Let $d$ be the discrete metric and $e:\mathbb N\times\mathbb N\to[0,\infty)$ defined as $$e(n,m)\equiv\left|\frac 1n-\frac 1m\right|\quad\forall n,m\in\mathbb N.$$
Then $e$ is a totally bounded metric, whereas $d$ is not totally bounded. However, both $d$ and $e$ induce the discrete topology.
