# Is total boundedness a topological property?

If a metrizable topological space is totally bounded with one metric, is it totally bounded with all others? A related, stronger question: if every metrization of a topological space is bounded, are they all totally bounded?

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To paraphrase GEdgar: Look for a counterexample for the first question. I answered the second question here: a metrizable space is compact if and only if every compatible metric is bounded. –  t.b. Oct 16 '11 at 23:37

The answer to your first question is "no". The real numbers are homeomorphic to the open interval $(-\frac{\pi}{2},\frac{\pi}{2})$, via the bijection given by $x\mapsto \arctan(x)$. Both spaces have a metrizable topology. But $(-\frac{\pi}{2},\frac{\pi}{2})$ is totally bounded, whereas $\mathbb{R}$ is not. Equivalently, endow $\mathbb{R}$ with the metric $d(a,b) = |\arctan(a)-\arctan(b)|$; this metric induces the same topology on $\mathbb{R}$ as the usual metric, but whereas $\mathbb{R}$ is bounded under $d$, it is not bounded under its usual metric.

Second question was answered inter alia elsewhere.

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Consider the positive integers with the discrete topology. On the one hand this topology is induced by the usual metric, which is certainly not totally bounded. On the other hand, it is also induced by the metric $$d(n,m)=\left|\frac1n-\frac1m\right|\;,$$ which clearly is totally bounded.

This is really just the observation that $\left\{\dfrac1n:n\in\mathbb{Z}^+\right\}$ and $\mathbb{Z}^+$ are homeomorphic as subspaces of $\mathbb{R}$ with the Euclidean topology, via the map $n\mapsto\dfrac1n$, where the first subspace has compact closure in $\mathbb{R}$.

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