Open interval $(0,1)$ is totally bounded

Question

Is true that an open interval $(0,1)\subseteq \mathbb{R}$ totally bounded? I think it is not true.

Since there is an homeomorphism from $(0,1)$ to $\mathbb{R}$ and $\mathbb{R}$ is not totally bounded.

Is my argument correct?

Answer 1

It is true that $(0,1)$ is totally bounded, and this is a property of the metric (or of the uniformity, if you know about those), not of the topology. So it is not necessarily preserved by homeomorphisms (but it is by isometries!), as your own example in fact shows.

That $(0,1)$ is totally bounded can be seen in different ways: for $r>0$ we can cover $(0,1)$ by at most $1+\frac{1}{r}$ (I think, but at least is finite) many open intervals of length $r$ (which are also open balls in the metric). Or we can apply to a general theorem that $[0,1]$ is compact and a subset of a compact set is totally bounded as well.

The notion is not topological, and neither is completeness (same example, but now $(0,1)$ is not complete in the usual metric, but $\mathbb{R}$ is), but totally bounded plus completeness as a combined property (for metric or uniform spaces, where these notions are defined) is equivalent to compactness, which is topological (in the sense that a homeomorphic space to a compact one is also compact).