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The Euclidean spaces $\Bbb R^n$ have a nice property that every bounded subset is totally bounded, while for infinite-dimensional Banach spaces, the closed unit ball is bounded but not totally bounded.

I wonder what makes a metric space have (or not have) the property that bounded set is totally bounded. Could it be whether the space can be isometrically embedded into some $\Bbb R^n$? Or are there some other characterisations?

edm
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    This is actually a property of the metric, not the topology. For instance consider the following two metrics on $\mathbb{R}$: $$d_1(x,y) = |x - y|,\quad d_2(x,y) = \min{|x - y|, 1}.$$ With $d_1$, bounded implies totally bounded but not with $d_2$. – Trevor Gunn Apr 27 '17 at 02:44

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Just to give a counterexample to the conjecture of "whether the space can be isometrically embedded into some $\mathbb{R}^n$:

Consider the set $P$ formed by the disjoint union of a point $O$ with a countably infinite number of quills $Q_i$, where $i\in \mathbb{N}$. Each $Q_i$ is taken to be a copy of the interval $(0, i^{-1}]$.

Take the metric on $P$ to be given by

  • $d(O,x) = x$ for $x\in Q_i$
  • $d(x,y) = |x-y|$ if $x,y\in Q_i$
  • $d(x,y) = x + y$ if $x\in Q_i$ and $y\in Q_j$ and $i\neq j$.

Note that $(P,d)$ is totally bounded: for every $\epsilon > 0$, the ball $B(O, \epsilon)$ contains all but finitely many of the quills. But $(P,d)$ does not have an isometric embedding into $\mathbb{R}^n$ for any $n$: an isometric embedding will require that every quill be parallel or antiparallel, which is not possible.

A remark: the obstruction I gave for embedding into $\mathbb{R}^n$ can be worked around partially by taking instead

  • $d(x,y) = \sqrt{x^2 + y^2}$ if $x\in Q_i$, $y\in Q_j$, and $i\neq j$.

Then in this case we can see that $P$ is "almost" isometrically embedded in $\mathbb{R}^n$, in the sense that for any $\epsilon > 0$, we have $P \setminus B(0,\epsilon)$ can be embedded in $\mathbb{R}^{n_\epsilon}$ for some $n_\epsilon < \infty$.

The other implication clearly holds, giving a sufficient condition.

In the case your metric space is complete, this would mean that it also satisfies the Heine-Borel property. This MSE question provides some additional discussion of these kinds of spaces. A particularly interesting class of examples are those of the nuclear spaces. In a way they are similar to the example I gave above, in that they are infinite dimensional spaces that behave almost as if they were finite dimensional.

Willie Wong
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