Lemma 2.5-2 and Theorem 2.5-3 in Kreyszig's Functional Analysis Book: Does compactness of every closed and bounded subset also imply ...?

Question

Here is Lemma 2.5-2 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig:

A compact subset $M$ of a metric space is closed and bounded.

But the converse is not true, as is shown by the set $$M = \left\{ \ (1, 0, 0, \ldots), \ (0, 1, 0, 0, \ldots), \ (0, 0, 1, 0, 0, \ldots), \ \ldots \ \right\}$$ in $\ell^2$.

But here is Theorem 2.5-3:

In a finite dimensional normed space $X$, any subset $M \subset X$ is compact if and only if $M$ is closed and bounded.

Now my question is as follows:

Let $X$ be a metric space such that every closed and bounded subset of $X$ is (sequentially) compact. Does this imply that $X$ is a finite dimensional normed space?

I have no idea of how to proceed, although I'm clear about the proofs in Kreyszig.

Answer 1

Say that $X$ is Heine-Borel if every closed and bounded subset is compact. I will interpret your question in the following way: if $X$ is a Heine-Borel metric space, is there a Bi-Lipschitz embedding from $X$ into $\mathbb R^N$ for some $N \in \mathbb N$? A map $f$ is Bi-Lipschitz if there exist positive constants $c,C$ such that $c d(x,y) \leq d(f(x),f(y)) \leq C d(x,y)$ for all $x,y \in X$.

The answer is no. In fact, we may take $X$ itself to be compact. Recall $$\ell^1 = \{(a_n)_{n=1}^\infty : \sum\limits_{n \in \mathbb N} |a_n| < \infty\},$$ $$\|(a_n)_{n=1}^\infty\|_{\ell^1} = \sum\limits_{\mathbb N} |a_n|.$$ By $\delta_n$, we mean $(0,0,...,0,1,0,0,...) \in \ell^1$, where the $1$ occurs in the $n$th entry. Let $$X = \{r \delta_n: n \in \mathbb N, r \in [0,{1 \over n}]\}.$$ To see that $X$ is sequentially compact, let $(x_k)_{k=1}^\infty \in X$. Let $X_n = \{r \delta_n : r \in [0,{1 \over n}]\}$. If every $x_k \in X_n$ for only finitely many $n$, then there is exists $n \in \mathbb N$ and a subsequence $(x_{k_j})_{j=1}^\infty$ such that $x_{k_j} \in X_n$ for all $j$, which gives a convergent subsequence. If there exists infinitely many $n$ such that $x_k \in X_n$ for some $n$, then there exists a subsequence $(x_{k_j})_{j=1}^\infty$ such that $x_{k_j} \in X_{n_j}$ for all $j$, where $n_j \to \infty$. Then $x_{k_j} \to 0$.

To see that $X$ does not Bi-Lipschitzly embed into any finite dimensional space, let us make the following definition. Say that a metric space $(Y,d)$ is doubling if there exists a constant $K \in \mathbb N$ such that, for every $y \in Y$ and $r>0$, there exist $y_1,y_2,...,y_K \in Y$ such that $$B(y,r) \subset \bigcup\limits_{k=1}^K B(y_k,{r \over 2}).$$ You should check four things:

1. Doubling is preserved under Bi-Lipschitz maps.

2. Every finite dimensional vector space is doubling.

3. Every subspace of a doubling metric space is doubling.

4. $X$ is not doubling.

There are likely other ways to see that $X$ does not embed Bi-Lipschitzly into $\mathbb R^N$, some possibly easier, but this is one route. Hints: For 2., it is straightforward to show that $\mathbb R^N$ with the $\sup$ norm is doubling, and since all norms on finite dimensional vector spaces are equivalent, 1. implies 2. For 4., note that $B(0,{1 \over n})$ requires roughly $n$ balls of radius ${1 \over 2n}$ to cover.

Answer 2

This is not true, and the basic reason is that most metric spaces are not normed vector spaces at all. For instance, no normed vector space of positive dimension can be compact. So if $X$ is any compact metric space at all with $|X|\neq 1$, $X$ cannot be homeomorphic to any normed vector space, but any closed subset of $X$ is compact.

In fact, it is possible that $X$ is not even homeomorphic to a subset of any finite dimensional normed vector space. For instance, let $X$ be the Hilbert cube, which is a compact metric space. Then $\mathbb{R}^n$ embeds topologically in $X$ for all $n$, since $X$ is homeomorphic to $[0,1]^\mathbb{N}$. It follows that $X$ cannot embed topologically in $\mathbb{R}^N$ for any $N$, since $\mathbb{R}^n$ only embeds in $\mathbb{R}^N$ if $n\leq N$ (this is a hard theorem in topology; it follows from invariance of domain, for instance).

(Incidentally, you should really be more precise about what you mean by "$X$ is a finite dimensional normed space". Do you mean it is isometric to a finite dimensional normed space, or homeomorphic, or something else? However, I think the examples above show pretty conclusively the answer will be no for any reasonable interpretation.)