I was thinking about the following: if $X$ is totally bounded, then we can always choose the space spanned by the centers of an $\varepsilon$-ball net so that $X$ is lies in a neighborhood of finite dimensional vector space. Is the converse true?

Let $V$ be a Banach space and $X \subseteq V$ a bounded subset. If for all $\varepsilon >0$, $$X \subseteq B_\varepsilon(W_{\varepsilon}):= \{ x \in V : ||x-y|| < \varepsilon, y \in W_{\varepsilon} \}$$ for some $W_{\varepsilon}$ finite dimensional subspace, then $X$ is totally bounded.

EDIT: I think I have a proof for the statment.

Suppose $X$ is not totally bounded. Exists $(x_i) \subseteq X$ such that $||x_i - x_j || \ge 2\varepsilon$ for all $i \not= j$.

  1. Exists $y_i \in W$ such that $||x_i - y_i || < \varepsilon/2$ for each $i$.
  2. $||y_i -y_j|| \ge ||x_i- x_j|| - \Big(||x_i-y_i||+||x_j-y_j||\Big) \ge \varepsilon$ for all $i \not= j$.

  3. Note that $||y_i||$ is bounded as $||y_i|| \le ||x_i||+\varepsilon<2 \le M +\varepsilon/2$ .

  4. But the closed ball $B_M(0) \subseteq W$ is closed, bounded, hence compact. Hence sequentially comapct. Contradiction.

Is this proof correct?

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    $\begingroup$ Yes, because all bounded subsets of $W$ are totally bounded, since $W$ is finite-dimensional. $\endgroup$ – Giuseppe Negro Apr 24 '18 at 9:12
  • $\begingroup$ Where can I get a reference for the proof of your statement? $\endgroup$ – CL. Apr 24 '18 at 9:18
  • $\begingroup$ Sorry, but I am also failing to see how the statement might follow directly from your claim. $\endgroup$ – CL. Apr 24 '18 at 9:26
  • $\begingroup$ I meant that, since $X$ is a bounded subset of $W$ which is a finite-dimensional space, it is totally bounded. This is the same as the proof you wrote, essentially. $\endgroup$ – Giuseppe Negro Apr 24 '18 at 9:28
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    $\begingroup$ In the present form, your condition is just a complicated way to say that $X\subset W$. $\endgroup$ – MaoWao Apr 24 '18 at 10:10

Your proof is correct. A more direct proof (without arguing by contradiction) would be:

Given $\epsilon > 0$, let $W$ be a finite-dimensional subspace such that $X$ is contained in the $(\epsilon/2)$-neighborhood of $W$. For each point of $X$, pick a point of $W$ at distance less than $\epsilon/2$ from it. Let $E$ be the set of all such points in $W$. Since $X$ is bounded, so is $E$. Being a bounded subset of a finite-dimensional linear space, $E$ is totally bounded. Thus, there exists a finite set $F$ such that $E$ is contained in the $(\epsilon/2)$-neighborhood of $F$. Consequently, $X$ is contained in the $\epsilon$-neighborhood of $F$, which shows it is totally bounded.

The above is taken from my blog where I mention a simple corollary: a subset of a Banach space is compact iff it is closed, bounded, and "flat" in the sense you described (contained in $\epsilon$-neighborhood of a finite-dimensional subspace, for every $\epsilon$).

  • $\begingroup$ Great, I only dislike the adjective "flat": for example, with this notation we have that $\mathbb S^{d-1}$ is flat, because it is compact. That's just a matter of taste. $\endgroup$ – Giuseppe Negro Apr 25 '18 at 9:13

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