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Searching material to know more about totally bounded sets, I found this property

Let $(M,d)$ be a metric space. So $A\subset M$ is totally bounded if and only if for every $\varepsilon >0$ exists a compact $K$ such that $d(x,K)<\varepsilon$ for all $x \in A$

I'd appreciate any hint for proovingg this result.

pablomerlet
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    a) finite sets are compact, that yields one direction, b) compact sets are totally bounded, that yields the other. Does that already help? – Daniel Fischer Sep 12 '13 at 00:21
  • @DanielFischer , I see one direction with a). But I don't get how use b) yet – pablomerlet Sep 12 '13 at 00:30
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    Try with $\varepsilon/2$. – Daniel Fischer Sep 12 '13 at 00:32
  • @DanielFischer, Would you tell me if I'm doing good? I take a compact $K$ such that $d(x,K)< \varepsilon\ \forall x \in A$, and get that exists $k_1,\ldots,k_n$ such that $A \subset \bigcup_{i=1}^n B(k_i,\varepsilon)$...but I don't see from where I should pick the $x_i$'s D: – pablomerlet Sep 12 '13 at 00:53
  • If $d(x,K) < \varepsilon$, then there is a $y \in K$ with $d(x,y) < \varepsilon$. Now, there is a $k_i$ with $d(y,k_i) < \varepsilon$, so... – Daniel Fischer Sep 12 '13 at 00:55
  • @DanielFischer I think I get it. If I already have that $A \subset \bigcup_{i=1}^n B(k_i,\varepsilon /2)$, for ${k_i}{i=1}^n\subset K$, then I just have to pick $x_i \in B(k_i,\varepsilon /2)$ for each $i$ to conclude that $A \subset \bigcup{i=1}^n B(x_i,\varepsilon)$ – pablomerlet Sep 12 '13 at 01:12
  • Ah, does your definition of totally bounded demand that the centres of the balls lie in $A$? That's inconvenient for proofs (it's of course equivalent to the one that doesn't demand the centres being in $A$). Yes, that's the idea, if you cover $K$ with finitely many balls of radius $r$, where $d(x,K) < r$ for all $x \in A$, then $A$ is contained in the union of the finitely many balls with radius $2r$ and the same centres. If you need the centres in $A$, you then cover $A$ by balls of radius $4r$. – Daniel Fischer Sep 12 '13 at 01:17
  • @DanielFischer thanks for your help – pablomerlet Sep 12 '13 at 01:49

1 Answers1

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Since the hinting was successful, let's write it up in full.

Let $A\subset M$ be totally bounded. By the definition of total boundedness, for every $\varepsilon > 0$, there are finitely many points $x_1,\,x_2,\, \dotsc,\, x_{n_\varepsilon}$ such that

$$A \subset \bigcup_{k=1}^{n_\varepsilon} B_\varepsilon(x_k).$$

The set $K_\varepsilon = \{ x_k : 1 \leqslant k \leqslant n_\varepsilon\}$ is finite, hence compact, and for every $x \in A$, there is a $k$ with $x \in B_\varepsilon(x_k)$, hence $d(x,K_\varepsilon) \leqslant d(x,x_k) < \varepsilon$.

Conversely, suppose that $A$ is such that for all $\delta > 0$ there is a compact $K_\delta$ such that for all $x \in A$ we have $d(x,K_\delta) < \delta$.

Let $\varepsilon > 0$. By assumption, there is a compact $K_{\varepsilon/4}$ with $d(x,K_{\varepsilon/4}) < \varepsilon/4$ for all $x \in A$. Since $K_{\varepsilon/4}$ is compact, there are finitely many points $y_1,\, y_2,\, \dotsc,\, y_{n_\varepsilon}$ such that

$$K_{\varepsilon/4} \subset \bigcup_{k=1}^{n_\varepsilon} B_{\varepsilon/4}(y_k).$$

Since $d(x,K_{\varepsilon/4}) < \varepsilon/4$ for all $x \in A$, that implies

$$A \subset \bigcup_{k=1}^{n_\varepsilon} B_{\varepsilon/2}(y_k).$$

Let $ I = \{ k : A \cap B_{\varepsilon/2}(y_k) \neq \varnothing\}$, and for $k \in I$ choose $x_k \in A \cap B_{\varepsilon/2}(y_k)$. Then $B_{\varepsilon/2}(y_k) \subset B_\varepsilon(x_k)$ and

$$A \subset \bigcup_{k \in I} B_\varepsilon(x_k).$$

So for all $\varepsilon > 0$, $A$ can be covered by finitely many balls of radius $\varepsilon$ with centre in $A$, i.e. $A$ is totally bounded.

Daniel Fischer
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