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Definition: Let $(E,d)$ be a metric space, $A \subseteq E$ a ⊂. $A$ is said to be totally bounded if $\forall \epsilon > 0 \ \exists x_1, ..., x_n \in A$, such that $A \subseteq \bigcup\limits_{i=1}^{n} B_d(x_i, \epsilon).$

Let now $A$ be a set such that $cl(A)$ is totally bounded. How do I show that $A$ itself is totally bounded? If I use the definition, I get open balls centered in $x_i \in cl(A), 1 \leq i \leq n$, which union contains $cl(A)$, and hence $A$, but I want that the balls are centered in points that lay in $A$.

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Since the $x_i$ are in $cl(A)$ the balls $B(x_i, \epsilon)$ have intersections with $A$ (call them $a_i$) then the balls $B(a_i, 2\epsilon)$ cover $A$

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  • $\begingroup$ Thank you! Does this argumentation also hold in the more general case that $x_i$ are in $E$ and $cl(A) \subseteq \cup B(x_i,\epsilon)$? $\endgroup$ – nbayo Aug 21 '16 at 19:04
  • $\begingroup$ @nbayo welcome! no it was necessary that $x_i$ are in the closure to use the definition that for any $e>0$ the intersection $B(x_i,e)\cap A$ is not empty $\endgroup$ – user290300 Aug 21 '16 at 19:07
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Suppose that $cl(A)$ is totally bounded, and let $\epsilon>0$ arbitrary. We are going to show that there finitely many points $a_1,\ldots,a_n\in A$ such that $A\subseteq \displaystyle{\bigcup_{i=1}^n B_d(a_i,\epsilon)}$.

Since $cl(A)$ is totally bounded, for the given $\epsilon'=\frac{\epsilon}{2}>0$ there exists finitely many points $x_1,\ldots,x_n\in cl(A)$ such that $cl(A)\subseteq \displaystyle{\bigcup_{i=1}^n B_d\left(x_i,\frac{\epsilon}{2}\right)}$. On the other hand, by definition of $cl(A)$ we have for each $x_i$ that, there is an element $a_i\in A$ such that $a_i\in B_d\left(x_i,\frac{\epsilon}{2}\right)$.

Claim: For each $i=1,2,\ldots,n$, $B_d\left(x_i,\frac{\epsilon}{2}\right)\subseteq B_d(a_i,\epsilon)$

Proof: Suppose $z\in B_d\left(x_i,\frac{\epsilon}{2}\right)$. Then by triangle inequality we have $$d(z,a_i)\leq d(z,x_i)+d(x_i,a_i)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ So, $z\in B_d(a_i,\epsilon)$.

Thus, using the Claim, we conclude that

$$A\subseteq cl(A)\subseteq \bigcup_{i=1}^n B_d\left(x_i,\frac{\epsilon}{2}\right)\subseteq \bigcup_{i=1}^n B_d(a_i,\epsilon),$$ and we can conclude that $A$ is totally bounded.

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  • $\begingroup$ Thank you for your answer! $\endgroup$ – nbayo Aug 21 '16 at 19:11

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