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Here's what I'm trying to prove:

Let $(X,d)$ be a metric space and $A$ be a subset of $X$. If every sequence $(x_n)$ of $A$ contains a Cauchy subsequence, then $A$ is totally bounded.


Here's the definition of "totally bounded" according to my textbook:

Let $(X,d)$ be a metric space. The subset $A$ of $X$ is said to be totally bounded if given $\varepsilon >0$ there exists a finite number of subsets $A_1$, $A_2$, ..., $A_n$ of $X$ such that $\text{diam } A_k <\varepsilon$ for $k=1,2, \ldots, n$ such that $ A\subset \cup_{k=1}^{n} A_k$.


Here's my proof:

Let $\varepsilon > 0$ be given. Pick any point $x_1 \in A$. If $A \subset B(x_1 , \varepsilon /4)$ then we are done. If not, pick another point $x_2 \in A$ such that $x_2 \not\in B(x_1 , \varepsilon /4)$. Now, if $A \subset B(x_1 , \varepsilon /4) \cup B(x_2 , \varepsilon /4)$ then we are done. Otherwise, pick a point $x_3 \in A$ such that $x_3 \not\in B(x_1 , \varepsilon /4) \cup B(x_2, \varepsilon /4)$. We claim that this repeated process of finding a cover for $A$ must end in finitely many steps. Suppose not. Then we would have ended up constructing a sequence $(x_n) \in A$ such that for $m>n$, we have $d(x_m , x_n) \ge \varepsilon /4$. But this would imply that $(x_n)$ has no Cauchy subsequence which is a contradiction to our assumption.


Is this proof correct? Alternative proofs are welcome.

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Your proof is fine. (You omitted 'totally' in the definition though.) To be very, very pedantic, your proof assumes that $A$ is non-empty. But it is trivial that the empty set is totally bounded.

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    $\begingroup$ I added "totally" in the definition. But in $\mathbb{R}^2$ (with the usual metric) the open ball $B(0, \varepsilon /2)$ has diameter equal to $\varepsilon$. I used $\varepsilon /4$ to avoid that. $\endgroup$ – Ashish K Oct 1 '19 at 14:59
  • $\begingroup$ You are right. I edited my answer to remove this wrong 'correction'. $\endgroup$ – Mark Wildon Oct 1 '19 at 15:02

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