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.


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.

  • 1
    $\begingroup$ 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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