# If $X$ is totally bounded then every sequence contains a Cauchy subsequence

I attempted the proof, I just want to see if it is correct:

Suppose $X$ is totally bounded and $(x_n)$ is a sequence in $X$. Then $(x_n)$ has a subsequence contained in a ball of radius $1/2$. This subsequence has a subsequence contained in a ball of radius $1/3$ and so on. Take the first term in each of these subsequences and call this sequence $(x_{n_k})$.

Then if $m>l$, $d(x_{n_m},x_{n_l})< \frac{2}{n+1}$. And since $\frac{2}{n+1}\rightarrow 0$ it follows that $(x_{n_k})$ is a cauchy sequence.

• There is already a question on this, see: math.stackexchange.com/questions/556150/… – Josh R May 22 '16 at 8:45
• I have taken a somewhat different approach, I am not asking for the problem to be solved, I am asking if my proof is correct. – fosho May 22 '16 at 8:46
• @Dman This question might be a bit old now, but I'm a bit confused about what the balls are centered at at each stage of this construction. – Alfred Yerger Aug 7 '16 at 6:35

It is a correct approach, except that it should be $\frac{2}{l+1}$. But note that your proof as stated uses the axiom of dependent choice, because you can't uniquely choose the subsequence in the ball of desired radius if there are more than one and you don't have any way of tie-breaking.
• @Dman: It's accepted by all mathematicians with very rare counter-examples. My reason for mentioning it is that it is glossed over by most teachers (who aren't even aware that they're using something strictly stronger than induction). Normal induction can only give you a finite sequence of any desired length $n$, but cannot give you the infinite sequence that you seek. Also, DC (dependent choice) cannot be proven in ZF alone, but can be proven in ZFC. – user21820 May 22 '16 at 13:34
• @Dman: No. In normal mathematical speaking and writing it needn’t be mentioned at all. What user21820 says is entirely correct, but the fact is that the use of dependent choice is glossed over because in most mathematical discourse it’s irrelevant: $\mathsf{DC}$ and even the full axiom of choice are simply taken for granted. Mentioning this highly technical foundational issue would be an unnecessary and fairly pointless distraction/complication in most undergraduate and many graduate courses. – Brian M. Scott May 22 '16 at 20:25