# 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

## 1 Answer

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.

• Is that a bad thing? – fosho May 22 '16 at 13:05
• @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
• @Dman: Sorry to say this but I disagree with Brian. It's fine to not mention the use of DC only if students already know it's being used. Otherwise it is a disservice to students. Most undergraduate students I've come across have such a weak foundation in logic that they aren't even aware of this issue, and more than half of their teachers aren't either. – user21820 May 23 '16 at 1:07
• @BrianM.Scott: If I'm the first one to make the asker aware of this, it just supports my point. Note that in my original answer I just mentioned it as a side note, and it's Dman who asked about it, indicating implicitly that he/she wasn't aware of it prior to my mention. I personally think it is necessary for students to be able to justify their reasoning. Indeed, if a student has a full grasp of induction he/she would know immediately upon trying to prove this result that induction isn't strong enough. And in modern mathematical practice, it is common to mention whether AC is used. – user21820 May 23 '16 at 1:16