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Basically, Rudin states that a Cauchy sequence in a compact metric space is convergent. However, his proof is a bit non-intuitive and different from what one comprehends at first sight.

Thus, I have put my own proof, but I want it to be checked if correct, since it seems to be way more intuitive.

If $p_n$ takes only a finite number of values but is not ultimately constant, then it is obvious that cannot be Cauchy. Thus, a Cauchy sequence has an infinite number of values.

Thus, a Cauchy sequence has a convergent subsequence $p_{n_k}$ that converges to $p$. Thus, there is $N_1$ such that $d(p_{N_1},p)<\frac{\varepsilon}{2}$

However, since it's a Cauchy sequence, $d(p_n,p_m)<\frac{\varepsilon}{2}\forall n,m>N_2$ for some $N_2<N_1$. (To obtain that, we choose a large enough $N_1$)

Thus, $\forall n>N_2$, $d(p_n,p)\leq d(p_n,p_{N_1})+d(p_{N_1},p)<\varepsilon$.

QED.

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  • $\begingroup$ Why can't a Cauchy sequence take a finite number of values? $\endgroup$ – Seven Aug 8 '15 at 14:30
  • $\begingroup$ $\{p_n\}$ with $p_n=0$ for all $n$ is obviously Cauchy and takes only finitely many values. $\endgroup$ – parsiad Aug 8 '15 at 14:30
  • $\begingroup$ Sorry, I meant a Cauchy sequence that is not ultimately constant. $\endgroup$ – Hasan Saad Aug 8 '15 at 14:30
  • $\begingroup$ Why should a Cauchy sequence have a convergent subsequence? The space being compact is not being used anywhere. $\endgroup$ – Seven Aug 8 '15 at 14:40
  • $\begingroup$ That is exactly where it's being used. Since I said that it has an infinite number of values, and since the space is compact, then it must have a convergent subsequence. Is that not true? $\endgroup$ – Hasan Saad Aug 8 '15 at 14:45
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The statement "a Cauchy sequence has a convergent subsequence $p_{n_k}$ that converges to $p$" uses the result that every sequence in a compact metric space has a convergent subsequence. If you proved this earlier, the proof looks fine.

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