Theorem 3.6 in Rudin's principles of mathematical analysis says the following thing:

(a) If $(p_n)$ is a sequence in a compact metric space $X$, then some subsequence of $(p_n)$ converges to a point of $X$

I will now provide the worked out proofs, and I wonder if someone could verify whether I understood the details left out correctly.

Before I start the proof, let me introduce the notation $N_r(p)$, which is the open ball (or neighborhood) with center $p$ and radius $r$


(a) Define $E:= \{p_n \in X\mid n \in \mathbb{N}\}$

We consider two cases:

$\boxed{1}$ $E$ is finite

In this case, since a sequence is infinite, it must follow that there is an element $p \in E$ that occurs an infinite amount of times in the sequence. So, there are integers $n_1 <n_2 <\dots$ such that $p_{n_i} = p$ for $i = 1,2,3, \dots$. So, the sequence $(p_{n_i})_{i }$ is a constant subsequence with limit $p \in X$

$\boxed{2}$ $E$ is infinite

By a theorem that is already proven (theorem 2.37), the set $E$ has a limit point $p \in X$, since $X$ is compact. Now, we construct a subsequence that converges to $p$.

Since $p$ is a limit point of E, for any neighborhood $N$ of $p$, there are infinitely many point in the intersection $E \cap N$.

Because of this, we can find an element in $N_1(p) \cap E$, meaning that we can pick an element of the sequence on distance less than $1$ from $p$. Call this element $p_{n_1}$.

Now, pick an element of the sequence on distance less than $1/2$ from $p$. Moreover, pick this element such that it occurs after $p_{n_1}$ in the sequence. This is certainly possible, because there are an infinite elements on such a distance and only a finite amount of elements occur before $p_{n_1}$ in the sequence. Call this element $p_{n_2}$

Continuing in this way (i.e. every time, we pick a point $p_{n_i}$ such that $d(p,p_{n_i}) < 1/i$, the argumentation to make this rigorous can be found in the previous paragraphe), we obtain the existence of a subsequence $(p_{n_i})_i$ with $d(p,p_{n_i}) < 1/i$.

We now prove that $p_{n_i} \to p$ . For this, let $\epsilon > 0$. By the archimedian property of the real numbers, there is a positive integer $I$ such that $1/I < \epsilon$. So, let $i > I$, then $1/i < 1/I < \epsilon$, such that $d(p,p_{n_i}) < \epsilon$. This shows that $p_{n_i} \to p \quad \square$


Have I filled up the details correctly? Is the last paragraph correct?

  • $\begingroup$ It seems you take it as the definition of compact so there is nothing to prove. Morally, if you have a finite union of $k$ balls $B_i$ and sequence $(p_n), p_n \in \bigcup_{i=1}^k B_i$ then there is a $i$ such that $p_n \in B_i$ infinitely often. Now repeat this recursively by dividing $B_i$ in $k$ smaller balls, you'll get that your sequence is infinitely often in an arbitrary small ball, thus it has an accumulation point $p$ and a subsequence converges to it (this is the motivation for the definition of compact sets). $\endgroup$
    – reuns
    Commented Aug 20, 2017 at 17:32
  • $\begingroup$ I think it sufficient to construct a subsequence satisfying $d(p,p_{n_k}) < {1 \over k}$. There is nothing wrong with invoking the Archimedean property, but it is sufficiently obvious in context and just serves to distract. $\endgroup$
    – copper.hat
    Commented Aug 20, 2017 at 17:40
  • $\begingroup$ You mean that what follows in my proof is trivial, so it can be left out? $\endgroup$
    – user370967
    Commented Aug 20, 2017 at 17:57
  • $\begingroup$ It is fine and anyone reading it would accept the argument. But a sequence contains terms, not elements. So, a better presentation is possible. In my next comment I give a suggestion to the paragraph that starts with 'Because of this,...'. $\endgroup$ Commented Sep 24, 2017 at 12:42
  • 1
    $\begingroup$ Thanks. I still have difficulties with the english (not native language for me), but it gets better; $\endgroup$
    – user370967
    Commented Sep 24, 2017 at 13:14

1 Answer 1


The OP certainly expanded on Rudin's terse style, but one paragraph lacks precision/clarity and needs work:

Because of this, we can find an element $q_1 \in N_1(p) \cap E$, so we can write $q_1 = p_{n_1}$ with $d(p,p_{n_1}) <1$. Now, assume $n_1 <n_2 < \dots <n_{k-1}$ are chosen such that $d(p,p_{n_i}) < 1/i$ for $i = 1,2,3 \dots, k-1$. Then, we can find $q_k \in N_{1/k}(p) \cap E$, meaning that we can write $q_k = p_{n_k}$ for some integer $n_k$ with $d(p,p_{n_k}) < 1/k$. Because the intersection above contains an infinite amount of points, we can certainly find an integer $n_k > n_{k-1}$.

The following is a rework that is more amenable to proof verification:

Define the relation $F_p$ on E by $a \, F_p \, b$ if

$\tag 1 d(b,p) \le (.5) \, d(a,p)$


$\tag 2 \text{IF } a = p_n \text{ AND } b = p_m \text{ THEN } m \gt n$

For any $x \in E$ there exist a $y \in E$ with $x \, F_p \, y$. By the axiom of dependent choice, there exist a sequence $q_n$ in $E$ satisfying $q_n \, F_p \, q_{n+1}$. But by (1) it converges to $p$ and by (2), this sequence can also be represented as a subsequence of $p_n$.

I don't think Rudin talks about the axiom of choice in his book.

We encourage the OP to rework his paragraph so that the presentation of variables/notation has a better flow. For example, a sequence is being recursively generated, but the OP's paragraph is kind of fuzzy on $q_2,q_3,\cdots,q_{k-1}$.

The OP might find Some Remarks on Writing Mathematical Proofs / J. M. Lee of interest.

  • $\begingroup$ Isn't this just an induction argument? So there seems to be no need to invoke the axiom of choice. First, choose an element such that the distance $d(p,p_{n_1})$ is smaller than one. Then, choose a second element such that $d(p,p_{n_2}) < 1/2$ such that it satisfies the conditions. Proceed in this way. As in my proof, this argument can be made rigorously using induction, to obtain the existence of the desired subsequence. I agree though, that the paragraph should be rewritten, such that it is easier to read. $\endgroup$
    – user370967
    Commented Sep 23, 2017 at 11:22
  • $\begingroup$ Your comment looks like a good start to a rewrite of the paragraph. The sequence you are creating does not require the full strength of AC, but still, at each inductive step, a choice is made that is not, say, a 'recursive formula'. For the purposes of real analysis, you can gloss over ${\displaystyle {\mathsf {DC}}}$ and finer points of set theory, $\endgroup$ Commented Sep 23, 2017 at 12:15
  • $\begingroup$ Yes. My argument is almost exactly the same as yours. I just used a link to make the presentation more succinct. $\endgroup$ Commented Sep 23, 2017 at 12:16
  • $\begingroup$ ... But instead of $1/n$ I kept cutting the distance to $p$ by at least $1/2$. $\endgroup$ Commented Sep 23, 2017 at 12:24
  • $\begingroup$ I will rewrite it soon (don't have time atm). I will award you the bounty then if you can see whether it is correct then. Will let you know when I edited my post. $\endgroup$
    – user370967
    Commented Sep 23, 2017 at 13:03

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