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 $r$
Proof:
(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 $p_{n_i} \to p \quad \square$
Questions:
Have I filled up the details correctly?
Is the last paragraph correct?
Best Answer
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
$\tag 1 d(b,p) \le (.5) \, d(a,p)$
$\text{and}$
$\tag 2 \text{IF } a = p_n \text{ AND } b = p_m \text{ THEN } m \gt n$
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.