Here's Theorem 3.7 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.
The subsequential limits of a sequence $(p_n)$ in a metric space $X$ form a closed subset of $X$.
And, here's Rudin's proof.
Let $E^*$ be the set of all subsequential limits of $(p_n)$ and let $q$ be a limit point of $E^*$. We have to show that $q \in E^*$.
Choose $n_1$ so that $p_{n_1} \neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove. ) Put $\delta = d(q, p_{n_1})$. Suppose $n_1, \ldots, n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $x \in E^*$ with $d(x, q) < 2^{-i} \delta$. Since $x \in E^*$, there is an $n_i > n_{i-1}$ such that $d(x, p_{n_i}) < 2^{-i} \delta$. Thus $$d(q, p_{n_i}) \leq 2^{1-i} \delta$$ for $i = 1, 2, 3, \ldots$. This says that $(p_{n_i})$ converges to $q$. Hence $q \in E^*$.
Now here's my reading of Rudin's proof.
If the set $E^*$ of all the subsequential limits of the sequence $(p_n)$ has no limit points, then the set of all the limit points of the set $E^*$ is empty and is therefore contained in $E^*$.
So let's suppose that $q$ is a limit point of the set $E^*$. If $p_n = q$ for all $n \in \mathbb{N}$, then the sequence $(p_n)$, being a constant sequence, converges to $q$, and so every subsequence of $(p_n)$ also converges to $q$; therefore the set $E^*$ consists of a single point $q$ and thus cannot have limit points. So there is a natural number $n$ for which $p_n \neq q$. Let $n_1$ be the smallest such natural number.
Let's put $\delta = d\left(q, p_{n_1}\right)$. Then $\delta > 0$. Now since $q$ is a limit point of the set $E^*$, there exists a point $x_1 \in E^*$ such that $$0 < d\left(q, x_1\right) < \frac{\delta}{4}.$$
Now since $x_1$ is a subsequential limit of the sequence $(p_n)$, there is a strictly increasing function $\varphi_1 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_1 = \lim_{n \to \infty} p_{\varphi_1(n)}.$$ So there exists a natural number $N_1$ such that $$d\left( \ x_1\ ,\ p_{\varphi_1(n)} \ \right) < \frac{\delta}{4} $$ for all natural numbers $n$ such that $n > N_1$.
We note that, for each $n \in \mathbb{N}$, the inequality $n \leq \varphi_1(n)$ holds.
Let $n_2$ be the natural number defined as $$n_2 \colon= \max \left( \ \varphi_1(n_1 + 1)\ , \ \varphi_1(N_1 + 1) \ \right).$$ Then $n_2 > N_1$ and so $$ d\left( \ q \ , \ p_{n_2} \ \right) \leq d\left( \ q,\ x_1 \ \right) + d\left(\ x_1\ , \ p_{n_2} \ \right) < \frac{\delta}{4} + \frac{\delta}{4} = \frac{\delta}{2}.$$
Now as $q$ is a limit point of $E^*$, there exists a point $x_2 \in E^*$ such that $$0 < d(q, x_2) < \frac{ \delta}{8}.$$
Moreover, since $x_2$ is a subsequential limit of the sequence $(p_n)$, there is a strictly increasing function $\varphi_2 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_2 = \lim_{n \to \infty} p_{\varphi_2(n)}.$$ So there is a natural number $N_2$ such that $$ d\left( \ x_2 \ , \ p_{\varphi_2(n)} \ \right) < \frac{\delta}{8}$$ for all natural numbers $n > N_2$.
Note that, for each $n \in \mathbb{N}$, we have $n \leq \varphi_2 (n)$.
Now let $$n_3 \colon= \max \left( \ \varphi_2(n_2 + 1) \ , \ \varphi_2 ( N_2 + 1) \ \right).$$ Then $n_3$ is a natural number greater than $N_2$ and so we have $$d\left( \ q\ ,\ p_{n_3} \ \right) \leq d\left( \ q \ , \ x_2 \ \right) + d\left( \ x_2 \ , \ p_{n_3} \ \right) < \frac{\delta}{8} + \frac{\delta}{8} = \frac{\delta}{4}.$$
We note that $$n_3 \geq \varphi_2(n_2 + 1) > \varphi_2(n_2) \geq n_2 \geq \varphi_1 (n_1 + 1) > \varphi_1 (n_1) \geq n_1.$$ That is, $n_1, n_2, n_3$ are all natural numbers such that $$n_1 < n_2 < n_3.$$
Continuing in this way, we obtain a subsequence $(p_{n_i})$, where $n_1, n_2, n_3, \ldots \in \mathbb{N}$ and $n_1 < n_2 < n_3 < \cdots$, such that $$d\left(\ q \ , \ p_{n_i} \ \right) \leq \frac{2\delta}{2^i}$$ for all $i \in \mathbb{N}$.
We now show that this subsequence $(p_{n_i})$ converges to $q$. Let $\varepsilon > 0$ be given. Then we can find a natural number $$K > \frac{2\delta}{\varepsilon}.$$ Then $$2^K > K > \frac{2\delta}{\varepsilon}.$$ So for any natural number $i > K$, we have $$2^i > 2^K > \frac{2\delta}{\varepsilon}$$ and so $$ d\left( \ q \ , \ p_{n_i} \ \right) \leq \frac{2 \delta}{2^i} < \varepsilon. $$
Thus every limit point $q$ of the set $E^*$ is also an element of $E^*$. Hence $E^*$ is closed.
Is my reading of Rudin's proof correct? If not, what is it that I'm missing?
Best Answer
The hard bit in this theorem to show that $E^* \subset X$. In his proof Rudin is trying to make a direct proof by complete induction that for any limit point $q$ of $E^*$ we are able to construct an infinite sequence of elements of $E^*$ that converges to $q$.
Rudin implicitly(!) assumes that $E^* \subset X$, and as we are able to construct a sequence some points of $X$ taken from $E^*$, the limit $q$ of this subsequence belongs to $E^*$ by the way we chose to construct this sequence with elements of $E^*$.
Complete induction goes from the base case where we are picking points (each called $x$) starting from an arbitrary point $p_{n1} \in E^* : p_{n1} \ne q$ from smaller and smaller neighborhoods around $q$. The next one has radius that is half as small as the previous neighborhood of $q$. Since $q$ is a limit point, by Theorem 2.20, there are infinitely many points in $q$'s neighborhood that belong to $E^*$. So we are done and we are happy. $\Box$
The point that concerned me about the proof is that Rudin did not restrict $X$ to be either compact, or $\mathcal{R}$ in order to use Heine-Borel as a leverage here. He just made a ton of very important implicit assumptions and went along.
Suppose, we have $\mathcal{Q} = X$ for this case. Since $\mathcal{Q}$ is countable for this case and we can construct subsequences with irrational numbers as their limits. We can go along with Rudin's proof and get nothing.
The crucial point in this case is to play with the assumption that subsequential limits are part of $X$ (or not). This makes one think a lot over Rudin's explanations in this case. I understand that the empty set is closed, but Rudin's wording basically fails.