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$\textbf{Theorem 3.7}$. The subsequential limits of $\{p_n\}$ is a closed subset of a metric space $X$.

$\textbf{Proof.}$ Let $E^* $ be the set of subsequential limits. Let $q$ be a limit point of $E^*$. Choose $n_1$ so that $p_{n_1}\neq q$ (if such $n_1$ doesn't exist, then we're done.). Let $\delta = d(q,p_{n_1})$. Suppose $n_1,\cdots,n_{i-1}$ 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 $\cdots$

I don't understand the highlit part: I understand that some $p_{k}$ with $d(x,p_k)<2^{-i}\delta$ must exist because $x$ is a limit of some subsequence, but how can he be sure that this $k=n_i$ is bigger than $n_{i-1}$?

E: There are two other questions about this theorem here, but this was not asked in them.

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    $\begingroup$ Since there is a subsequence converging to $x$, there are infinitely many $k$ with $d(x,p_k) < 2^{-i}\delta$. $\endgroup$ – Daniel Fischer Feb 20 '16 at 20:58
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Yo can choose some $n_i > n_{i_1}$ because you have an infinite number of candidates.

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The basic idea that underlies all what Rudin does in his book is this.

Warm-up. Any sequence contains an infinite number of points, as well as all its subsequences. Any neighborhood of a limit point contains infinite number of points.

Now, we have a set $E^*$ of subsequential limits of the original sequence $\{p_n\}$. We need to see what happens if there is a limit point $q$ of $E^*$ such that $q \notin E^*$.

Proof by contradiction, suppose such point $q$ exists.

Each point of $E^*$ is a limit of some subsequence of $\{p_n\}$. The points of $E^*$ do not need themselves to be points of $\{p_n\}$, subsequences may "approach" them infinitely close. So any point of $E^*$ is a limit point for points of $\{p_n\}$ and any point of $E^*$ in each of its neighborhoods has infinite bunch of points of $\{p_n\}$.

Now, if there exists a point $q$, which is a limit point for points of $E^*$, then any neighborhood of $q$ "captures" infinite number of points of $E^*$, each of which in any their neighborhood in their turn "capture" infinite number of points of $\{p_n\}$. The story takes place in one metric space, so any neighborhood of $q$ eventually "captures" infinite number of points of $\{p_n\}$.

In other words, a hypothetical $q$ would pull in any of its neighborhoods infinite bunch points of $E^*$, which would pull in the same neighborhood of $q$ an infinite bunch of points of $\{p_n\}$. Because of the latter fact, $q$ is a limit of some subsequence of $\{p\}$, and as such must belong to $E^*$ by its definition, a contradiction. So $q \in E^*$.

Picture:

any $N(q)$ <- infinite number of points of $E^*$ <- any $N($ any point of $E^*)$ <- infinite number of points of $\{p_n\}$,

so any $N(q)$ includes infinite number points of $\{p_n\}$

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  • $\begingroup$ This is a really great intuitive explanation; it really helps. Thanks so much :) $\endgroup$ – Abdu Magdy Jan 17 '18 at 9:08
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That a subsequence $p_{n_i}$ tends to $p_n$ means that for every $\epsilon$ there is just finitely many $i$ that $d(p_{n_i},x)\geq\epsilon$. It means, that are arbitry big $i$s that $d(p_{n_i},x)<\epsilon$.

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