# Rudin theorem $3.7$

$$\textbf{Theorem 3.7}$$. 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 $$n_{i-1}$$?

• Since there is a subsequence converging to $x$, there are infinitely many $k$ with $d(x,p_k) < 2^{-i}\delta$. Feb 20, 2016 at 20:58

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 $\{p_n\}$

• This is a really great intuitive explanation; it really helps. Thanks so much :) Jan 17, 2018 at 9:08

Yo can choose some $n_i > n_{i_1}$ because you have an infinite number of candidates.

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 $d(p_{n_i},x)<\epsilon$.