$\bf 3.7\ \ $ Theorem $\ \ $ The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$.
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 $q\in E^*$.
$\qquad$ 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 $n_1,…,n_{i-1}$ 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 $i=1,2,3,…$. This says that $\{p_{n_i}\}$ converges to $q$. Hence $q\in E^*$.
Can someone explain how Rudin is selecting the $n_i$?
Best Answer
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}\neq q$ and set $\delta = d(p_{n_1}, q)$. Then he consider $\delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < \delta /2$$
Inductively, he choose $n_k>n_{k-1}> \cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< \frac{\delta}{2^i}.$$