Baby Rudin Theorem 3.7 Clarification

Question

$\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 $n_i$?

Answer 1

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

$$d(p_{n_i}, q)< \frac{\delta}{2^i}.$$

Answer 2

If $E^* = \emptyset$, then $E^*$ is a closed subset of $X$.

If $E^* = \{q\}$ for some $q \in X$, then $E^*$ is a closed subset of $X$.

We assume that $\#E^* \geq 2$.

Let $q$ be a limit point of $E^*$.

If $p_n = q$ for all $n \in \{1,2,\cdots\}$, then any subsequence of $\{p_n\}$ converges to $q$. So $E^* = \{q\}$. But we assumed that $\#E^* \geq 2$. So there is $n_1 \in \{1,2,\cdots\}$ such that $p_{n_1} \neq q$.

Because $q$ is a limit point of $E^*$, there exists $x_2 \in E^*$ such that $d(x_2, q) < \frac{\delta}{2^2}$.

Because $x_2$ is a subsequential limit of $\{p_n\}$, there exists $n_2 \in \{1,2,\cdots\}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < \frac{\delta}{2^2}$.

By the triangle equality in $X$, $d(p_{n_2}, q) \leq d(p_{n_2}, x_2) + d(x_2, q) < \frac{\delta}{2^2} + \frac{\delta}{2^2} = \frac{\delta}{2}$.

Because $q$ is a limit point of $E^*$, there exists $x_3 \in E^*$ such that $d(x_3, q) < \frac{\delta}{2^3}$.

Because $x_3$ is a subsequential limit of $\{p_n\}$, there exists $n_3 \in \{1,2,\cdots\}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < \frac{\delta}{2^3}$.

By the triangle equality in $X$, $d(p_{n_3}, q) \leq d(p_{n_3}, x_3) + d(x_3, q) < \frac{\delta}{2^3} + \frac{\delta}{2^3} = \frac{\delta}{2^2}$.

$\cdots$