proof understanding : $\lim x_{n_k} = \limsup x_n$

Question

It is a bit hard for me to understand the proof.

1. In the first yellow line, does this mean that we make a subsequence $\{x_{n_1},x_{n_2},....,x_{n_k}\}$ = $\{x_1,x_2,....x_k\}$ ??

2. I don't understand why $a_{(n_k+1)}=$sup $\{x_n:n\ge n_k+1\}$

Thank you in advance!

Answer 1

Here is a different proof.

Note that there exists a subsequence of $(x_n)_n$ that converges to $x$ iff whenever $J$ is an open interval containing $x,$ the set $\{n: x_n\in J\}$ is infinite.

Suppose by contradiction that no subsequence of $(x_n)_n$ converges to $x.$ Then let $r,s>0$ such that $\{n: x_n\in (-s+x,r+x)\}$ is finite.

Now the set $\{n:x_n\geq r+x\}$ is also finite. Otherwise $a_m=\sup_{n\geq m}x_n> r+x $ for all $m,$ implying $x=\lim_{m\to \infty}a_m\geq r+x>x,$ which is absurd.

So $x_n\leq -s+x$ for all but finitely many $x.$ This implies that for some $M\in \Bbb N$ we have $n\geq M\implies x_n\leq-s+x,$ and hence we have $n\geq M\implies a_n\leq -s+x,$ but this implies $x=\lim_{n\to \infty}a_n\leq -s+x<x, $ which is absurd.

Answer 2

Point 1: No, it doesn't -- it just means that $n_1 = 1$. We could have $n_2 = 30$, and $n_3 = 25295$, for example. We just have to pick the $n_k$ so that $x_{n_{k+1}}$ gets very close to $a_{n_k+1}$.

Point 2: the author's notation is a bit annoying here. In the first line of the proof, $a_n$ is defined as

$$ a_n = \sup\{x_k:k \ge n\} $$

Now swap the symbol "$n$" with "$n_k + 1$" and the symbol "$k$" with "$n$". This gives us

$$ a_{n_k+1} = \sup\{x_n:n\ge n_k+1\} $$

and for some reason the author puts parentheses on the subscript of $a_{n_k+1}$... Clearly they're equivalent, though. In other words, we're just letting $n := n_k+1$ and using the name "$n$" instead of "$k$" in the expression above so that we aren't confusing it with the $k$ in $n_k$ (which is something completely different).