Can anyone please explain the theorem/proof in the picture attached? There is a theorem that states that if $X$ converges to $x$, then all subsequences of $X$ converge and converge to $x$. However, in this theorem, it is just considering the subsequences that are convergent. Moreover, I fail to understand the last two lines of the proof, can anyone explain those, maybe by breaking it down a bit?
Thank you.
On
(1).Let $S$ be an infinite subset of $\Bbb N .$ Then $(y_n)_{n\in S}$ converges to $y$ iff, whenever $J$ is an open interval containing $y,$ the set $\{n\in S:y_n\not \in J\}$ is finite.
(2). Suppose $A=(a_n)_{n\in \Bbb N}$ is bounded and does not converge to $x.$ Take $r,s>0$ such that the set $U=\{n\in \Bbb N: a_n\not \in (-s+x,r+x)\}$ is infinite.
Then $B=(a_n)_{n\in U}$ is a subsequence of $A$ and is a bounded sequence, so $B$ has a convergent subsequence $C=(a_n)_{n\in V}$ (where $V$ is some infinite subset of $U$) with $C$ converging to $y.$ But since $a_n\not \in (-s+x,r+s)$ for every $n\in V,$ we have $y\not \in (-s+x,r+x),$ so $y\ne x.$
So if the bounded sequence $A$ does not converge to $x$ then $A$ has a subsequence converging to some $y\ne x.$
Remarks. I think that is a better way to state the result, as the usual def'n of a subsequence allows $A$ to be a subsequence of $A.$ Consider any sequence $X$ to be some (any) function $f$ whose domain is an infinite subset of $\Bbb N,$ and a subsequence of $X$ to be $f$ restricted to some infinite subset of the domain of $f$.
You have that every bounded sequence has a convergent subsequence by Bolzano Weirstrass. Let's call the limit $x$. The contradiction from the theorem just formalizes the idea that, in fact, the whole sequence can be thought as a subsequence itself, so it has to converge to the same place. If not, you then are able to find a subsequence that diverges from the original limit where all the sequences are going.