This is related to a previous post of mine (link) regarding how to show that for any sequence $\{x_{n}\}$, the limit superior of the sequence, which is defined as $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}$, is equal to the to supremum of limit points of the sequence. Below I think I have found a counter-example to this (although I know I am wrong but just don't know where!).
Define the sequence $x_{n}=\sin(n)$. We have $\text{sup }_{k\geq n} x_{k}=1$ for any $k\geq 1$ (i.e. for any subsequence). Thus $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}=1$. Now the sequence has no limit points since it does not converge to anything so the supremum of all the subsequence limits is the supremum of the empty set which presumably is not equal to 1 (I actually do not know what it is). So we have a sequence where the supremum of limit points does not equal $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}=1$?
Any help with showing where this is wrong would be much appreciated.
Best Answer
The problem is that you’ve misunderstood the notion of limit point of a sequence. I prefer the term cluster point in this context, but by either name it’s a point $x$ such for any nbhd $U$ of $x$ and any $n\in\mathbb{N}$ there is an $m\ge n$ such that $x_n \in U$. Thus $1$ and $-1$ are both cluster points of the sequence $\langle (-1)^n:n\in\mathbb{N}\rangle$, though this sequence doesn’t converge to anything.
Equivalently $x$ is a cluster point of $\langle x_n:n\in\mathbb{N}\rangle$ iff for each nbhd $U$ of $x$, $\{n\in\mathbb{N}:x_n \in U\}$ is infinite. If the space is first countable, $x$ is a cluster point of $\langle x_n:n\in\mathbb{N}\rangle$ iff some subsequence of $\langle x_n:n\in\mathbb{N}\rangle$ converges to $x$.