Why there exists a subsequence $\{x_{n_i}\}$ of $\{x_{n}\}$ and $x\in X$ such that $ x_{n_i}\underset{i}{\to} x $

Question

Let $X$ be a Banach space and $f:X\to [0,+\infty]$ be sequencially inf-compact function (for all $r\in\mathbb{R}$ the set $\{ x\in X: f(x)\leq r\}$ is sequencially compact) . Let $\{x_n\}$ be a sequence such that: $$ \limsup_nf(x_n)<\infty $$ Why there exists a subsequence $\{x_{n_i}\}$ of $\{x_{n}\}$ and $x\in X$ such that $$ x_{n_i}\underset{i}{\to} x $$

Answer 1

If $M =\lim \sup f(x_n)$ then there exist $n_0$ such that $f(x_n) \leq M+1$ for all $n \geq n_0$. Since $\{x_{n_0}, x_{n_0+1},...\}$ is sequence in the sequentially compact set $\{x: f(x) \leq M+1\}$ it has a convergent subsequence. This subsequence is also a subsequence of $(x_n)$.