It is a proof dealing with sequence/ subsequence and infimum. Question: Suppose that $x_{n}$ is a bounded sequence and $x_{n_k}$ is a sub sequence. Prove that $\lim_{n\to\infty} \inf(x_n) \leq \lim_{k\to\infty} \inf(x_{n_k})$.
What I know: If $x_{n}$ is a convergent sequence then any sub sequence $x_{n_k}$ is also convergent and $\lim_{n\to\infty} x_n = \lim_{k\to\infty} x_{n_k}$. So that means that for every $\epsilon>0$ we have an $M\in\mathbb{N}$ such that for all $n\ge M$ I get $|x_{n}-x|<\epsilon$. So how do I even start to put this together to prove the above?
 A: You don't need to work with convergence at all (and I'm not sure how you would). Here is a quick proof. Feel free to ask if anything is unclear.
Suppose for the sake of argument that $\liminf_{k\to\infty}X_{n_{k}}<\liminf_{n\to\infty}X_{n}$. Write $A$ for $\liminf_{k\to\infty}X_{n_{k}}$ and $B$ for $\liminf_{n\to\infty}X_{n}$. Let $\epsilon=B-A$. By definition, for every $K\in\mathbb{N}$ there exists $k\geq K$ such that $X_{n_{k}}\in[A,A+\epsilon/2)$. But this tells us that for every $N\in\mathbb{N}$ there exists $n\geq N$ such that $X_{n}<B-\epsilon/2$, and this contradicts $B$ being the limit inferior of $\{X_{n}\}_{n\in\mathbb{N}}$. It will be very intuitive/easy once you get it.
A: By definition, you have
$$
\liminf_{n\to\infty} (x_n) = x \iff \forall\epsilon\exists N\, (n>N \Rightarrow x_n > x-\epsilon)
$$
You can use this plain definition on the subsequence, and prove that if $\liminf_{k\to\infty} (x_{n_k}) < x$ then the $\liminf x_n$ couldn't be $x$ in the first place.
A: By definition, the liminf of a sequence is just the infimum of its subsequential limit set (the collection of subsequential limits). So, consider the subsequential limit sets of the two sequences. The subsequential limit set of  $\left\{ { a }_{ n } \right\} $ contains the subsequential limit set of  $ \left\{ { a }_{ { n }_{ k } } \right\} $. Thus the inequality is automatically true.
