prove the limit inferior of $(x_n)$ where $n \in\mathbb{N}$ The problem states let $(x_n)$ be a bounded sequence for each $n \in\mathbb{N}$. Let $t_n=inf\{x_k: k\geq n\}$. Prove that $(t_n)$ is monotone and convergent.
After a little research because I was confused what limit inferior really was this was the proof I tried coming up with. Proof: Since $(x_n)$ is bounded then it has a supremum and info um. Let $t_n=inf\{x_n: k\geq n\}=inf\{x_n,x_{n+1},...\}$. It follows that $t_{n+1}=inf\{x_{n+1}: k\geq n+1\}=inf\{x_{n+1},x_{n+2},...\}$ If we let set $A= \{x_{n+1},...\}$ and set $B=\{x_n,...\}$ (Note* I assume that set A and B have greatest lower bounds since these are subsets of $(x_n)$ and $(x_n)$ is bounded). Then it follows that $A \subset B$ and $inf A\geq inf B$ Otherwise if $inf A<inf B$ then $inf A \leq x_n$ a contradiction since $x_n \notin B$.  It follows that $x_{n+1} \geq x_n$ otherwise if $x_n> x_{n+1}$ then $inf A \leq x_n$ a contradiction from above. Hence $t_n$ is increasing and by the Monotone convergence theorem $t_n$ is monotone and convergent. Not sure If I have all this correct.  

It is well known that an increasing sequence bounded above converges. 
Since $(x_n)$ is bounded above $(t_n)$, always being a lower bound of some subset of $\{x_n : n \in \mathbb{N}\}$, is also bounded above.
Let $A_n = \{x_k : k \ge n\}$.  We have $A_{n+1} \subseteq A_n$, and so for any $x \in A_{n+1}$, $t_n \le x$.  Therefore,  $t_n$ is a lower bound of $A_{n+1}$, so by definition,
$$t_n \le \inf A_{n+1} = t_{n+1}$$
