Show the set of the limits of all the subsequence of a bounded set contains sup and inf. Let $(X_n)$ be a bounded sequence, and let $E$ be the set of subsequential limits of $(X_n)$. Prove that $E$ is bounded and contains $\sup E$ and $\inf E$.
Does this ask us to prove limsup and liminf exists? 
Could you help me ?
 A: Since $(x_n)$ is a bounded sequence by the Bolzano–Weierstrass Theorem contain a convergent subsequence, then $E$ is non-empty. 
Now we have to show that $E$ is bounded. Let $M$ be a bound for the entire sequence, i.e., $|x_n|< M$ for all $n\in \mathbb{N}$, and suppose for sake of contradiction that $E$ is unbounded. Then we must have $|x|>M$ for all but finitely many $x\in E$. Choose one, say $x_0\in E$ and $|x_0|>M$, then  since $x_0\in E$ and by definition of $E$, we can find a subsequence $(x_{n_i})$ which converges to $x_0$ and in particular for $i\ge n_0$ we have $|x_{n_i}-x_0|< |x_0|-M$, so $|x_{n_i}|\ge |x_0|-|x_{n_i}-x_0|>M$, a contradiction. Hence $E$ is non-empty and bounded.
Let $s= \sup E$, we shall show that lies in $E$. Only we check the claim for the least upper bound, the other part is similar. 
First we will  show that is a point of accumulation (a limit point). Given $\varepsilon>0$ and $n_0\ge 0$. Then $s-\varepsilon/2$ is not the least upper bound and so there is $x\in E$ such that $s-\varepsilon/2<x\le s$, but since $x\in E$, we have  $|x_{n_i}-x|< \varepsilon/2$ for some $n_i\ge i\ge n_0$. Thus $|x_{n_i}-s|\le |x_{n_i}-x|+|x-s|< \varepsilon$. Hence $s$ is a limit point. 
Define recursively the sequence: $n_0=0$ and $n_k = \min  \{n>n_{k-1}:|x_n-s|<1/k \}$. Notice that since $s$ is a limit point $\{n>n_{k-1}:|x_n-s|<1/k \} \not= \varnothing$ for all $k$, since otherwise contradicts what we have shown in the above paragraph. Then $(x_{n_k})$ is a subsequence of $x_n$, and $s-1/k<x_{n_k}<s+1/k$, by the squeeze theorem we conclude $x_{n_k} \to s$. Thus $s\in E$.
as was to be shown. 
A: No you are not required to prove the existence of the limsup or liminf. You are required to prove that $E$ is bounded and the supremum and infimum of the set $E \subseteq \Bbb R$ is contained in $E$. Here is my humble suggestion:
Suppose $x \in E$. Then $x = \lim (a_n)$ where $(a_n)$ is a subsequence of $(x_n)$. $x \gt x_n$ for every $n \in \Bbb N \implies x \gt a_n \;\; \forall n \in \Bbb N$. Then there is a neighbourhood of $x$ which contains no elements of the sequence $(a_n)$ leading to a contradiction. Therefore $x \le x_m$ for some $m \in \Bbb N$. A similar proof shows that $x_r \le x$ for some $r \in \Bbb N$. Therefore $|x| \le |x_t|$ for some $t \in \Bbb N$. But since $(x_n)$ is a bounded sequence there is a positive real number $M$ such that $|x_n| \le M$ for every $n \in \Bbb N$, $|x| \le M$ for every $x \in E$. This proves that $E$ is bounded. 
Now, $\sup E$ and $\inf E$ exist since $E$ is bounded and non-empty. Let Consider the sequence $(r_n)$ such that $r_n \in \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 n\} \cap (x_n)$ for each $n \in \Bbb N$ such that $r_n \neq r_i$ for $i \in \{1,2,.. n - 1\}$. Suppose $r_m$ does not exist for some $n \in \Bbb N$. Then either there are no elements of $(x_n)$ in $ \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies $ No subsequence in $(x_n)$ can converge to a limit in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies c - \frac 1 m$ is an upper bound for $E$ leading to a contradiction. 
OR
For some $m \in \Bbb N$ Every element in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \cap (x_n)$ is in $\{r_1, r_2,.. r_{m - 1}\}$. Let $M = \text{Max}  \{r_1, r_2,.. r_{m - 1} \}$. If $M \lt c$ then picking $ l = \frac {c - M} 2 \gt 0$ we have that $c - l$ is an upper bound for $E$ leading to a contradiction. If $M \ge c$ then either the elements in $\{r_1, r_2,.. r_{m - 1}\}$ form a subsequence of $(x_n)$ or they do not. If they do then their limit is either equal to $c$ or is greater than it. If they do not form a subsequence of $(x_n)$ they are irrelevant. 
This is the best I could do. I apologise for the immense number of cases and for clustered language.  
The proof for $\inf E \in E$ is similar. 
Hope this helped. 
