liminf and limsup of a sequence (II) I came across the following problem about limit supremums and infimums of sequences:

Let $(a_n)$ be a bounded sequence in $\mathbb{R}$.

*

*Prove that $b = \text{lim sup} \ a_n \implies (\forall \epsilon >0) \ a_n < b+\epsilon$ ultimately and $a_n > b-\epsilon$ frequently.


*Show that the condition first condition characterizes the limit superior, in the sense that if $b \in \mathbb{R}$ satisfies the condition then necessarily $b = \text{lim sup} \ a_n$.

Here is my attempted solution:

*

*Proof. Let $(a_n)$ is a bounded sequence in $\mathbb{R}$. Suppose $b = \text{lim sup} \ a_n$. By definition, $$b = \text{lim sup} \ a_n = \inf\limits_{n \geq 1} \left(\sup\limits_{k \geq n} \ a_k \right)$$ Let $A = \left\{\sup\limits_{k \geq n} \ a_k \right\}$. Then by definition of infimum, $b+ \epsilon >a_n$ for every $a_n \in A$. So $(\exists N) \ \ni n \geq N \implies a_n < b+ \epsilon$. Thus $a_n < b+ \epsilon$ frequently. Since $b$ is a lower bound of $A$, $b \leq a_n$ for all $n$. Thus $b- \epsilon < a_n$ for all $n$. Hence $a_n > b- \epsilon$ frequently.


*Proof. Suppose $(\forall \epsilon >0) \ a_n < b+ \epsilon$ ultimately and $a_n > b- \epsilon$ frequently. Then $b \leq \text{lim sup} \ a_n$ and $b \geq \text{lim sup} \ a_n$. Hence $b = \text{lim sup} \ a_n$ by antisymmetry.
Are these on the right track?
 A: As soon as you write "Then by the definition of infimum", you are saying incorrect things.
First: if $b$ is the infimum of $A$, then for every $\epsilon\gt 0$ there exists at least one $a\in A$ such that $a\lt b+\epsilon$. However, you claim this holds for all elements of $A$, which is false. Take $A = (0,1)$. Then $\inf A = 0$; and it is indeed true that for every $\epsilon\gt0$ there exists at least one $a\in A$ with $a\lt\epsilon$; but if $\epsilon=\frac{1}{10}$, it is certainly false that every $a\in A$ is smaller than $\epsilon$.
Second: the elements of $A$ are not $a_n$s! They are suprema of infinite sequences of $a_n$s, and as such cannot be assumed to be $a_n$s.  For example, if $a_n = 1-\frac{1}{n}$, then $A=\{1\}$, and no $a_n$ is equal to any element of $A$. 
So that sentence is not just wrong, it's doubly wrong. The rest of course is now nonsense.
The second part does not seem to be proving anything; you are just asserting things. Why do the conditions imply the inequalities? What properties of the limit superior are you using? It's a mystery.
Rather: let $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$. Prove that $A_n$ is a decreasing sequence: $A_{n+1}\leq A_n$ for all $n\in \mathbb{N}$. Your set $A$ is precisely the set of $A_n$s. 
Now let $b= \inf A = \inf\{ A_n\}$. By the definition of infimum, for every $\epsilon\gt 0$ there exists $N$ such that $b\leq A_N\lt b+\epsilon$. Since the sequence of $A_n$s is decreasing, then  for all $n\geq N$ we have $b\leq A_n\leq A_N\lt b+\epsilon$, so in fact we have that $A_n\lt b+\epsilon$ ultimately. Moreover, since $a_n\leq A_m$ for all $n\geq m$, this implies that $a_n\lt b+\epsilon$ ultimately, as required. 
For the second clause of the first part, let $\epsilon\gt 0$. Then $b-\epsilon\lt A_n$ for all $n$. Now remember what $A_n$ is.  $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$; since $b-\epsilon\lt A_n$, there exists $k\geq n$ such that $b-\epsilon\lt a_k\leq A_n$. That is: for all $n$, there exists $k\geq n$ such that $a_k\gt b-\epsilon$, so $a_k\gt b-\epsilon$ frequently.
For part (2), let $b$ be a real number that satisfies the given properties. Since for every $\epsilon\gt 0$ we have that $b-\epsilon \lt a_n$ frequently, that means that $b-\epsilon$ is not an upper bound for $\{a_k\mid k\geq n\}$ for any $n$. Therefore, $\sup\{a_k\mid k\geq n\} = A_n\gt b-\epsilon$. This holds for all $A_n$, so $\liminf a_n = \inf\{A_n\mid n\in\mathbb{N}\} \geq b-\epsilon$. This holds for all $\epsilon\gt 0$, so $\liminf a_n \geq b$.
Now, since $b+\epsilon\gt a_n$ ultimately, then $b+\epsilon$ is an upper bound for $A_m$ all sufficiently large $m$; since the $A_m$ are decreasing, that means that $\inf A_m \lt b+\epsilon$, hence $\limsup a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\limsup a_n \leq b$.
Now that we have established the inequalities (rather than merely asserting them), we have $b\leq \limsup a_n \leq b$, hence $b=\limsup a_n$, as claimed. QED
