Proof that convergent, increasing sequence's limit = supremum I am trying to prove that if $\{a_n\}$ is an increasing (non-decreasing) sequence with $a_n\to a$, for $n \to \infty$ then $\{a_n\}$ is bounded above and $a=\text{sup}\{a_n:n \in \mathbb{N}\}$.
my proof:
By convergence, for any $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that $|a_n-a|<\varepsilon$ for $n\geq N$.  Let $a_k>a$ for some $k \in \mathbb{N}$.  Because a limit of a covergent sequence is unique, and $a_k$ is not the limit, there does not exist an $m \in \mathbb{N}$  such that, for all $n\geq m$, $|a_n-a_k|<\varepsilon$, some $\varepsilon>0$.  Thus $|a_n-a_k|\geq\varepsilon$.  So $a_k$ cannot be the supremum of $\{a_n:n \in \mathbb{N}\}$.  Thus, $a=\text{sup}\{a_n:n \in \mathbb{N}\}$.  By definition of the supremum, for all $n \in \mathbb{N}$ $an\leq a$, so $\{a_n:n \in \mathbb{N}\}$ is bounded above.
Is this on the right track?  Any comments/improvements are appreciated.  
 A: 
By convergence, for any $\varepsilon > 0$, there exists $N \in \mathbb N$ such that $|a_n - a| < \varepsilon$ for $n \ge N$.

Correct.

Let $a_k > a$ for some $k \in \mathbb N$.

I would normally use the word "suppose" here rather than "let", because you are not saying that such an $a_k$ exists, you are supposing it exists and aiming for a contradiction to show that it cannot exist (which will then imply that $a$ bounds $\{a_n : n \in \mathbb N\}$ above).

Because a limit of a covergent sequence is unique, and $a_k$ is not the limit, there does not exist an $m \in \mathbb N$ such that, for all $n \ge m$, $|a_n - a_k| < \varepsilon$, some $\varepsilon > 0$. Thus $|a_n - a_k| \ge \varepsilon$.

The way you're phrasing this makes it a bit unclear what you're saying. What you want to say is "$a_k$ is not the limit, so there is a $\varepsilon > 0$ such that for every $m \in \mathbb N$, there is an $n \ge m$ with $|a_n - a_k| \ge \varepsilon$." Using logical notation makes it clear what's going on here. To say that $a_k$ is the limit, you'd write
$$ (\forall \varepsilon > 0) (\exists m \in \mathbb N) (\forall n \ge m) (|a_n - a_k| < \varepsilon) $$
To negate this statement, you go along the initial sequence of $\forall$ and $\exists$ clauses, replacing $\forall$ with $\exists$ and $\exists$ with $\forall$; then you negate the statement at the end after all these quantifiers:
$$ (\exists \varepsilon > 0) (\forall m \in \mathbb N) (\exists n \ge m) (|a_n - a_k| \ge \varepsilon) $$

So $a_k$ cannot be the supremum of $\{a_n : n \in \mathbb N\}$.

Yes, that makes sense: to be the supremum $a_k$ would have to have members of that set as close as you like to that, but we've shown that it's at least $\varepsilon$ away from all of them.

Thus, $a = \sup \{a_n : n \in \mathbb N\}$.

This is too strong a conclusion at this point. All you've shown is that if you assume $a_k > a$, then $a_k$ is not the supremum of $\{a_n : n \in \mathbb N\}$. So all this tells us about the supremum of $\{a_n : n \in \mathbb N\}$ is that it's not a term of that sequence which is greater than $a$. It could still be a term of that sequence which is less than or equal to $a$, and it could still be not in the sequence and greater than $a$. Remember, suprema are not, in general, greatest elements: the supremum of a set will not necessarily be in the set. 
Now, we do know that the sequence is increasing, so if the supremum was a term of the sequence, then every term after it would be equal to the supremum, and hence the sequence would be eventually constant at the supremum, which would mean its limit was the supremum. But you should probably explain this; it's a missing step.
And it's not obvious to me, if I'm already doubting the theorem in question, that the supremum can't be greater than the limit of the sequence while not being a term of the sequence. So that step definitely needs expanding.
There's another missing step, too: how do you know a supremum even exists? A set of real numbers can have no supremum, if it is not bounded above. Before you can talk about the supremum, you must prove that the set is bounded above.

By definition of the supremum, for all $n \in \mathbb N$ $a_n \le a$, so $\{a_n : n \in \mathbb N\}$ is bounded above.

The fact that this was so easy could be seen as a clue that something went wrong here. Yes, of course any set with a supremum is bounded above. A supremum is a kind of upper bound. If it was possible to prove the existence of the supremum before proving boundedness, the question wouldn't have said "prove this set is bounded above, AND that its supremum is this", it would have just said "prove that the supremum is this".
