Closure containing the supremum and infimum

My classmate of mine did the following. I cannot help but feel that isn't what you are supposd to be at all.

Classmate work

Since A is bounded, both infA and supA are bounded

To show $inf(A) \in \bar{A}$, for every n, $\exists x_n\in A$ s.t

$$\inf A \leq x_n \leq \inf(A)+\frac{1}{n}$$

Then $(x_n)_{n\mathbb{N}}$ is a sequence in A with $x_n \to infA$

Hence $\inf A \in \bar{A}$ is a limit point of A

The show that supA is also in $\bar{A}$, the proof is similar.

I am guessing this person took the sequence that is in A and wrote out how the sequence is bounded. Then he assumed (he didn't really write this) that the sequence is decreasing (then increasing for the sup case) and it converges.

Why does the sequence having needed to converge is necessary here?

I don't have a correct proof, but the proof idea I think is we should do something really simple like

Assume $A$ is a bounded subset of $\mathbb{R}$, that is $\exists B > 0 \in \mathbb{R}$ s.t. $|A| \leq B$ and in particular let $\sup A = \text{least upper bound}$ and $\inf A = \text{greatest lower bound}$ and the closure is the set that is the smallest set that contains the limit point and everything in $A$. And then somehow we relate sup and inf to "B" and show that it is in the closure

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– wj32 Oct 25 '12 at 3:16
Oh thanks. Does a bounded set imply it is closed? – Hawk Oct 25 '12 at 3:21
No, it doesn't. I was just trying to point out a related question. – wj32 Oct 25 '12 at 3:22
Oh yeah, that too. I am not sure if my comment will get to anyone on that question though. I am not sure how one of the answer manages to construct an increasing sequence though. Is this always possible? – Hawk Oct 25 '12 at 3:28
Since bounded there is an inf $p$ and a sup $q$. Deal with sup. If $q$ is in $A$ we are finished. Otherwise, for any $n$ there is an $a_n\in A$ such that $q-1/n\lt a_n\lt q$. Thus $q$ is a limit point of $A$. Basically your friend's proof. – André Nicolas Oct 25 '12 at 3:31

Your classmate's proof looks OK. The fact that $x_n \rightarrow \inf A$ doesn't come from the monotone convergence theorem. As you noted, the sequence may not be monotonically increasing/decreasing. Instead, to see why it converges you should rewrite the inequality as $$0 \le x_n - \inf A \le \frac{1}{n}.$$
As for your idea - I don't think it really works. I could pick the bound $B$ to be extremely large, and this would have almost nothing to do with the $\sup$.
Here's another proof of the theorem. Assume $\inf A \notin A$, for otherwise $\inf A \in \overline{A}$ already. If $N$ is an open ball of radius $r$ around $\inf A$ then there exists a point $x \in A$ with $\inf A < x < \inf A + r$, for otherwise $\inf A + r$ would be a lower bound of $A$. So $x \in N$, which proves that $\inf A$ is a limit point of $A$.