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I want to proof that if $M\subseteq \mathbb R$ is bounded, then so is $\overline{M}$ or more precise that if $s$ is the supremum of $M$, then it is the supremum of $\overline{M}$. I came up with a proof but I am not sure if it is correct:

Let $s := \sup M$. If $M$ is closed we are done, so suppose $M$ is not closed. Suppose $s$ was not the supremum of $\overline{M}$. Then $\exists x\in \overline{M}$ with $x>s$.

By the definition of the closure there exists a sequence $(a_n)_{n\in \mathbb N_0}$ in $M$ with $\lim_{n\to \infty} a_n = x$, i.e. for all $\epsilon> 0, |a_n -x| < \epsilon$ for all big enough $n\in \mathbb N_0$.

But that means $$-\epsilon < a_n -x < \epsilon \Longleftrightarrow x < a_n + \epsilon \leq s+\epsilon$$ This is a contradiction to $x> s$.

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    It's not clear to me why $a_n \leq s$. Couldn't we have a sequence $\{a_n\}$ approaching $x$ from above? In this case, we would have $a_n + \epsilon > s + \epsilon$. – 2017-02-13
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    Well, the sequence has only values in $M$, so all $a_n$ must be smaller or equal to the supremum, right? – 2017-02-13
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    @Daniel Xiang: The fact that $a_{n}\leq s$ for all $n$ is clear since $s=\sup M$ and $a_{n}\in M$ for all $n$. – 2017-02-13
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    wasn't the assumption that $s$ was not the supremum? – 2017-02-13
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    Not the supremum of $\overline{M}$, but it of course is still the supremum of $M$ and all $a_n$ are in $M$, only the limit is not. – 2017-02-13

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Very good. To finish off, you may want to say something like: "since $x\leq s+\epsilon$ for all $\epsilon>0$, it follows that $x\leq s$, a contradiction".

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    I will add that. Thanks! – 2017-02-13
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I think you show that $s$ is an upper bound of $\overline{M}$, but you don't actually show that it's the least upper bound of $\overline{M}$.

It should, however, be straightforward to show that for any upper bound $z$ of $\overline M$, $s \le z$. This should follow from the fact that any such $z$ will also be an upper bound of $M$.

You also might want to think carefully about whether you really needed a proof by contradiction here. (See, for example, Tim Gowers' thoughts on it here.) I don't think you do, and dropping the attempt entirely but maintaining the rest of the argument still works.

Consider this reframing of your proof (with some details you've supplied omitted):

Suppose $M$ is not closed, and consider $x \in \overline{M} \cap M^c$. There must then exist a sequence $\{ a_n \} \subset M$ such that $a_n \to x$. This means that for any $\varepsilon > 0$, $x < s + \varepsilon$, which implies that $x\le s$. $x$ was arbitrary, so $s$ is an upper bound of $\overline M$.

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    Thanks, I'll think about it. I only needed in upper bound in my proof, but knowing that I don't quite proof that it's the supremum too is good. – 2017-02-13
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    @lappen68 Yes, I did notice that you seemed to want to prove two different things in your post, one stronger than the other. I decided to discuss the stronger claim, as it does imply the weaker one (naturally). Glad that helped! – 2017-02-13
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I think I have a clear and simple solution using the definition of closure of M as $M\cup$ Boundary(M):

To show that it's bounded, try and latch onto this concept: suppose x is in M: i.e. let $x\in M$. Then x is in [inf(M), Sup(M)]. Now suppose x is in Boundary(M). Could we be so lucky as to have any boundary point x also be in [inf(M), Sup(M)]? That would be really easy to construct a proof out of. So suppose there was a boundary point b outside of this interval:

We'd have either $b<$inf(M) or $sup(M)

The formal proof would go something like:

Let x be in cl(M).

Then x is in $M\cup bd(M)$.

Since x is in M or x is in Bd(M), then....

and you'd proceed to consider each case in the concept above...ending with the threat of contradiction in case two to conclude that regardless of which case it is, x is in [inf(M),sup(M)] and thus cl(M) is bounded.

Let me know if you have any questions.

Adam V. Nease