# Proof. Maximum of the closure is equal to the supremum of the set

The exercise is:

Show that if $$A \subset \mathbb{R}$$ is bounded and $$A \neq \varnothing$$ then $$sup(A)=max(\overline{A} ).$$

Now, I wanted to ask you whether my proof is watertight:

Let $$A \subset \mathbb{R}$$ be a non-empty and bounded set.

Then $$A$$ has a finite supremum $$sup(A) \equiv \widetilde{x}$$, which is the least upper bound on $$A$$.

Further define $$\overline{x} \equiv max(\overline{A} )$$.

Assume, for the sake of contradiction, that $$\overline{x} \neq \widetilde{x}$$, which implies that there exists a distance $$d(\overline{x}, \widetilde{x} ) \equiv \epsilon > 0$$. Given that $$\widetilde{x} \geq x, \forall x \in A,$$ we have that $$B_{\epsilon /2} (\overline{x}) \cap A = \varnothing$$.

This is a contradiction of the definition of closure.

Therefore, $$sup(A)=max(\overline{A} )$$.

"Further define $$\overline{x}=max(\overline{A})$$." This is not allowed, as you need to prove that this is well-defined. Not every set has a maximum. Since the standard proof of this is done by showing that it equals the supremum, you end up in a circular reasoning.
Some hints for how you should do this: Show that $$\tilde x:=\sup{A}\in\overline{A}$$ and that $$x\leq \tilde x$$ for all $$x\in\overline{A}$$.
First show that $$\sup(A)$$ (which exists by boundedness of $$A$$ and completeness of $$\mathbb{R}$$) is in $$\overline{A}$$. This follows from the definition of sup and the fact that we have the order topology on the reals. Then at least $$\sup(A)$$ is a candidate for being $$\max(\overline{A})$$.