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

Question

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:

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

Answer 1

"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 $x\in\overline{A}$.

Answer 2

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 $\max(\overline{A})$.