The exact problem goes like:
Let some $A$ be a bounded and non-empty set. There is a sequence ${x}$ which converges to some point $x'$ in $A$ and I have to prove that $x'$ is the max($A$) given that the sequence ${x}$ > sup($A$)
My work: Since ${x}$ converges to some element $x'$ that belongs in $A$, we can say that for for all $r > 0$, there exists some $n>N$ s.t. $d(x',x) < r$. Let $a = $sup($A$) and by the definition of supremum, $\forall a' \in A, a-r<a'$. Since $x' \in A$, it implies that $a - r < x'$. We know that $a$ is the lowest upper bound but should I also try and approach it via contradiction? Take some element $x''\in A$ that is max($A$).
I would really appreciate some help because I feel like I'm so close but I'm stuck.
By definition, your sequence $x_n$ satisfies $x_n>\sup(A)$ for all $n$. Taking the limit, you obtain $x'\ge \sup(A)$. This means that $x'\ge y$ for all $y\in A$. As $x'\in A$, then $x'=\max A$.