Verification - does $x < \sup A$ necessarily mean $x \in A$? Suppose we have a non-empty and bounded above set $A$, and some $x \in \mathbb{R}$ such that $x < \sup A$.
Do we then have that $x \in A$?
Since we assume $\sup A$ exists, it follows that $A \subset \mathbb{R}$.
If we take $x = \sqrt{2}$ and define $A = \{...,-3,-2,-1,0,1,2\}$
Then $A$ is non-empty, bounded above and has $\sup A = 2$. Furthermore $x < \sup A$, but $x \notin A$.
I know this seems rather elementary, but I just wanted to make sure there was absolutely no flaw in my example.
Is there an even simpler case for which you could prove this?
thanks.

There is no problem in your example. In fact you don't need to consider an infinite set as an example. You can just consider any finite set and $x < \min(A) \leq \max(A)$ but $x \notin A$.
