The question is as follows:
Let $A$ be an open subset of $\mathbb{R}$ with $x \in A.$ Prove that if $B \equiv \left \{ b \in (x, \infty) \mid [x,b) \subset A \right \}$ is bounded, then $\sup B \notin A.$
I thought this question seemed difficult initially, but now I think it is quite trivial.
Clearly, $x<b.$
We know that the set $A$ has a supremum since it is nonempty and bounded. If we take $x \in A$ to be very close to $\sup A$, the number $b$ can only be ever be as big as $\sup A.$ If $b = \sup A + \epsilon$ for any $\epsilon >0$, then $[x,b)$ would not be a subset of $A.$
Therefore, $\max{b}=\sup A$, which implies that $\sup A = \sup B.$ Since $A$ is open, $\sup A \notin A$. Thus we can conclude that $\sup B \notin A.$ $\blacksquare$
I think I am looking at the question incorrectly. Would anyone know how to complete this proof?
Thanks in advance!