Please check my answer.
……….Question………
Suppose $A \subseteq \mathbb{R}$ is closed and nonempty. Show that if $A$ is bounded above, then it contains its supremum, and if it is bounded below, then it contains its infimum.
……..My answer………..
If $sup(A) \notin A$, then it must be in the compliment of $A$, $sup(A) \in \mathbb{R}-A$. A ball around $sup(A)$ would then not be contained in the compliment of $A$. This implies that the compliment of A is not open, and thus A is not closed. RAA.
The same argument holds for $inf(A)$
Best Answer
Suppose $A \subset \Bbb{R}$, non-empty, bounded above, and closed. Then, $\sup(A)$ exists.
Suppose in order to derive a contradiction that $\sup(A) \notin A$.
Then, $\sup(A) \in \Bbb{R} \setminus A$.
Since $A$ is closed, $\Bbb{R} \setminus A$ is open.
Thus, $\Bbb{R} \setminus A$ entirely contains $B(\sup(A), \varepsilon)$, the ball of center $\sup(A)$ and of some radius $\varepsilon > 0$.
Then, $B(\sup(A), \varepsilon)$ contains the number $\sup(A) -\frac{\varepsilon}{2}$.
Since $\sup(A) - \frac{\varepsilon}{2} \in B(\sup(A), \varepsilon)$, $\sup(A) - \frac{\varepsilon}{2} \geq x, \forall x \in A$.
But $\sup(A) - \frac{\varepsilon}{2} < \sup(A)$.
This contradicts our assumption that $\sup(A)$ was the least upper bound.
So, we must have $\sup(A) \in A$.