Let $X \subseteq R$ a bounded set.
Prove $\inf X = \inf\overline{X}$ and $\sup X = \sup\overline{X}$.
I don´t know how to prove these two statements. I already proved that $A \subseteq B \implies \inf B \leq \inf A$ and $\sup B \geq \sup A$, so I already have $\inf \overline{X} \leq \inf X$ and $\sup \overline{X} \geq \sup X$.
But I don´t know how to prove $\inf X \leq \inf \overline{X}$ or $\sup\overline{X} \leq \sup X$ to get the equalities by antissimetry.
Any other way to prove the two statements would be accepeted as well.
$\overline{X}$ is the closure of $X$.
Thanks.
Best Answer
Let's show that $\sup \overline X\le\sup X$. Suppose otherwise. Then $\sup \overline X> \sup X$. This means that $\sup X$ is not an upper bound of $\overline X$, so there is some $z\in \overline X$ such that $$ \sup X <z\le \sup \overline X $$ Pick $r>0$ such that $\sup X<z-r$. As $z\in\overline X$, $X\cap (z-r,z+r)\neq\emptyset$. Fix $x\in X\cap (z-r,z+r)$. Then $x>z-r>\sup X$, which is impossible because $\sup X$ is an upper bound of $X$.