$\inf X = \inf\overline{X}$ and $\sup X = \sup\overline{X}$ 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.
 A: 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 $X$.
A: A sequential proof that $\sup\bar X\le\sup X$:
Suppose $x\in \bar X$. Then there is a sequence $(x_n)$ of points from $X$ such that $x_n\to x$. Since each $x_n\in X$, by definition of sup we have
$x_n\le \sup X$
for every $n$. But the inequality is preserved in the limit, so
$x\le \sup X$.
Since this holds for every $x$, this means that $\sup X$ is an upper bound for $\bar X$. Since the sup of a set is the least upper bound, conclude
$\sup\bar X\le \sup X$.
A: The key part to understand here is that the numbers $\inf X$ and $\sup X$ are members of $\overline{X} $.
Further these numbers are the minimum and maximum elements of $\overline {X} $ so that $$\sup\overline{X} =\max\overline{X} =\sup X$$ and $$\inf\overline {X} =\min\overline{X} =\inf X$$ Thus the problem boils down to the proof of those two obvious properties given by the italicized statements above. I hope you can prove these statements easily (they are almost self-evident). 
