# $\inf X = \inf\overline{X}$ 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.

• By $\overline X$ you mean the closure of $X$? May 18 '20 at 0:22
• Yes the closure. sorry, I forgot to clarify May 18 '20 at 0:23
• If $\inf X$ is not an element of $X$, then you can show that there is a sequence in $X$ that converges to it, so it’s a limit point, and hence in the closure of $X$, and show that no value less than $\inf X$ is a limit point of $X$.
– Joe
May 18 '20 at 0:23

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 $$\sup X Pick $$r>0$$ such that $$\sup X. 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 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 $$\sup\bar X\le \sup X$$.
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).