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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.

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  • $\begingroup$ By $\overline X$ you mean the closure of $X$? $\endgroup$
    – Reveillark
    May 18 '20 at 0:22
  • $\begingroup$ Yes the closure. sorry, I forgot to clarify $\endgroup$
    – User1618
    May 18 '20 at 0:23
  • $\begingroup$ 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 $X$. $\endgroup$
    – Joe
    May 18 '20 at 0:23
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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 $X$.

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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 $\sup\bar X\le \sup X$.

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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).

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