Prove $\inf \{\overline{m}-x: x\in S\}=0$

[alexmahone]

Let $S$ be a non-empty bounded set of real numbers, and $\overline{m}=\sup S$. Prove that $\inf \{\overline{m}-x: x\in S\}=0$.

[Use only the definitions of supremum and infimum, and not identities like $\inf(A+B)=\inf A+\inf B$ and $\inf(-S)=-\sup(S)$.]

 

[Evgeny.Makarov]

Show that 0 is a lower bound of $\{\overline{m}-x\mid x\in S\}$ and that any positive number is not a lower bound.

 

[alexmahone]

Show that 0 is a lower bound of $\{\overline{m}-x\mid x\in S\}$ and that any positive number is not a lower bound.

$\overline{m}=\sup S$

$x\le\overline{m}$ for all $x\in S$.

$\overline{m}-x\ge 0$ for all $x\in S$.

So, 0 is a lower bound of $\{\overline{m}-x: x\in S\}$. ------ (1)

Assume, for the sake of argument, that $a>0$ is a lower bound of $\{\overline{m}-x: x\in S\}$.

$\overline{m}-x\ge a$ for all $x\in S$.

$x\le\overline{m}-a$

This contradicts the fact that $\overline{m}$ is the least upper bound of $S$.

So, any $a>0$ is not a lower bound of $\{\overline{m}-x: x\in S\}$.

Together with (1), this implies that $\inf\{\overline{m}-x: x\in S\}=0$.

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Is that ok?

 

[Evgeny.Makarov]

Yes, this is fine.