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

---------------------------------------------------------------------------

Is that ok?

 

[Evgeny.Makarov]

Yes, this is fine.

 

[blue_raver22]

First, let us recall the definitions of supremum and infimum. The supremum of a set $S$ is the smallest number that is greater than or equal to every element in $S$. In other words, it is the least upper bound of $S$. On the other hand, the infimum of a set $S$ is the largest number that is less than or equal to every element in $S$. It is the greatest lower bound of $S$.

In this problem, we are given that $S$ is a non-empty bounded set, meaning that it has both an upper and lower bound. Let $\overline{m}=\sup S$. This means that $\overline{m}$ is the smallest number that is greater than or equal to every element in $S$. Now, we need to prove that $\inf \{\overline{m}-x: x\in S\}=0$.

To prove this, we will use the definition of infimum. We need to show that $0$ is the largest number that is less than or equal to every element in $\{\overline{m}-x: x\in S\}$. In other words, we need to show that $0$ is the greatest lower bound of $\{\overline{m}-x: x\in S\}$.

Let $y\in \{\overline{m}-x: x\in S\}$. This means that $y=\overline{m}-x$ for some $x\in S$. Since $\overline{m}$ is the smallest number that is greater than or equal to every element in $S$, we have $y=\overline{m}-x\leq \overline{m}$ for all $x\in S$. This implies that $y\leq \overline{m}$. Since $\overline{m}$ is the smallest upper bound of $S$, we also have $y\leq \overline{m}$ for all $y\in \{\overline{m}-x: x\in S\}$. This means that $0$ is indeed the greatest lower bound of $\{\overline{m}-x: x\in S\}$.

Therefore, we can conclude that $\inf \{\overline{m}-x: x\in S\}=0$. This proves the statement that $\inf \{\over