# set theory (concept of infinity)

Let $S$ be a set. $f$ be a function on $S$ into real line $\mathbb{R}$. Let us define $Af$ as a function from $\mathbb{R}$ into power set of $S$, $PS$ $Af(x) = \{s\mid f(s) \leq x\}$

Question: Is $S$ in the range of $Af$?

(as an example we can chose $S=\mathbb{R}$ and $f(x) = x$)

Thanks PS: I'm having some technical difficulties in proving few results. It boils down to above. (and related in infinity)

• What do we know about $S$, is it finite or something? – user302982 Dec 4 '16 at 17:03
• for finite it is trvial. im unable to understand when S is infinite – aman_cc Dec 4 '16 at 17:09

$S$ is in the range of $Af$ iff $f$ is bounded from above.
If $f$ isn't bounded from above, there is for every $x\in \mathbb{R}$ a $s_x\in S$ with $f(s_x)>x$ and so $s_x\notin Af(x)$ and so $S\neq Af(x)$ which means $S$ isn't in the range.
If $f$ is bounded from above, there is a $x\in \mathbb{R}$ with $f(s)\leq x$ for all $s\in S$ and so $S=Af(x)$ which means $S$ is in the range.
• +1. A concrete example which might help the OP: let $f(x)=x$, and let $S=\mathbb{R}$. Then $Af(x)=\{s: s\le x\}$, and this is never all of $\mathbb{R}$, since $\mathbb{R}$ has no maximal element. By contrast, if we took $S=(0, 1)$, then $Af(1)=S$. So this is what boundedness gets us. – Noah Schweber Dec 4 '16 at 17:26