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)

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

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    $\begingroup$ +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. $\endgroup$ – Noah Schweber Dec 4 '16 at 17:26
  • $\begingroup$ thanks. i was trying to define probability spaces starting from cdfs and random variables. above puts that in risk as in some cases i will not have sample space in the sigma algebra. so seems i cant go much further. thanks $\endgroup$ – aman_cc Dec 4 '16 at 17:35
  • $\begingroup$ but if i assume /infinity/ belongs to R then I can get over this issue. Because then Af (infinity) = S and hence S will be in the image of Af. Am I going in right direction plz? $\endgroup$ – aman_cc Dec 5 '16 at 2:01

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