Let $S$ be a set. How is $f(S) = S$ different from $f(s) = s$ for all $S$?

Question

Usually, $f$ denotes a function, $f(x)$ is an image of $x$ under $f$. But what's $f(X)$ if $X$ is a set?

edit: Please, disregard the body of this question. I had to put something here to be able to post the question.

Answer 1

Say $S=\{1,2\}$ and $f(1)=2, f(2)=1$. Then $f(S)=S$, but $f$ is not the identity. The big letter means all of the things, the small letter means bit by bit. Another good way to think of it is in analogy with $1+2+3=2+2+2$ the sums are the same (large structure) but the individual pieces are different.

It's a weaker statment because $f(s)=s$ for every $s$ implies

$$f(S)=\{f(s): s\in S\}=\{s: s\in S\}=S$$

but say we're in $X=\{1,2,3\}$ and $S=\{1,2\}$ and $f(1)=2,\, f(2)=1, f(3)=3\}$. Then $f(S)=S$ but $f(s)$ is not necessarily equal to $s$ for every $s\in S$, so $f(S)=S$ does not imply $f(s)=s$, that's what's means by "weaker."

Answer 2

It is $f(X)=\{f(x): x\in X\}.$

Answer 3

If $f(x)=x$ for all $x$ then $f(S)=\{s:\exists x\text{ s.t. }f(x)=s\}$ equals to $S$. The other direction generally doesn't hold. Consider $S=\{0,1\}$ and $f(x)=1-x$. Then, $f(S)=S$ but in general $f(x)\neq x$.

Answer 4

If $f(S)=S$, we say $f$ fixes the set $S$ setwise. If $f(s)=s$, for all $s \in S$, then we say $f$ fixes the set $S$ pointwise. The terminology is intuitive and makes it clear that the latter condition is stronger than and implies the former. But the former does not imply the latter; for eg, take $f$ to be the bijection from $\{1,2,3,4,5\}$ to itself, defined by the permutation $f:=(1,2,3)(4)(5)$. Then $f$ fixes the set $\{1,2,3\}$ setwise, but $f$ does not fix the set $\{1,2,3\}$ pointwise because $f$ moves 1 to 2.