For any function $f$ and set $S$, $f(s) \in f(S) \not\implies \Leftarrow s \in S$

Question

This already contains many counterexamples, so I'm not seeking any more of them;
I'm interested in learning about my errors with the notation and definitions.

Richard Hammack P213 Defintion 12.9: Suppose $f : A \rightarrow B$ is a function.
I. If $S \subseteq A$, then the image of S is defined as the set $f(S) = \{ \, f (s) : s \in S \, \} \quad \subseteq B.$

$1.$ $f(s) \in f(S)$ means: There exists $s \in S$ that effects $f(s)$. So why does this not imply $s \in S$ ?

In view of user2357112's answer, I still don't perceive the falsity. Even if I write it as "There exists $t \in S$ such that $f(t) = f(s)$", then I'm still operating on something (now $t$ instead of $s$) in $S$?

$2.$ Do we simply apply $f$ to both sides of: $\qquad s \in S$,
to effect $f(s) \in f(S)$? Is there a deeper explanation? What qualifies one to "apply f"?

Answer 1

Cosider the constant function $f\colon \mathbb N\to\mathbb N$ given by $f(n)=42$. Let $S=\{1,2,3\}$ and $s=4$. Then $f(s)=42$, $f(S)=\{42\}$, i.e. $f(s)\in f(S)$, but $s\notin S$.

For the other direction, if $s \in S$, then by definition $f(s)\in\{\,f(s):s\in S\,\}$ - this is precisely what the notation implies!

Answer 2

$1.$ $f(s) \in f(S)$ means: There exists $s \in S$ that effects $f(s)$. So why does this not imply $s \in S$ ?

You reused the name $s$. What you should be saying is that there exists some $t \in S$ that effects $f(s)$, or, to use more standard terminology, there exists $t \in S$ such that $f(t) = f(s)$. $t$ may be some object completely unrelated to $s$. When you reuse the name $s$, you effectively introduce a hidden assumption that $s$ is in $S$.

Let's make things more concrete. Suppose $S$ is the set of current Major League Baseball players, and $f$ is "the age of". Thus $f(Steve Clevenger)$ is the age of Baltimore Orioles catcher Steve Clevenger.

Suppose $f(me) \in f(S)$. That means there is some player $t \in S$ such that $f(t) = f(me)$ - some current Major League Baseball player whose age is the same as mine. $t$ is a new variable we've introduced - it does not necessarily refer to me. Your misunderstanding of the definition would say that there is some $me \in S$ with my age, implicitly asserting that I play in Major League Baseball.

$2.$ Do we simply apply $f$ to both sides of: $\qquad s \in S$,
to effect $f(s) \in f(S)$? Is there a deeper explanation? What qualifies one to "apply f"?

If you're trying to show that $s \in S$ implies $f(s) \in f(S)$, this follows from the definition of $f(S)$ as the set of all outputs of $f$ when $f$ is applied to elements of $S$. If you're trying to go in the other direction, showing that $f(s) \in f(S)$ implies $s \in S$, you can't, because that's not true.