In Tao's Analysis I, the definition of an image of some subset $S$ is given as follows:
If $f:X\to Y$ is a function from $X$ to $Y$, and $S$ is a set in $X$, we defined $f(S)$ to be the set:
$f(S):=\{f(x):x\in S\}$
Now, when I first read this, I thought I had a very clear picture of what this conveyed. However, after running through some of the exercises at the end of the chapter, I realized that I misunderstood what was entailed by this definition.
My initial interpretation of this definition (in plain english) was: "If $x \notin S$, $y \notin f(S)$." The reasoning for this was that I thought $\{f(x):x\in S\}$ effectively meant $y: [(y =f (x) )\land (x \in S)]$ where "$(y =f (x) )\land (x \in S)$" is some proposition of the form $P(x,y)$. Therefore, if $x \notin S$, this statement is false, and thus $y \notin f(S)$.
After running through the exercises, I realized that my interpretation was wrong. Let's consider the statement $x^* \notin S$.
When we say $y \in f(S)$, what we are really saying is that IF $y \in f(S)$ then there must exist an $ x \in S$ that generates this $y$. However, it could certainly be the case that there are OTHER elements, such as $x^*$, which are NOT in $S$, that also map to $f(S)$. As such, just because an element like $x^*$ is NOT in $S$, it is not necessarily true that $f(x^*) \notin f(S)$.
So it seems like my initial interpretation should actually have been "If $(\neg\exists x \in S $ s.t.$ y=f(x)) \land f(x^*) =y$, then $y \notin f(S)$".
My question is as follows:
Why was my reasoning wrong, which led to the initial misinterpretation? Is there a more symbolically heavy way of describing the set $f(S)$ to avoid the confusion that I encountered?
Your initial misinterpretation was due to the fact that you did not include quantifiers in it. That is, you wrote that $f(S)$ should be
$$\{y| y=f(x)\land x\in S\}$$ however the $x$ in this definition is unquantified. The correct version of this would be $$\{y| \exists x: (y=f(x)\land x\in S)\}$$
while you understood it more as
$$\{y| \forall x: (y=f(x)\implies x\in S)\}$$ which is incorrect.