Why is the definition of an image of a subset use existential quantifier rather than universal? According to my textbook (Discrete Mathematics and Its Applications by Rosen), the definition of the image under function $f $ $(f:A\rightarrow B)$ of the subset $S$ $(S\subseteq A)$ is $$f(S) = \{t | \exists s \in S (t = f(s)) \}$$
What I'm confused about this is that I think it should be a universal quantification since we're talking about the set containing all the images of elements in set $S$, right?
 A: Just because a definition in spoken English uses the word 'all' doesn't mean there should be a universal quantifier. Here's another example: suppose you have a bunch of sets $X_1, X_2, \dots$. The set of all elements appearing in these sets is
$$\bigcup_{n \in \mathbb{N}} X_n = \{ x \mid \exists n \in \mathbb{N},\, x \in X_n \}$$
which is defined using an existential quantifier, not a universal quantifer. This is because the set of 'all the elements' of these sets is precisely the set of elements which appear in 'at least one' of these sets.
In this case, $f(S)$ is the set of all the elements in the image of $S$ under $f$. That is, it's the set of elements which are of the form $f(s)$ 'for some' $s$... that's why we have an existential quantifier.
If we had a universal quantifier instead, the set would be
$$\forall_f(S) = \{ t \mid \forall s \in S,\, t = f(s) \}$$
That is, $t \in \forall_f(S)$ if every element of $S$ maps to $t$ under $f$. If $S \ne \varnothing$ then this set is either empty (if $f$ is non-constant on $S$) or has one element (if $f$ is constant on $S$)... so in general not very interesting.
A: Take, for example, two sets, $A=\{1,2,3\}$, and $B=\{0,1,2,3,4,5\}$. Let your function $f$ be defined as $f(n)=n+1$, so $f$ maps $1$ to $2$, it maps $2$ to $3$ and it maps $3$ to $4$.
Now, intuitively, what is the image of $A$ under the function $f$?
I hope that, intuitively, you agree that $f(A) = \{2,3,4\}$.


*

*The set $\{t| \exists a\in A: t=f(a)\}$ is precisely the set $\{2,3,4\}$

*The set $\{t| \forall a\in A: t=f(a)\}$ is an empty set. This is because no element of $B$ is equal to $f(a)$ for all values $a\in A$.


Why is the second set empty and why is the first set what you need? Well, let's see their definition. 

The second set is equal to, in words:

All such elements $t$ for which it is true that for all elements $a\in A$, the element $t$ is equal to $f(a)$.

Now, is $2$ in this second set? Well, let's see. Is it true that $2$ is equal to $f(a)$ for all elements $a\in A$? It is certainly true that $2=f(1)$, so for $a=1$, it is true that $2=f(a)$. However, for $a=2$, $f(a)$ is equal to $3$, so $f(a)$ is not equal to $f(a)$ for this value of $a$, and thus, it is not equal to $f(a)$ for all values $a\in A$.

Let's now check the first set. it is equal to, in words:

All such elements $t$ for which there exists some element $a\in A$ such that $t$ is equal to $f(a)$.

Is $0$ an element of it? No. There is no such thing in $A$ that $f(t)$ would equal $0$. Is $2$ an element of it? Yes: if $a$ is equal to $1$, then $f(a)=2$. We found some element $a$ for which $2=f(a)$, so at least some element exists, meaning $2$ is in this set.
A: There's an "invisible $\forall$" hidden in the set builder notation -- namely $\{x\mid \phi(x)\}$ means the set $A$ such that $\forall x: x\in A\Leftrightarrow \phi(x)$. That give you the universal quantification you feel should be there.
In other word, $\{x\mid \exists s\in S\, (x=f(s))\}$ is the set whose members is every $x$  that where there is something in $S$ it is the image of.
You can also write this set as $\{f(s)\mid s\in S\}$ -- in that case there is no visible quantifier, but the meaning is that for every $s$ in $S$, $f(S)$ is a member of the set we're speaking of (and nothing else is).
A: The natural-language definition of image is: "the set of all points $x$ such that $f(a)=x$ for some $a\in A$."
Before reading further, convince yourself that this is in fact the image of $f$.

The use of the word "all" in the above may make you think that there should be a "$\forall$" in the formal definition. And there is! Remember that "$\{x: \varphi(x)\}$" means "the set of all $x$ such that $\varphi(x)$," or $$I=\{x: \varphi(x)\}\iff \forall x (\varphi(x)\iff x\in I).$$ Now that universal quantifier is visible!
So in the expression "$\{x: \exists a\in A(f(a)=x)\}$," the universal quantifier you expect is hidden in the set-builder notation (that "$\{ . . . \}$" business).
So where does the "$\exists$" come from? Well, look again at the natural-language definition:

"The set of all points $x$ such that $f(a)=x$ for some $a\in A$."

That "for some" is an existential quantifier: "$f(a)=x$ for some $a\in A$" is the same as "there exists an $a\in A$ such that $f(a)=x$."
So that's why! Reading the natural-language definition, we see two quantifiers: "all" and "some." And indeed, the formal definition has a universal quantifier (build into the set-builder notation) and an existential quantifier (visible inside the scope of the set-builder notation).
A: I've never seen this notation ($s\in S(t = f(s))$), is it really exactly like this in your textbook? It simply means, that the image of $S$ under $f$ is the set of all points $t \in B$, such that there is a $s \in S$ with $f(s) = t$. You can also write is like this:
$$ f(S) = \{ f(s) \; | \; s \in S \} \subset B $$
