Meaning of "$\exists$" in "$\{y \in Y : \exists x \in X \text{ such that }f(x) = y\}$"

Question

I came across this definition of the range of a function:

For a function $f : X → Y$, the range of $f$ is $$\{y \in Y : \exists x \in X \text{ such that }f(x) = y\},$$ i.e., the set of $y$-values such that $y = f (x)$ for some $x \in X.$

I have a little doubt regarding the use of “$\exists x \in X$”. Shouldn't it be “$\forall x \in X$” instead, since the range of a function is the set of all values that $y$ can acquire, which is by mapping all $x$'s in $X$ to $Y?$

Answer 1

For a function $f : X → Y$, the range of $f$ is $$\{y \in Y : \exists x \in X \text{ such that }f(x) = y\},$$ i.e., the set of $y$-values such that $y = f (x)$ for some $x \in X.$

Th given set is more accurately read “the set of elements of $Y$ such that each, for some element $x$ of $X,$ equals $f(x)$” or, more simply, “the set of elements of $Y$ such that each equals some output of $f$”.

I have a little doubt regarding the use of “$\exists x \in X$”. Shouldn't it be “$\forall x \in X$” instead

On the other hand, your suggested set $$\{y \in Y : \forall x \in X,\;\, f(x) = y\}$$ is read “the set of elements of $Y$ such that each equals every output of $f$”.

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Here's a different example providing a similar contrast:

1. $$A=\{n\in\mathbb Z: \exists a\in\mathbb Z\;\,n=2a\}\\ =\text{the set of integers such that each one is double }\textit{some }\text{ integer}\\ =\{\ldots,-6,-4,-2,0,2,4,6,8,\ldots\}\\ =2\mathbb Z.$$

   Set $A$ is populated precisely with the even integers:

   • take some (any) integer, then double it; the result is a member of set $A$;
   • repeat infinitely.

2. $$B=\{n \in\mathbb Z: \forall a\in\mathbb Z\;\,n=2a\}\\ =\text{the set of integers such that each one is double }\textit{every }\text{ integer}\\ =\emptyset.$$

   Since no integer is simultaneously twice of $-5,$ twice of $0,$ twice of $71,$ etc., the set $B$ has no member.

Answer 2

The last line of your writing is very true, but that doesn't mean it's okay to write $\forall$.

Let me explain the reason with one very simple and specific example.

Let $f:\{0,1\}\to\mathbb R$ as $f(x)=x$. Then $\{y \mid \forall x \in \{0,1\}$ such that $f(x) = y\}$ is $\emptyset$, because $y$ cannot be 1 at the same time as 0.