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

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 $$Y?$$

• You might, in the future, find slightly different uses of this word, the range of $f:X\to Y$ could also be defined as all of $Y$, and the set you wrote is then the image of $f$. Oct 27 at 6:57

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 one, for some element $$x$$ of $$X,$$ equals $$f(x)$$” or, more simply, “the set of elements of $$Y$$ such that each one 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 one equals every output of $$f$$”.

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.

• I was putting too much belief in the 'language'(which isn't precise) of the statement and not the mathematical notation(which is always precise). Thanks @ryang ! I was confused about how to read a set, which was very trivial. Oct 27 at 2:39
• @Prakhar 1. To be fair, the natural-language description of the set can be made precise, as shown. 2. Describing a set (translating logical and set symbols into natural language) is not always trivial, and your confusion was very understandable. Oct 27 at 6:47

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.

• I don't understand how the set you defined is ∅?! How does $\forall$ implies that y is simultaneously 0 and 1?? Oct 26 at 21:12
• No, never. @Prakhar. $\{y \in Y : \exists x \in X$ such that $f(x) = y\}$ is $\mathrm{Im}f$(range of $f$), but $\{y \in Y : \forall x \in X$ such that $f(x) = y\}$ is generally $\emptyset$. Oct 26 at 21:14
• i'm unable to grasp this. But thanks @Nightflight Oct 26 at 21:31
• The set $\{y∈Y: f(x)=y \ ∀x∈X\}$ is the set of elements of $Y$ that ALL of $X$ maps to. This would only be non-empty if f was a constant map. i.e. mapping all elements to a single element of Y.
– Lev
Oct 26 at 21:51
• What @Lev said. That set is empty unless $f$ is constant, in which case it's a singleton containing the constant value. There can't be more than one $y\in Y$ such that $\forall x\in X, f(x) = y$ (unless $X$ is empty!) Oct 27 at 5:37