1
$\begingroup$

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 $Y?$

$\endgroup$
1
  • $\begingroup$ 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 $f$. $\endgroup$ Oct 27 at 6:57
5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Prakhar
    Oct 27 at 2:39
  • $\begingroup$ @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. $\endgroup$
    – ryang
    Oct 27 at 6:47
4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I don't understand how the set you defined is ∅?! How does $\forall$ implies that y is simultaneously 0 and 1?? $\endgroup$
    – Prakhar
    Oct 26 at 21:12
  • $\begingroup$ 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 $\emptyset$. $\endgroup$ Oct 26 at 21:14
  • $\begingroup$ i'm unable to grasp this. But thanks @Nightflight $\endgroup$
    – Prakhar
    Oct 26 at 21:31
  • 1
    $\begingroup$ 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. $\endgroup$
    – Lev
    Oct 26 at 21:51
  • 2
    $\begingroup$ 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!) $\endgroup$
    – BrianO
    Oct 27 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.