Suppose $f:\mathbb R \rightarrow\mathbb R$, defined by $f(x)=x^2$, then what is $f(\emptyset\ )$? $\emptyset$ is the empty set.

Either $f(\emptyset)=(-\infty,0)\,$ or $\,f(\emptyset)=\emptyset\,$ makes no sense to me.

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    $\begingroup$ How many numbers have the form $x^2$ for some $x$ in the empty set? $\endgroup$ Feb 11 '17 at 9:15
  • 1
    $\begingroup$ Actually how did you get the image to be $(-\infty,0)$? $x^2 \geq 0$ Don't even think about the $x^2$ part $\endgroup$
    – IAmNoOne
    Feb 11 '17 at 9:35
  • $\begingroup$ In order to get $f(\varnothing)$ you are to collect all elements $f(x)$ where $x$ runs over $\varnothing.$ Since there are no any elements in $\varnothing,$ the set $f(\varnothing)$ is likewise empty. Nothing begets nothing, so to speak. $\endgroup$
    – Olod
    Feb 11 '17 at 9:49
  • $\begingroup$ But $f^{-1}(-\infty,0) = \emptyset$, right? There is obviously no real number such that $x^2<0$, then how come $f(\emptyset)=\emptyset$ makes sense? $\endgroup$
    – jason wong
    Feb 11 '17 at 10:54
  • $\begingroup$ It is true that $f^{-1}(-\infty,0)=\varnothing,$ and $f^{-1}( (-5,-4) )=\varnothing,$ etc. But $f^{-1}$ isn't a notation of a function, it is notation for the full preimage of a set: $$ f^{-1}(Y) =\{ x : f(x) \in Y\}. $$ $\endgroup$
    – Olod
    Feb 11 '17 at 11:58

In a first/strict sense, the expression $\:f\big(\emptyset\big)$ is not meaningful because $\,\emptyset\,$ is not an element in $f$'s domain of definition, but rather a subset of it: $\:\emptyset\subset\mathbb{R}$.

There is an obvious & often used way to define/extend a given function to the subsets of its domain: $$f(\,S\,):=\{\,f(x)\mid x\in S\,\}\qquad\text{where }S\subset\operatorname{Domain}(\:f\,)$$ Values are then subsets of the range of $\,f$.

  • This understood the statement $\,$"$f(\emptyset)=\emptyset$"$\,$ makes sense and is correct.

  • "$f(\emptyset)=(-\infty,0)$"$\,$ makes sense too, and it's simply wrong.


$f[A]$ for $A \subset \mathbb{R}$ makes sense and is just $\{f(x) ; x \in A\}$, the set of all images of points in $A$. As $\emptyset$ has no points to take an image of, we get $f[\emptyset] = \emptyset$, don't use $f(\emptyset)$, as this would suggest that $\emptyset$ is an element of the domain, which it is not.

The question suggests to me that you are confused with the true statement $f^{-1}[(-\infty, 0)] =\emptyset$ .This is true, as $f^{-1}[(-\infty, 0)]$ is defied as all $x$ in the domain that map into $(-\infty, 0)$, so $$f^{-1}[(-\infty, 0)] = \{x \in \mathbb{R} : f(x) \in (-\infty, 0)\} = \{x \in \mathbb{R}: x^2 < 0 \}$$ and no such real $x$ exist, so the result is $\emptyset$.

  • $\begingroup$ I am confused with $f[\emptyset]=\emptyset$ and $f^{-1}[(-\infty,0)]=\emptyset$. $\endgroup$
    – jason wong
    Feb 11 '17 at 10:56
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    $\begingroup$ @jasonwong both are true and follow straightforwardly from the definitions. $\endgroup$ Feb 11 '17 at 10:57

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