The empty function $f: \emptyset \rightarrow X$ is a function from the empty set to an arbitrary set $X$. Since $\emptyset$ is the only subset of $\emptyset \times X = \emptyset$, $f$ must be equal to $\emptyset$. Is it possible to invoke the empty function?

To clarify what I mean, consider a function $g: \{ 0 \} \rightarrow \{ 100 \}$ which is defined as $g(0) = 100$. Note that I am allowed to write $g(0)$ to obtain an image of $0$, which is $100$. What is not clear for me is whether it is possible to use the same notation for the empty function; even if you were to invoke one as $f()$, how do you know which element of the co-domain $X$ you obtain?

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    $\begingroup$ I'd argue that this not different from the definition of a variable. How do you know what $g(0)$ is? You need a definition for that. So define a function $f$ so that $f()=5$ but that of course is not different than being a variable $f:=5$. I'm not too versed in math, so I'm not sure if this actually qualifies as a function or just a map (I suspect it's not a function, just a map). $\endgroup$ Commented May 28 at 8:25
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    $\begingroup$ @infinitezero maps and functions are used synonymously, and what you said is not a function, nor a relation. You’re just specifying one element in the target space. OP: here the image is again the empty set, so you don’t get any element of the codomain. $\endgroup$
    – peek-a-boo
    Commented May 28 at 8:30
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    $\begingroup$ Mathematics generally avoids this issue by having no concept of "invoking" a function. For any $x$ and $y$, $f(x) = y$ would be false if $f$ is the empty function. $\endgroup$ Commented May 28 at 19:48
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    $\begingroup$ It is meaningless to "invoke" a function without applying the function to some element of the domain. If the domain contains no elements (because the domain is the empty set) then there is no element to which one can apply the function, so the function cannot be "invoked". The notation f() has no meaning in the usual (mathematical) sense of a function f. $\endgroup$ Commented May 28 at 20:32
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    $\begingroup$ I wouldn't say that $f(x)$ has no meaning, just that $f(x)$ is undefined for all objects $x$. It's similar to how if $g:\mathbb{R}\to \mathbb{R}$ is defined as $g(x):=x^2$ for all $x \in \mathbb{R}$, the value of $g((v_1,v_2))$ is undefined for every vector $(v_1,v_2) \in \mathbb{R}^2$, because $(v_1,v_2)$ is not in the codomain of $g$. $\endgroup$
    – 1Rock
    Commented May 29 at 5:44

7 Answers 7


$f$ is a function iff for all $x$ in the domain there exists exactly one $y$ in the codomain such that $(x,y)\in f,$ and in that case we write $f(x)$ for the uniquely identified $y.$ Note the in that case: the notation $f(x)$ presupposes that $x$ belongs to the domain. For the empty function no $x$ can belong to the domain and therefore the notation $f(x)$ cannot acquire meaning.

The notation $f()$ would suggest a unary relationship; functions are by definition binary relationships.

  • $\begingroup$ Does it imply that the empty function has no concept of an image? $\endgroup$ Commented May 28 at 10:09
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    $\begingroup$ It does have an image set, but that set is the empty subset of the codomain. It does not have any images of individual points, by definition. $\endgroup$
    – Lieven
    Commented May 28 at 11:55
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    $\begingroup$ You can write $\forall x \in \emptyset,\, f(x) = 7$, but you cannot write $\exists x \in \emptyset,\,...$ and you cannot write $f(3)$ and you cannot write $f()$ $\endgroup$
    – Stef
    Commented May 28 at 18:11
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    $\begingroup$ You can certainly write something like $\exists x \in \emptyset, P(x)$ for any unary predicate $P$. It's just that this statement is always false. $\endgroup$
    – Lee Mosher
    Commented May 28 at 22:28
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    $\begingroup$ @VladMikheenko Any proposition of the form ∀𝛼∈∅, P is vacuously true (even when P = ⊥). Like Stef said above, f(x) is indeed well-formed if x is quantified over the empty set. (Disclaimer: I'm speaking from the perspective of constructive logic, but I think this all holds in classical logic as well.) $\endgroup$ Commented May 29 at 6:03

When $f : X \to Y$ is an arbitrary function, you can invoke it on any given $x \in X$ - the rule is the same no matter what $X$ and $Y$ are. However, when $X$ happens to be the empty set, there is simply no $x \in X$ on which you could invoke the function. So theoretically invoking the function is possible, only there is no valid input to invoke it on.

Just as with any function, you can also compute the image of any subset $A \subseteq X$. With $X = \varnothing$ there actually is one such subset, namely the empty set. Therefore it makes perfect sense to write $f[\varnothing]$, and the result is - what else? - the empty set.


It seems perfectly legitimate to consider the empty function. It is a binary relation that happens to be empty. You can compose it with another function and get another empty function. You can even square it, and obtain itself. If $X$ and $Y$ are different spaces, the functions $\varnothing\to X$ and $\varnothing\to Y$ would apparently be the same entity from the set-theoretic viewpoint.


Is it possible to invoke the empty function?

In calculus, the concept of "invoking" a function (as in imperative programming languages, or something like the lambda calculus from Computer Science) does not exist. Since the formalization of calculus via set theory, $f$ is identified with a set; specifically, a binary relation. We can define many functions by a calculation (i.e. $f(x)=x^2$), but that is just an intuitive/naive definition (which is more than good enough in very many cases). Under the hood, it's sets all the way down.

What is not clear for me is whether it is possible to use the same notation for the empty function; even if you were to invoke one as 𝑓(), how do you know which element of the co-domain 𝑋 you obtain?

If you wish to use the function, you would never write it as $f()$, but as either $f$ (to talk about the function as a whole) or $f(x)$ (to talk about which value $f$ maps the variable $x$ to). $x$ here is akin to the variable name you would use in the "head" of a function definition in a programming language. Even if you know that there can never be a legal value supplied when calling your programming-language-function, syntactically you would still have to write it down. The same is true for the function syntax in maths.

$f(x)$ is well-defined for all $x\in\emptyset$, and undefined for all $x\not\in\emptyset$. Talking about $x\in\emptyset$ is perfectly fine in the context of logic statements like $\forall x \in \emptyset ...$ or $\exists x\in\emptyset ...$.

Check out the formal definition of functions on Wikipedia. In each of the three steps, every set is empty (i.e., the domain $X=\emptyset$, therefore the relation $\emptyset=R \subset X\times Y=\emptyset$), all $\forall$'s are automatically "true" and all $\exists$'s automatically "false" (by definition/axiom).


Per definition, a function $f: A \rightarrow B$ is a subset of $ A\times B$, which happens to verify the following condition: if (a,b) and (a,b') belongs to f, then b = b'.

With that in mind and the fact that $\varnothing \times B = \varnothing$, it is easy to follow that your empty function is nothing else that the empty set.

  • $\begingroup$ With this I mean that it is so legitimate to "invoque" the empty function as it is to do it with the empty set. $\endgroup$
    – Javi Ruiz
    Commented May 28 at 20:12

Let us skip the "is a function a triple, or its graph" debate and assume that the team defending "a function is its graph" has won.

For all sets $x, y, f$, let us denote $image(f,x,y)$ the sentence: $(x,y) \in f$.

A set $f$ is called a function if for all $z \in f$, $\exists x,y,\ z = (x,y)$ and if $\forall x,y,y',\ (image(f,x,y) \wedge image(f,x,y')) \Rightarrow y = y'$.

It is an easy exercise to check that the empty set is a function.

Moreover, $f(x)$ is an abbreviation. That is, every time you write a statement $P$ with $f(x)$ appearing somewhere, what you are really writing is actually the statement $\forall y, image(f,x,y) \Rightarrow P[f(x) \to y]$, where $P[f(x) \to y]$ means "every time you see $f(x)$ in the definition of $P$, replace it by $y$", assuming $y$ does not appear anywhere in $P$ (if it does, choose another letter that does not appear in $P$).

Using this rule, you can write $f(x)$ in every formula you like, even if $f$ is the empty function, and even if $f$ isn't a function at all, and any $x$. For example, let $f$ be the empty function, and let $x$ by any set. Let us prove the following sentence: $\forall z,\quad f(x) \neq z$. Indeed, by the rewriting rule I just gave, this sentence is equivalent to $\forall y,\ \forall z, \ image(f,x,y) \Rightarrow y \neq z$. To prove this, let $y$ and $z$ be sets. Then $(x,y) \not \in f$, since $f = \emptyset$. Therefore, we have the negation of $image(f,x,y)$. From this, we can deduce $y \neq z$, so the assertion is proved.

So, writing $f(x)$ is completely harmless, from a strictly logical point of view. The only thing is that for $f$ that are not functions, or for $f$ that are functions and when $x$ is a set not in their domain, you can prove anything from the assumption $\exists y, f(x) = y$.

The word "invoke" has no meaning in mathematics.


Is it possible to invoke the empty function?

Yes. Using the commonly used definition of a function, ordinary set theory, and a form of natural deduction, we can formally prove the existence of function $f$ such that $\forall a: [a\in \emptyset \implies f(a)\in X]$ (60 lines, available on request). Hint: The graph set is $g=\emptyset \times X$.

Even if you were to invoke one as $f()$, how do you know which element of the co-domain $X$ you obtain?

In this case, $f(x)$ is indeterminate/undefined for every $x$ since that $x$ will not be an element of the domain of $f$.

  • $\begingroup$ Can you please show how you formally proved it? I am interested in the process itself. Or at least point me ina direction because right now I am lost. $\endgroup$ Commented May 28 at 21:17
  • $\begingroup$ As requested: dcproof.com/EmptyFunctionArbCodomain.htm $\endgroup$ Commented May 29 at 3:10
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    $\begingroup$ Every function $f$ satisfies the condition, not just the empty function. Anyway, what do formals proofs or set theory have to do with the question at all? It's a simple question for which the naive set theory is more than enough. $\endgroup$
    – Adayah
    Commented May 29 at 11:03
  • $\begingroup$ @Adayah A formal proof eliminates the need for any "hand-waving" arguments about simply extending the usual notions of functions to empty sets. To many readers including the OP, it may not be so "intuitively obvious" . $\endgroup$ Commented May 29 at 18:48
  • $\begingroup$ If that is your motivation, you should be formally proving that it is impossible to invoke the empty function (which you can't do anyway, because it is not a valid mathematical theorem). Meanwhile you suggest formally proving the existence of $f$ such that $(\forall a) \big( a \in \emptyset \implies f(a) \in X \big)$ (which is inaccurate and not what the OP asked), then you proceed to "hand-wave" the answer to the actual question. $\endgroup$
    – Adayah
    Commented May 30 at 8:28

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