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It is easy enough to prove in set theory, but it seems counter-intuitive to me that an empty set could be the domain of a function. Is there any literature requiring that functions have non-empty domains?


I mean definitions something like...

For all sets $B$ and non-empty $A$, $f$ is a function mapping $A$ to $B$ iff

1) $f\subset A\times B$

2) $\forall x\in A: \exists y\in B: (x,y)\in f$

3) $\forall x,y_1,y_2:[ (x,y_1)\in f \land (x,y_2)\in f \implies y_1=y_2]$

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Mathematics is in desperate need of an anti-psychologist trend. – Git Gud Jan 12 '14 at 15:41
You should edit "an empty set could be a function" to "an empty set could be a domain of a function." – Joel Reyes Noche Jan 13 '14 at 4:23
@Joel: The usual set-theoretic way to represent a function is by its graph. Any function whose domain is the empty set has a graph that is also the empty set. – Hurkyl Jan 13 '14 at 4:26
Of course, if you disallow the empty function, then the empty set could not be said to have the same cardinality as itself, unless you further complicate the definition of the equipotence relation to explicitly mention this case. – Arthur Fischer Jan 13 '14 at 4:33
@JoelReyesNoche Done. Thanks. – Dan Christensen Jan 13 '14 at 5:23

1 Answer 1

I honestly can't think of any literature off the top of my head with that restriction. Note that by allowing the empty set to be the domain of functions, you make the empty set an initial object in the category of sets.

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Interesting. A number of other key notions seem to hinge on it as well. Best not to open that Pandora's box, I guess. – Dan Christensen Jan 13 '14 at 3:40
Even worse, if you forbid functions with empty domains, you don't have a category of sets, because such a thing would need an identity function from the empty set to itself! – Hurkyl Jan 13 '14 at 5:37
@Hurkyl That would be a catastrophe! Wouldn't it? ;^) – Dan Christensen Jan 13 '14 at 6:27
@Hurkyl There I go forgetting the obvious, again. – Nick Jan 13 '14 at 14:18

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