# Does the map $f:\emptyset\longrightarrow \{0\}$ exist?

What function $f$ maps in the following way: $$f:\emptyset\longrightarrow \{0\}$$

-
Does the cardinality of the domain count as a function? –  Ataraxia Jul 31 '13 at 18:28

There is exactly one such function, which is the empty set.

Explanation:

By definition, functions $f : A \to B$ are certain subsets of the cartesian product $A\times B$ (for the defining property, see below). You can imagine this definition as identiying $f$ with its graph. In your case, $A = \emptyset$, $B = X$ and $A \times B = \emptyset \times X = \emptyset$, whose only subset is the empty set.

The defining property is that for every $a\in A$ there is a unique element of $f$ having $a$ as its first component. In our case $f = \emptyset$, this condition is vacuously true.

-

The only subset of the cartesian product $\,\emptyset\times X\;,\;\;X$ any set, is the only function $\,\emptyset\to X\,$

-