Cardinality of Set of All Mappings to Empty Set

Theorem

Let $S$ be a set.

Let $\O^S$ be the set of all mappings from $S$ to $\O$.

Then:

$\card {\O^S} = \begin{cases}

1 & : S = \O \\ 0 & : S \ne \O \end{cases}$

where $\card {\O^S}$ denotes the cardinality of $\O^S$.

Proof

From Null Relation is Mapping iff Domain is Empty Set, the null relation:

$\RR = \O \subseteq S \times T$

is not a mapping unless $S = \O$.

So if $S \ne \O$:

$\card {\O^S} = 0$

If $S = \O$:

$\card {\O^S} = 1$

Hence the result.

$\blacksquare$

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 8$: Functions: Exercise $\text{(ii)}$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory: Exercise $1.9 \ \text{(c), (d)}$