Cardinality of Singleton

Theorem

Let $A$ be a set.

Then $\left\vert{A}\right\vert = 1$ if and only if $\exists a: A = \left\{ {a}\right\}$

where $\left\vert{A}\right\vert$ denotes the cardinality of $A$.

Proof

Sufficient Condition

Assume that

$\left\vert{A}\right\vert = 1$

By definition of cardinality of finite set:

$A \sim \N_{< 1} = \left\{ {0}\right\}$

where $\sim$ denotes set equivalence.

By Set Equivalence is Equivalence Relation:

$\left\{ {0}\right\} \sim A$

By definition of set equivalence there exists a bijection:

$f: \left\{ {0}\right\} \to A$

By definition of bijection:

$f$ is s surjection.

Thus

\(\displaystyle A\)

\(=\)

\(\displaystyle f^\to\left({\left\{ {0}\right\} }\right)\)

Definition of Surjection

\(\displaystyle \)

\(=\)

\(\displaystyle \left\{ {f\left({0}\right)}\right\}\)

Image of Singleton under Mapping

$\Box$

Necessary Condition

Assume that:

$\exists a: A = \left\{ {a}\right\}$

Define a mapping $f: A \to \left\{ {0}\right\}$:

$F\left({a}\right) = 0$

It is easy to see by definition that

$f$ is an injection and a surjection.

By definition

$f$ is bijection.

By definition of set equivalence:

$A \sim \left\{ {0}\right\} = \N_{< 1}$

Thus by definition of cardinality of finite set:

$\left\vert{A}\right\vert = 1$

$\blacksquare$

Sources

• Mizar article CARD_2:42