# Cardinality of Singleton

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## 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

\(\ds A\) | \(=\) | \(\ds f^\to\left({\left\{ {0}\right\} }\right)\) | Definition of Surjection | |||||||||||

\(\ds \) | \(=\) | \(\ds \left\{ {f\left({0}\right)}\right\}\) | 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