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

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