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