Cardinality of Singleton
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Theorem
Let $A$ be a set.
Then $\card A = 1$ if and only if $\exists a: A = \set a$
where $\card A$ denotes the cardinality of $A$.
Proof
Sufficient Condition
Assume that
- $\card A = 1$
By definition of cardinality of finite set:
- $A \sim \N_{< 1} = \set 0$
where $\sim$ denotes set equivalence.
By Set Equivalence behaves like Equivalence Relation:
- $\set 0 \sim A$
By definition of set equivalence there exists a bijection:
- $f: \set 0 \to A$
By definition of bijection:
- $f$ is s surjection.
Thus
\(\ds A\) | \(=\) | \(\ds \map {f^\to} {\set 0}\) | Definition of Surjection | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map f 0}\) | Image of Singleton under Mapping |
$\Box$
Necessary Condition
Assume that:
- $\exists a: A = \set a$
Define a mapping $f: A \to \set 0$:
- $\map f a = 0$
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:
- $A \sim \set 0 = \N_{< 1}$
Thus by definition of cardinality of finite set:
- $\card A = 1$
$\blacksquare$
Sources
- Mizar article CARD_2:42