Cardinality

Informally the cardinality of a set is the number of things in it.

The cardinality of a set $A$ is denoted $$|A|$$

Equipotent cardinality ( $$|A|=|B|$$ )

$A$ and $B$ are equipotent (have the same cardinality, $$|A|=|B|$$ ) if there is a one-to-one (injective) function with domain $A$ and range $B$, note it need not be a bijective function, for example if $$B\subset C$$ then $$f:A\rightarrow C$$ can still be injective, but would not be surjective if $$\exists x(x\in C\wedge x\notin B)$$ , thus not bijective.

This is an equivalence relation

Less than or equal to ( $$|A|\le|B|$$ )

There is an injective mapping from $A$ into $B$, it differs from equality in that the range need not be the entire of $B$

Cantor-Bernstein Theorem ( $$[|A|\le |B|\wedge |B|\le |A|]\implies|A|=|B|$$ )

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TODO: Cantor-Bernstein Theorem

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Addition

We define the sum of cardinals $a$ and $b$ to be:

$$a+b=|A\cup B|$$ where $$a=|A|$$ , $$b=|B|$$ and $$A\cap B=\emptyset$$

To be sure this definition is unique (that we can add cardinals if the intersection is empty) we require the following theorem:

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Proof that if $$A,B,A',B'$$ are such that $$|A|=|A'|$$ and $$|B|=|B'|$$ and $$A\cap B=A'\cap B'=\emptyset$$ that $$|A\cup B|=|A'\cup B'|$$

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TODO: Easy - as cardinality of A and A' are equal there's a 1:1 function, we take the union, which is 1:1 from $$A\cup B\mapsto A'\cup B'$$

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