Identity of Cardinal Product is One

Theorem

Let $\mathbf a$ be a cardinal.

Then:

$\bsone \mathbf a = \mathbf a$

where $\bsone \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$.

Let $\mathbf a = \card A$ for some set $A$.

From the definition of (cardinal) one, $\bsone$ is the cardinal associated with a singleton set, say, $\set \O$.

Consider the mapping $f: \set \O \times A \to A$ defined as:

$\forall a \in A: \map f {\O, a} = a$

Let $a_1, a_2 \in A$ such that:

$\map f {\O, a_1} = \map f {\O, a_2}$

Then:

$\ds \map f {\O, a_1}$

$=$

$\ds \map f {\O, a_2}$

$\ds \leadsto \ \ $

$\ds a_1$

$=$

$\ds a_2$

Definition of $f$

$\ds \leadsto \ \ $

$\ds \tuple {\O, a_1}$

$=$

$\ds \tuple {\O, a_2}$

Thus $f$ is an injection.

Let $a \in A$.

By definition of $f$:

$a = \map f {\O, a}$

Thus $f$ is a surjection.

So $f$ is both an injection and a surjection, and so by definition a bijection.

Thus a bijection has been established between $\set \O \times A$ and $A$.

It follows by definition that $\set \O \times A$ and $A$ are equivalent.

The result follows by definition of cardinal.

$\blacksquare$