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 $\mathbf a$.

That is, $\bsone$ is the identity element of the product operation on cardinals.

Proof
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, $\set \O$.

We have by definition of product of cardinals that $\bsone \mathbf a$ is the cardinal associated with $\set \O \times A$.

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:

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.