Cardinal Number Plus One Less than Cardinal Product

Theorem
Let $x$ be an ordinal such that $x > 1$.

Then:


 * $\card {x + 1} \le \card {x \times x}$

Where $\times$ denotes the Cartesian product.

Proof
Since $x > 1$, then $0 < x$ and $1 < x$.

Define the function $f: x + 1 \to x \times x$ as follows:


 * $\map f y = \begin{cases}

\tuple {y, 0} &: y < x \\ \tuple {0, 1} &: y = x \end{cases}$

If $\map f y = \map f z$, then $y = z$ by cases.

Case 1: $y = x$
If $y = x$, then $\map f y = \tuple {0, 1}$ by the definition of $f$.

So if $z < x$, then $\map f y \ne \map f z$.

This contradicts the hypothesis, so $z = x$ and therefore $z = y$.

Case 2: $y < x$
If $y < x$, then $\map f y = \tuple {y, 0}$ by the definition of $f$.

$z = x$ yields the contradictory statement $\tuple {y, 0} = \tuple {0, 1}$.

Therefore, $z < x$ and $\tuple {y, 0} = \tuple {z, 0}$.

By Equality of Ordered Pairs, it follows that $y = z$.

It follows that $f$ is an injection.

By Injection iff Cardinal Inequality, it follows that $\card {x + 1} \le \card {x \times x}$.