# Cardinal of Union Equal to Maximum

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## Theorem

Let $S$ and $T$ be sets that are equinumerous to their cardinal number.

Let $\left|{ S }\right|$ denote the cardinal number of $S$.

Suppose $S$ is infinite.

Then:

$\left|{ S \cup T }\right| = \max\left({ \left|{ S }\right| , \left|{ T }\right| }\right)$

## Proof

Let $x$ denote $\max \left({ \left|{ S }\right| , \left|{ T }\right| }\right)$.

$x = \left|{ S }\right|$ if $\left|{ T }\right| \le \left|{ S }\right|$.

$x = \left|{ T }\right|$ if $\left|{ S }\right| \le \left|{ T }\right|$

By Relation between Two Ordinals, $x = \left|{ S }\right|$ or $x = \left|{ T }\right|$.

In either case, it follows that $x \le \left|{ S \cup T }\right|$ by Subset of Union.

$\Box$

If $\left|{ T }\right| = 1$ or $\left|{ T }\right| = 0$, it follows that $\left|{ S \cup T }\right| = \left|{ S }\right| = x$ by the definition of an infinite set.

If $1 < \left|{ T }\right|$, then $\left|{ S \cup T }\right| \le \left|{ S \times T }\right|$ by Cardinal of Union Less than Cardinal of Cartesian Product.

But by Cardinal Product Equal to Maximum, it follows that $\left|{ S \cup T }\right| \le x$.

$\Box$

Combining the two lemmas, it follows that $x = \left|{ S \cup T }\right|$

$\blacksquare$