Condition for Set Union Equivalent to Associated Cardinal Number

Theorem
Let $S$ and $T$ be sets.

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

Let $\sim$ denote set equivalence.

Then:


 * $S \cup T \sim \left|{S \cup T}\right| \iff S \sim \left|{S}\right| \land T \sim \left|{T}\right|$

Necessary Condition
Let $S \cup T \sim \left|{S \cup T}\right|$.

By definition of set equivalence, there exists a bijection $f: S \cup T \to \left|{S \cup T}\right|$.

Since $f$ is a bijection, it follows that:


 * $S$ is equivalent to the image of $S$ under $f$.

This, in turn, is a subset of the ordinal $\left|{S \cup T}\right|$.

$\left|{S \cup T}\right|$ is an ordinal by Cardinal Number is Ordinal.

By Condition for Set Equivalent to Cardinal Number, it follows that $S \sim \left|{S}\right|$.

Similarly, $T \sim \left|{T}\right|$.

Sufficient Condition
Suppose $S \sim \left|{S}\right|$ and $T \sim \left|{T}\right|$.

Let $f: S \to \left|{S}\right|$ and $g: T \to \left|{T}\right|$ be bijections.

By definition of set equivalence, these bijections are known to exist.

Define the mapping $F$ to be:


 * $F \left({x}\right) = \begin{cases}

f \left({x}\right) & : x \in S \\ \left|{S}\right| + g \left({x}\right) & : x \notin S \land x \in T \end{cases}$

It follows that $F: S \cup T \to \left|{S}\right| + \left|{T}\right|$ is an injection.

Therefore, $S \cup T$ is equivalent to some subset of the ordinal $\left|{S}\right| + \left|{T}\right|$.

By Condition for Set Equivalent to Cardinal Number, it follows that $S \cup T \sim \left|{S \cup T}\right|$.