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 $F : S \cup T \to \left|{S}\right| + \left|{T}\right|$ 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 \end{cases}$

Suppose $F \left({x}\right) = F \left({y}\right)$.

Let $x, y \in S$.

Then:
 * $f \left({x}\right) = f\left({y}\right)$

Since $f$ is a bijection, it follows that:
 * $x = y$

Let $x \in S, y \in T$.

Then:
 * $f \left({x}\right) = \vert S \vert + g\left({x}\right)$

But this is a contradiction, since $f\left({x}\right)$ has to be an element of $\left\vert{S}\right\vert$.

Finally, let $x, y \in T$.

Then:
 * $\left\vert{S}\right\vert + g \left({x}\right) = \left\vert{S}\right\vert + g\left({y}\right)$

From Ordinal Addition is Left Cancellable:
 * $g \left({x}\right) = g\left({y}\right)$

By the definition of a bijection:
 * $x = y$

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

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

Also see

 * Set Equivalent to Cardinal, which requires the axiom of choice.