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

Then, if $x,y \in S$, then $f\left({x}\right) = f\left({y}\right)$, so since $f$ is a bijection, it follows that $x = y$.

If $x \in S,y \in T$, 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, if $x,y \in T$, then $\left\vert{ S }\right\vert + g\left({x}\right) = \left\vert{ S }\right\vert + g\left({y}\right)$.


 * $g\left({x}\right) = g\left({y}\right)$ follows by Ordinal Addition is Left Cancellable.


 * $x = y$ follows by the definition of a bijection.

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.