Union of Disjoint Singletons is Doubleton

Theorem

Let $\set a$ and $\set b$ be singletons such that $a \ne b$.

Then:

$\set a \cup \set b = \set {a, b}$

Proof 1

Let $x \in \set a \cup \set b$.

Then by the Axiom of Unions:

$x = a \lor x = b$

It follows from the Axiom of Pairing that:

$x \in \set {a, b}$

Thus by definition of subset:

$\set a \cup \set b \subseteq \set {a, b}$

$\Box$

Let $x \in \set {a, b}$.

Then by the Axiom of Pairing:

$x = a \lor x = b$

So by the Axiom of Unions:

$x \in \set a \cup \set b$

Thus by definition of subset:

$\set {a, b} \subseteq \set a \cup \set b$

$\Box$

The result follows by definition of set equality.

$\blacksquare$

Proof 2

Straightforward from Union of Unordered Tuples.

$\blacksquare$

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 4$: Unions and Intersections