# Union of Disjoint Singletons is Doubleton

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## Theorem

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

Then:

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

## Proof

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.

$\blacksquare$

## Sources

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