Union of Singleton

Theorem

Consider the set of sets $\mathbb S$ such that $\mathbb S$ consists of just one set $S$.

Then the union of $\mathbb S$ is $S$:

$\ds \mathbb S = \set S \implies \bigcup \mathbb S = S$

Proof

Let $\mathbb S = \set S$.

Then from the definition of set union:

$\ds \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$

from which it follows directly that:

$\ds \bigcup \mathbb S = \set {x: x \in S}$

as $S$ is the only set in $\mathbb S$.

That is:

$\ds \bigcup \mathbb S = S$

$\blacksquare$

Also see

• Intersection of Singleton

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 4$: Unions and Intersections
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Unions and Intersections
• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.2: \ \text{(i)}$
• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.5. \ \text {(a)}$