Union of Singleton

Theorem

Consider the set of sets $A$ such that $A$ consists of just one set $x$:

$A = \set x$

Then the union of $A$ is $x$:

$\bigcup A = x$

Proof

Let $A = \set x$.

From the definition of set union:

$\bigcup \set x = \set {y: \exists z \in \set x: y \in z}$

from which it follows directly that:

$\bigcup \set x = \set {y: y \in x}$

as $x$ is the only set in $\set x$.

That is:

$\bigcup A = x$

$\blacksquare$

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 $5.5. \ \text {(a)}$