Union of Singleton
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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 $S$:
- $\displaystyle \mathbb S = \set S \implies \bigcup \mathbb S = S$
Proof
Let $\mathbb S = \set S$.
Then from the definition of set union:
- $\displaystyle \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly that:
- $\displaystyle \bigcup \mathbb S = \set {x: x \in S}$
as $S$ is the only set in $\mathbb S$.
That is:
- $\displaystyle \bigcup \mathbb S = S$
$\blacksquare$
Also see
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 $5.5. \ \text {(a)}$