Empty Set is Subset of All Sets/Proof 1
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Theorem
The empty set $\O$ is a subset of every set (including itself).
That is:
- $\forall S: \O \subseteq S$
Proof
By the definition of subset, $\O \subseteq S$ means:
- $\forall x: \paren {x \in \O \implies x \in S}$
By the definition of the empty set:
- $\forall x: \neg \paren {x \in \O}$
Thus $\O \subseteq S$ is vacuously true.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 3$: Unordered Pairs
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.3$. Intersection
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 6.4$: Subsets
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text A$: Sets and Functions: $\text{A}.1$: Sets
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Theorem $\text{A}.5$