Subset of Itself
From ProofWiki
Theorem
Every set is a subset of itself:
- $\forall S: S \subseteq S$
Thus, by definition, the relation is a subset of is reflexive.
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x \in S\) | \(\implies\) | \(\displaystyle x \in S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Law of Identity | A statement implies itself. | |
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle S\) | \(\subseteq\) | \(\displaystyle S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of subset |
$\blacksquare$
Sources
- Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 1$: The Axiom of Extension
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 1.2$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 1$
- Richard A. Dean: Elements of Abstract Algebra (1966): $\S 0.2$
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercises $\text{B i}$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 6.1$