Set is Equivalent to Itself
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Theorem
Let $S$ be a set.
Then:
- $S \sim S$
where $\sim$ denotes set equivalence.
Proof
From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ is a bijection from $S$ to $S$.
Thus there exists a bijection from $S$ to itself
Hence by definition $S$ is therefore equivalent to itself.
$\blacksquare$
Sources
- 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.7$. Similar sets
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.1$
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $1 \ \text{(a)}$
- 1999: András Hajnal and Peter Hamburger: Set Theory ... (previous) ... (next): $2$. Definition of Equivalence. The Concept of Cardinality. The Axiom of Choice: Theorem $2.1$