Set is Equivalent to Itself

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 $2.1$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.