Equality of Sets

Theorem

Let $S$ and $T$ be sets.

For $S$ and $T$ to be equal, $S$ must be a subset of $T$ and $T$ must be a subset of $S$:

$S = T \iff \left({S \subseteq T}\right) \land \left({T \subseteq S}\right)$

Thus, by definition, the relation is a subset of is antitransitive.

This can also be written as:

$S = T \iff \left({S \subseteq T}\right) \land \left({S \supseteq T}\right)$

or:

$S = T \iff S \subseteq T \subseteq S$

Proof

• First, suppose $S = T$. Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle S = T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in S \iff x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of set equality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in S \implies x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Material Equivalence
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle S \subseteq T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a subset

Similarly:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle S = T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in S \iff x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of set equality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in T \implies x \in S}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Material Equivalence
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle T \subseteq S\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a subset

Thus we have:

$S = T \implies S \subseteq T \land T \subseteq S$.

• Now, suppose $S \subseteq T \land T \subseteq S$.

First:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle S \subseteq T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in S \implies x \in T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a subset

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle T \subseteq S\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x \in T \implies x \in S}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a subset

Thus by definition of Material Equivalence:

$\left({S \subseteq T \land T \subseteq S}\right) \implies \left({x \in S \iff x \in T}\right)$.

and so by definition of set equality, $S = T$.

• So, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S = T\)	\(\implies\)	\(\displaystyle S \subseteq T \land T \subseteq S\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \subseteq T \land T \subseteq S\)	\(\implies\)	\(\displaystyle S = T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S = T\)	\(\iff\)	\(\displaystyle S \subseteq T \land T \subseteq S\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Material equivalence

Hence our result $S = T \iff S \subseteq T \land T \subseteq S$.

$\blacksquare$

Also see

The Axiom of Extension can be seen to be equivalent to:

$S = T \iff \left({S \subseteq T}\right) \land \left({T \subseteq S}\right)$

Applications

When determining the equality of sets, a standard technique is to determine whether or not they are subsets of each other.

Sources

• Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts (1951): Introduction $\S 1$
• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 1$: The Axiom of Extension
• W.E. Deskins: Abstract Algebra (1964): $\S 1.1$: Theorem $1.1$
• Steven A. Gaal: Point Set Topology (1964)... (previous)... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
• 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): Introduction
• Ian D. Macdonald: The Theory of Groups (1968): Appendix
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 3$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1$
• Gary Chartrand: Introductory Graph Theory (1977): Appendix $\text{A}.1$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 6.3$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.1$: Exercise $1.1.2$
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.1$