Equivalence of Definitions of Set Equality

Theorem

The following definitions of the concept of Set Equality are equivalent:

Definition 1

$S$ and $T$ are equal if and only if they have the same elements:

$S = T \iff \paren {\forall x: x \in S \iff x \in T}$

Definition 2

$S$ and $T$ are equal if and only if both:

$S$ is a subset of $T$

and

$T$ is a subset of $S$

Proof

Definition 1 implies Definition 2

Let $S = T$ by Definition 1.

Then:

$\ds S$

$=$

$\ds T$

$\ds \leadsto \ \ $

$\ds \leftparen {x \in S}$

$\iff$

$\ds \rightparen {x \in T}$

Definition of Set Equality

$\ds \leadsto \ \ $

$\ds \leftparen {x \in S}$

$\implies$

$\ds \rightparen {x \in T}$

Biconditional Elimination

$\ds \leadsto \ \ $

$\ds S$

$\subseteq$

$\ds T$

Definition of Subset

Similarly:

$\ds S$

$=$

$\ds T$

$\ds \leadsto \ \ $

$\ds \leftparen {x \in S}$

$\iff$

$\ds \rightparen {x \in T}$

Definition of Set Equality

$\ds \leadsto \ \ $

$\ds \leftparen {x \in T}$

$\implies$

$\ds \rightparen {x \in S}$

Biconditional Elimination

$\ds \leadsto \ \ $

$\ds T$

$\subseteq$

$\ds S$

Definition of Subset

Thus by the Rule of Conjunction:

$S \subseteq T \land T \subseteq S$

and so $S$ and $T$ are equal by Definition 2.

$\Box$

Definition 2 implies Definition 1

Let $S = T$ by Definition 2:

$S \subseteq T \land T \subseteq S$

First:

$\ds S$

$\subseteq$

$\ds T$

$\ds \leadsto \ \ $

$\ds \leftparen {x \in S}$

$\implies$

$\ds \rightparen {x \in T}$

Definition of Subset

Then:

$\ds T$

$\subseteq$

$\ds S$

$\ds \leadsto \ \ $

$\ds \leftparen {x \in T}$

$\implies$

$\ds \rightparen {x \in S}$

Definition of Subset

Thus by Biconditional Introduction:

$\forall x: \paren {x \in S \iff x \in T}$

and so $S$ and $T$ are equal by Definition 1.

$\blacksquare$

Sources

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