# 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.

$\blacksquare$