Equivalence of Definitions of Set Equality

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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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