📂Set Theory

Extensionality Axiom

Axioms ¹

$$ \forall A \forall B ( \forall x ( x \in A \iff x \in B) ) $$ If every element of two sets $A$ and $B$ are the same, then the two sets are considered equal, which is denoted as $A = B$.

Explanation

On the other hand, if $A$ and $B$ are not the same, it is denoted as $A \ne B$.

The equality of two sets is an axiom as well as a definition itself. Extensionality does not mean extension but externality, implying that a set is distinguished not by vague descriptions like ‘a certain set’ but solely by the visibility and relation of its elements. Given that the concept of an element is ‘an object that is distinctively separated from each other as targets of our intuition or thought’, such an approach can be considered valid.

In simpler terms, the axiom of extensionality disregards the ’essence of elements’. To our intuition, $a$ and $A$ might be the same characters with just different case, but no matter what, if stated as $a=A$, they are the same, and if stated as $a \ne A$, they are viewed as different.

In this definition of ’equality,’ elements of a set do not have order or duplication. For example, the following equations hold: $$ \left\{ 9, 6, 0, 1, 2, 5 \right\} = \left\{ 0, 1, 2, 5, 6 ,9 \right\} $$

$$ \left\{ y, y, x, y \right\} = \left\{ y , x \right\} $$

$$ \left\{ 1, 5, 0, 4, 2, 1 \right\} = \left\{ 0, 1, 2, 4, 5 \right\} $$

Introducing the following useful theorem along with the relationship of set inclusion. This property is widely used throughout mathematics, but especially in pure mathematics such as abstract algebra or topology, it is nearly indispensable. As the axiom of extensionality itself seems too commonsensical, it is occasionally underestimated, but since it’s truly fundamental, let’s not do that.

Theorem

$$ A = B \iff A \subset B \land B \subset A $$

Proof

$$ \begin{align*} A = B \iff & \forall x ( (x \in A) \iff (x \in B)) \\ \iff & \forall x \left( ( x \in A \implies x \in B) \land ( x \in B \implies x \in A) \right) \\ \iff &A \subset B \land B \subset A \end{align*} $$

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