Math 2.0

statement axiom

Axiom of extensionality

Statement

Let AA and BB be sets. If AA and BB have the same elements, then A=BA = B: ∀A ∀B (∀x (x∈A↔x∈B)→A=B).\forall A \, \forall B \, \bigl( \forall x \, (x \in A \leftrightarrow x \in B) \to A = B \bigr).

Remarks

It is one of the axioms of Zermelo–Fraenkel set theory. The converse is a rule of logic: equal sets have the same elements. In practice, A=BA = B is often proved with A⊆BA \subseteq B and B⊆AB \subseteq A. See subset.

Explanations

L1 human

A set is nothing more than its elements. If two sets contain exactly the same things, they are the same set, however you describe them. For example, {x∈R∣x2=1}\{x \in \mathbb{R} \mid x^2 = 1\} and {−1,1}\{-1, 1\} are equal.

What this node needs for W2
  • edge about out: the concepts that the statement is about (0 of 1)
  • edge proves in: at least one proof node (0 of 1)
  • edge instance_of or illustrates in: one example that shows the statement at work (0 of 1)
  • explanation L2 (0 of 1)

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Sources

  1. Web https://en.wikipedia.org/wiki/Axiom_of_extensionality, lead section