6

The Axiom of extensionality states that two sets are equal if they contain the same elements:

$\forall A \forall B [\forall x (x \in A \Leftrightarrow x \in B) \Rightarrow A = B]$

Can it be replaced by the statement that two sets are equal if they are contained by the same elements?

$\forall A \forall B [\forall x (A \in x \Leftrightarrow B \in x) \Rightarrow A = B]$

Roman
  • 523
  • Potential issues with infinity aside, this just doesn’t seem very practical - the axiom of extensionality implicitly suggests a strategy for proving set equivalence, whereas to prove set equivalence using the second statement you will need to prove a universal statement for an infinite number of possibilities – not quite as nice. – Chubby Chef Nov 27 '20 at 13:46
  • @ChubbyChef, why is that? In the first case we need to loop over all sets (x) and for each of them we need to check if it belongs to A and B. But we could say that instead of looping over all sets, we can just loop over sets belonging to A and B (it is what you are saying). But we can take this approach also for the second definition. Instead of looping over all sets and checking if they contain A and B, we can just "ask" A and B what sets they belongs to. In other words, we assume that sets "link" to their elements but we can also assume that elements "link" to sets containing them. – Roman Nov 27 '20 at 14:23
  • 1
    The question in the title is quite different from the question in the body. In the title you ask whether the second statement is also true, whereas in the body you ask whether the second statement can be used as a replacement for the axiom of extensionality. Which one do you want to ask? – Eric Wofsey Nov 27 '20 at 14:35
  • @EricWofsey, good point. Thank you for emphasising the difference. I would like to know the answers on both questions (if the statement is true and, if it is the case, if it can be used as a replacement of the extensionality axiom). – Roman Nov 27 '20 at 15:30

1 Answers1

6

The Axiom of Extensionality cannot be replaced by the statement that two sets are equal if they are contained by the same sets (this latter principle is a form of Leibniz's law, and I will refer to it by that name).

First, let me prove that, unlike Extensionality, $\forall A,B. (\forall x. A \in x \leftrightarrow B \in x) \rightarrow B = A$ follows from the other axioms of set theory, without invoking either Extensionality or Leibniz's law in the proof.

Take any two sets $A,B$, and assume that $\forall x. A \in x \leftrightarrow B \in x$. Define the set $\overline{A} = \{ y \in \mathcal{P}(A) \:|\: y = A \}$ using the Powerset axiom and the Separation schema. Clearly $A \in \overline{A}$, hence by our assumption $B \in \overline{A}$ as well. But by the defining property of $\overline{A}$, the following holds: $\forall y. y \in \overline{A} \leftrightarrow y \in \mathcal{P}(A) \wedge y = A$. In particular, since $B \in \overline{A}$, we can conclude $B = A$.

This proves Leibniz's law in set theory, and since the proof never invokes Extensionality, this implication holds even in set theory without Extensionality.

So all that remains to be shown is that the axiom of Extensionality is not a redudant axiom, i.e. that Extensionality, unlike Leibniz's law, does not follow from the other axioms of set theory. So we have successfully reduced your question to a previous Math.SE question.

Z. A. K.
  • 11,359