Leibniz's law states that $$ \forall x\forall y[\forall P(Px \equiv Py) \rightarrow x = y]. $$ This seems awfully similar to the Axiom of Extensionality, $$ \forall A \forall B[\forall x(x\in A \equiv x \in B)\rightarrow A = B]. $$ Within the framework of set theory, are these logically equivalent? Is Extensionality a strictly stronger statement? Are they even comparable?
Asked
Active
Viewed 263 times
4
-
1Equivalent in what sense? To some extent, yes, because the only $P$'s you can create in the language of set theory are based off equality and $\in$ anyway. So by induction on the structure of $P$ you should be able to prove something like that. Nevertheless, Extensionality simple states that if two sets are different, then this can be detected via the $\in$ relation within the universe. – Asaf Karagila May 09 '17 at 17:32
-
2Possible duplicate of Replacing Axiom of Extensionality with a logical formalism – Pedro Sánchez Terraf May 10 '17 at 15:34
2 Answers
3
It depends on how you interpret Leibniz' law.
Leibniz' law follows from Extensionality if we allow predicates of the form "contains $x$" for arbitrary $x$. But arbitrary $x$ are not in general definable in any nice way, so these predicates wouldn't be expressibile in any nice way. To give an example of why that's a problem, consider the predicate "$=x$" - this is expressible in terms of $x$, and serves to distinguish any $y\not=x$ from $x$, since $x$ satisfies the predicate "$=x$" but such a $y$ doesn't satisfy the predicate "$=x$".
So a too-broad interpretation of what predicates can be trivializes Leibniz' law. But it is exactly that kind of broad interpretation which allows extensionality to imply Leibniz' law in general - there can be indiscernible elements which are different but can't be distinguished by any formula in whatever fixed system we're working in.
What about the other way around? This also fails, and a good example of this happening in an interesting situation can be provided by certain set theories with urelements. In these set theories, we start with a bunch of "atoms" (or "urelements") in addition to the empty set. These urelements may come with additional structure, algebraic or topological or whatever - for instance, we could start with a ring of urelements. In this case we may be able to distinguish urelements from each other in a nice, definable way. But any two urelements have the same elements, namely none.
So really, neither implies the other; the most that can be said is that a too-liberal interpretation of Leibniz' law is trivially true, and hence a fortiori follows from extensionality.
Noah Schweber
- 245,398
0
In first order logic without equality:
$$\forall x(x\in A \equiv x \in B)$$
Does not imply the following and vice versa:
$$\forall y(A\in y \equiv B\in y)$$
Edit 12.02.2019:
The provisio without equality is indispensable when we have pairing. When we have equality and pairing then ∀y(A∈y≡B∈y) implies A=B. Just use y={A}. And then A=B implies ∀x(x∈A≡x∈B). I didn't check which other ZFC axioms can do that in the presence of equality.
-
1It doesn't even imply it with equality. The point is that "$\in$" could be interpreted as an arbitrary binary relation: "$A$ and $B$ have the same 'parents'" and "$A$ and $B$ have the same 'children'" are simply unrelated in the absence of further axioms governing the binary relation in question. – Noah Schweber Feb 12 '19 at 14:33
-
1If you have equality and pairing then the second implies the first. See my edit. – Feb 12 '19 at 18:02