it's called the axiom of extensionality because it defines extensional equality (two things are judged equal by their external=observable properties, rather than their origin or intrinsic identity)
I did some research before posting the video and found something to this effect. I wasn't really satisfied with the answer and didn't think I could explain it well to people unfamiliar with this, so I left that line in. For one, this explanation leaves open the question of why this is even called extensional equality in the first place. Also, the use of internal and external here is confusing in the case of sets because it's tempting to think of the axiom of extensionality as stating that sets are equal if they have the same internal properties (their elements). But here, we're clearly using internal and external differently, since we're saying sets are equal based on their external properties. All in all, I'm still confused about why the axiom is called that.
Is there a reason you didn’t mention the axiom of empty set and the axiom of pairing? I know you can drive them from the other axioms, but it’s a bit cumbersome.
More or less for that reason. I was trying to keep the statements of the axioms as simple as possible, so I didn't include axioms that are redundant. I intend to prove these statements later on in the series.
Your axiom of “refinement” is too inclusive. If it only allowed 1 object to map to 0-1 objects, you would indeed get specification + replacement axioms. But since it allows 1-to-many relations, it also entails the full axiom of choice. Take a family of nonempty sets. Take the relation F(x,y) to be “y is in x” Now apply the “axiom of refinement” and you get a set with one element from each set in the family. Which is exactly what the axiom of choice does.
The axiom of choice gives you a set with *exactly* one element from each set, but in the statement of the axiom of refinement, I say that "*at least one* of the objects it relates to lies in the new set". It's possible to prove refinement in ZF. The idea is to replace each set with the set of everything it relates to of minimal possible rank in the von Neumann hierarchy of sets (en.wikipedia.org/wiki/Von_Neumann_universe), then take the union. This gives a set satisfying refinement.
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your videos are very helpful!pls continue!
Set theory is so cool
Which book could you recommend that uses the same formulation? Do you have an idea how one could "implement" these axioms with Lambda calculus?
it's called the axiom of extensionality because it defines extensional equality (two things are judged equal by their external=observable properties, rather than their origin or intrinsic identity)
I did some research before posting the video and found something to this effect. I wasn't really satisfied with the answer and didn't think I could explain it well to people unfamiliar with this, so I left that line in. For one, this explanation leaves open the question of why this is even called extensional equality in the first place.
Also, the use of internal and external here is confusing in the case of sets because it's tempting to think of the axiom of extensionality as stating that sets are equal if they have the same internal properties (their elements). But here, we're clearly using internal and external differently, since we're saying sets are equal based on their external properties.
All in all, I'm still confused about why the axiom is called that.
Is there a reason you didn’t mention the axiom of empty set and the axiom of pairing?
I know you can drive them from the other axioms, but it’s a bit cumbersome.
More or less for that reason. I was trying to keep the statements of the axioms as simple as possible, so I didn't include axioms that are redundant. I intend to prove these statements later on in the series.
Your axiom of “refinement” is too inclusive.
If it only allowed 1 object to map to 0-1 objects, you would indeed get specification + replacement axioms.
But since it allows 1-to-many relations, it also entails the full axiom of choice.
Take a family of nonempty sets. Take the relation F(x,y) to be “y is in x”
Now apply the “axiom of refinement” and you get a set with one element from each set in the family. Which is exactly what the axiom of choice does.
The axiom of choice gives you a set with *exactly* one element from each set, but in the statement of the axiom of refinement, I say that "*at least one* of the objects it relates to lies in the new set".
It's possible to prove refinement in ZF. The idea is to replace each set with the set of everything it relates to of minimal possible rank in the von Neumann hierarchy of sets (en.wikipedia.org/wiki/Von_Neumann_universe), then take the union. This gives a set satisfying refinement.