(Axiomatic Set Theory, 16) Proof of the Recursion Theorem for well-ordered sets. скачать в хорошем качестве

(Axiomatic Set Theory, 16) Proof of the Recursion Theorem for well-ordered sets.

We prove that we can give definitions on well-ordered sets. Note: After the proof (around 25:00 or so) I slightly misspoke. My intention was to emphasize that the current form of this recursion theorem requires you to state B (the set where your recursively defined sequence lives) before the recursion is actually performed. In situations where you're defining sequences of natural numbers, etc, this doesn't present an issue. However, sometimes it's not clear in advance that the set B exists until after you construct your sequence by recursion. What I was trying to say is that there are different forms of the recursion theorem that get around this concern because they don't require you to explicitly state B. Informally, given some definable way to generate your recursive sequence, you could prove that the codomains of the z-approximations will be sets by the Axiom of Replacement and so our set B would exist as well. This seemed needlessly technical in a video series giving an introduction to axiomatic set theory, so I opted to avoid stating such results. I mentioned "recursion on classes" (where a class is a collection of objects that don't form a set) but I think that was slightly misplaced. In situations, say, where you construct an ordinal-length sequence whose codomain is a proper class, the set B wouldn't exist (because it would be a proper class). However, in these settings, it would still be the case that the codomains of all the z-approximations would still be sets.