Talk:Universe

The set of all hereditary finite sets is a universe, but not a model of ZF (since ZF stipulates the axiom of infinity). Boris Tsirelson (talk) 20:33, 12 October 2017 (CEST)

That does indeed appear to be the case. There is a definitional choice: (0) allow the empty set to be a universe; (1) require a universe to have an element (equivalently to have the empty set as an element); (2) require a universe to have an infinite set as an element (such as the natural numbers). Allowing the hereditarily finite sets to be a universe makes $\aleph_0$ the first inaccessible cardinal. Richard Pinch (talk) 20:58, 12 October 2017 (CEST)

Yes. On Wikipedia, only uncountable cardinals are classified into accessible and inaccessible. I have no appropriate books on my shell now, thus I do not know, whether that is the consensus, or not. Boris Tsirelson (talk) 21:39, 12 October 2017 (CEST)

One book found: T. J. Jech "Lectures in set theory with particular emphasis on the method of forcing" Springer 1971. There, only uncountable cardinals may be called strongly inaccessible. Universe probably does not appear there. Boris Tsirelson (talk) 22:41, 12 October 2017 (CEST)

Just an observation... If I am not mistaken, Gödel constructive sets provide an example of a transitive model of ZF (even ZFC) but (possibly) not a universe. It happens because the first axiom of universe accepts arbitrary families. And nevertheless, on Wikipedia I see "Gödel's constructible universe". I guess, "universe" is a rather fuzzy idea, without a consensus about the definition (and on WP it is treated as fuzzy). Boris Tsirelson (talk) 22:18, 12 October 2017 (CEST)