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)

Oops! I am afraid, this definition is utterly stupid. Every universe must be empty. Here is why. If V is a nonempty universe, then it contains $\emptyset$ and $ I = \{\emptyset\} $. Being a set (not a proper class) it misses some $x$. I take $ X_i = x $ for the sole $ i \in I $, take the union and get $ x \in V $. Boris Tsirelson (talk) 22:03, 12 October 2017 (CEST)