Difference between revisions of "Talk:Universe"
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The set of all [https://en.wikipedia.org/wiki/Hereditarily_finite_set hereditary finite sets] is a universe, but not a model of ZF (since ZF stipulates [[Infinity, axiom of|the axiom of infinity]]). [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 20:33, 12 October 2017 (CEST) | The set of all [https://en.wikipedia.org/wiki/Hereditarily_finite_set hereditary finite sets] is a universe, but not a model of ZF (since ZF stipulates [[Infinity, axiom of|the axiom of infinity]]). [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk: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. [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 20:58, 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. [[User:Richard Pinch|Richard Pinch]] ([[User talk: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. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 21:39, 12 October 2017 (CEST) | ||
Revision as of 19:39, 12 October 2017
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)
How to Cite This Entry:
Universe. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Universe&oldid=42056
Universe. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Universe&oldid=42056