View source for Talk:Cardinal number

==Comparability of cardinals==
The article deduces from the Schroder–Berstein theorem ($\mathfrak{a} \le \mathfrak{b}$ and $\mathfrak{b} \le \mathfrak{a}$ implies $\mathfrak{a} = \mathfrak{b}$) that cardinals are totally ordered.  This seems wrong: all it proves is that $\le$ is indeed a partial order on cardinals.  That any two cardinals are comparable is, I believe, a form of the axiom of choice.  A similar assumption is made a little later when it is asserted that "Any cardinal number $\mathfrak{a}$ can be identified with the smallest ordinal number of cardinality $\mathfrak{a}$".  Again this requires that any set can be well-ordered.  [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 18:50, 10 January 2015 (CET)

:Maybe. But maybe all this article assumes all axioms of ZFC; this is the default, isn't it? I did not find in this article any discussion of what may happen without the choice axiom . [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 20:24, 10 January 2015 (CET)