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. Richard Pinch (talk) 18:50, 10 January 2015 (CET)