# Talk:Cardinal number

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## 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)

- 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 . Boris Tsirelson (talk) 20:24, 10 January 2015 (CET)

- I think that at the very least an article in the area of set theory should say whether or not AC is being assumed, especially if assertions depend on it. But for this specific topic the article really needs to separate out what does and does not depend on AC. Richard Pinch (talk) 23:37, 10 January 2015 (CET)

- No one objects if you want to separate it out. But probably it is quite a piece of work. The article is not short... Also, many statements are provable from weaker forms of AC (such as Dependent Choice), but not provable with no AC at all. An example: the countable union of countable sets is countable. I guess, only experts in AC know exactly, how strange things look in the absence of any form of AC. Are you ready to this burden? Boris Tsirelson (talk) 07:55, 11 January 2015 (CET)

**How to Cite This Entry:**

Cardinal number.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Cardinal_number&oldid=36236