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Is it possible to show that every cardinality has an ordinal with this cardinality (without the axiom of choice)? If so, how?

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The statement "every cardinality has an ordinal with this cardinality" is equivalent to axiom of choice (and hence is unprovable without it). It is so because, if you think about it a little, your statement is equivalent "every set has an ordinal of the same cardinality", and so "for every set there is a bijection between it and some ordinal" and thus "every set can be well-ordered", which is well-known to be equivalent to axiom of choice.

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I'm no expert, but I think no. IIRC, ordinals are linearly ordered without using any choice. If we could show your result, it would imply cardinals are linearly ordered, and if memory serves me, this implies the Axiom of Choice.

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  • $\begingroup$ You are absolutely right. Thank you. $\endgroup$ – availanche Jan 9 '16 at 17:56
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As suggested by Wojowu and Henno, there are two prime reasons why you cannot show that every cardinality has an ordinal with this cardinality without the use of AC:

  1. The principle of linearity of cardinals is equivalent to AC, and if every cardinality has an ordinal with this cardinality, then, given two cardinalities, there are two ordinals with those cardinalities, one is a subset of the other and hence there is injection between representatives of these cardinalities and hence these cardinalities can be compared (thanks to Henno for this answer).

  2. If A is partially ordered set and there exists bijection f:A->B for some set B, then f defines partial order on B with the same properties as that on A, therefore if you could find for every cardinality an ordinal with this cardinality, since every set is presumed to have cardinality then you could find bijection between a given set and some ordinal, ordering well this set. But the principle of ordering sets well is equivalent to AC.

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