An Axiom to Derive Continuum Hypothesis

Prerequisite. No more than a passing acquaintance with the cardinals of Cantor is required to read what follows. However, the notations here are a little clumsy since Internet still lacks math symbols.

Axiom of Combinatorial Sets. This blog entry is about the most fascinating problem I know in metamathematics, the Continuum Hypothesis (CH) of Cantor. CH states as a guess that \aleph_1, the cardinal next to \aleph_0, is equal to 2^\aleph_0. It is known that neither CH, nor its negation, can be derived from Zermelo-Fraenkel theory (ZF) and hence it is necessary to introduce an axiom into ZF, if we want to derive CH. The suggestion we want to make here is that the The Axiom of Combinatorial Sets (ACS) can serve as an appropriate axiom for the purpose. Here, by combinatorial set, \aleph_0 \choose \aleph_0, is meant the set of all subsets of \aleph_0 of cardinality \aleph_0. Click here for details.

Can we hope for a better axiom? It is easy to see that ACS can accomplish the job, but the real issue is whether we can have any hope of getting a simpler or more elegant axiom for the same purpose. In other words, is it possible to generate an infinite subset of 2^\aleph_0 substantially different from the combinatorial set which has cardinality less than or equal to that of 2^\aleph_0? Or, is there a simpler way to generate a superset of \aleph_0 from \aleph_0, other than the combinatorial set, whose cardinality is less than or equal to that of 2^\aleph_0?

A fact that may be of some interest here is that the value of, \aleph_0 \choose k, is respectively 1, \aleph_0, 2^\aleph_0, for k=0, finite, \aleph_0.

Internet Principia Mathematica. For a Principia Mathematica type proof of the equivalence of combinatorial set and power set of \aleph_0, see Metamath Proof Explorer (MPE). In my opinion, this MPE is indeed a spectacular achievement of mathematical logic since Principia Mathematica. For more of my comments, see my review.

For some: proof is music to their ears. Did you ever feel that there is something musical about mathematical proofs? If you did, you are absolutely right. To hear a musical version of the proof mentioned above click here (courtesy: Norm Megill).