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That is wrong or right about this question and answer?

Question: Is there a cardinality which is greater than the continuum?

Answer: Yes and No. If there is a Universe where a given cardinal kappa is greater than the size of the continuum, then there is a Generic-Extension of this Universe where the size of the continuum is greater than kappa.

Edit: This question is basically about the size of the continuum, which has been discussed several times on mathoverflow. It is also about the philosophical position of whether there is the reality of the multiverse.

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  • $\begingroup$ If there is no cardinality greater than the continuum, then what would the cardinality of the power set of the reals be? $\endgroup$
    – M Turgeon
    Nov 3, 2011 at 0:51
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    $\begingroup$ I think she's pointing out (asking?) that for any particular cardinal in your current model of ZFC, you can pass to an extension where the extension's continuum is bigger than this set (in the extension). Is this right? $\endgroup$ Nov 3, 2011 at 0:57
  • $\begingroup$ The point seems to be that modal operators like "it is provable in ZFC that" or "it is true in all forcing extensions of the universe that" don't commute with "there is". In particular, it is true that, "in all forcing extensions of the universe, there is a cardinal greater than the continuum", but it is false that "there is a cardinal that is, in all forcing extensions, greater than the continuum." My answer refers to the former (in the stronger version with "it is provable in ZFC that"), while Richard Rast's comment refers to the latter. $\endgroup$ Nov 3, 2011 at 13:41

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In ordinary set theory, "Yes" is right and "No" is wrong. Even after you generically extend the universe to make the cardinal of the continuum bigger than a given $\kappa$, there are plenty of other cardinals that are even bigger than your new continuum. As M Turgeon says, to avoid cardinals larger than the continuum, you'd have to revoke the axiom of power set.

Before he became a science fiction writer, Rudy Rucker did some work on the set theory that you get by revoking power set and adding Martin's axiom for all cardinals. That makes the continuum a proper class.

There has also been a little work on a set theory that allows only countably infinite sets, like what you'd get by generically collapsing all cardinals. But all these ways of getting a "No" answer to your question are far from the usual set theory and are probably best understood as being about some notion other than "set".

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  • $\begingroup$ Thank you for pointing out the problem with the question. And, the problem with the solution because it doesn't really address the question. Also, I see your out of the box way of answering "No". The question and answer are interesting to me as a pair. $\endgroup$
    – user10290
    Nov 3, 2011 at 1:48
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    $\begingroup$ a thorough answer on a question about a question and answer $\endgroup$ Nov 3, 2011 at 10:16