Is $\frak c = \aleph_1$?

Asked Jun 07 '17 at 04:40

Active Jun 07 '17 at 04:40

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My textbook requires me to find out the cardinality of $I(\Bbb N)$, which is a set of all infinite subset of $\Bbb N$

What I found is cardinality of $I(\Bbb N)$ equals to $2^{\aleph_0} = \frak c$ then is it also okay to denote that $\frak c =\aleph_1$?

asked Jun 07 '17 at 04:40

Daschin

It is undecidable in ZFC, so my guess is that you are not supposed to say it. – Jun 07 '17 at 04:42
2

This is the Continuum Hypothesis. – Angina Seng Jun 07 '17 at 04:44
What's the definition of $\aleph_1$ and why do you think that $|I(\mathbb N)| = \aleph_1?$ If you don't think that why were you motivated to conclude $c=\aleph_1?$ As others have alluded to, whether or not that equality holds is a somewhat famous question and has been proven to have no easy answer. – spaceisdarkgreen Jun 07 '17 at 04:56

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