Question about cardinality and CH

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Under the continuum hypothesis (CH), the cardinality of the set of cardinals less than aleph_1 is countable. This remains true regardless of whether CH holds, as the set is always countable. However, the cardinality of the set of cardinals less than 2^{aleph_0} can vary significantly if CH is false, potentially equating to any cardinal, including aleph_{666}. There are constraints on the possible cardinalities of the continuum, particularly regarding the cofinality of the continuum, which cannot be countable. The discussion highlights the complexities of cardinality under different assumptions in set theory.
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Under the continuum hypothesis, we readily see that that |{a < \aleph_1 : \textrm{a is a cardinal}}| = \aleph_0. What happens under the negation of CH? Is this equality still true or not? If the latter, always under the negation of CH, are there any infinite cardinals lambda for which the inequality |{a < \lambda : \textrm{a is a cardinal}}| = \lambda fails?
 
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First, what do you mean with |a<\aleph_1 : a \text{ is a cardinal}|. Do you simply mean the cardinality of the set \{a<\aleph_1 : a \text{ is a cardinal}\}??

In that case, it is always true that

\{a<\aleph_1 : a \text{ is a cardinal}\}=\mathbb{N}\cup \{\aleph_0\}

So wheter CH holds or not, this set is always countable.
In general we have that (by definition almost)

\{a<\aleph_\alpha : a \text{ is a cardinal}\}=\mathbb{N}\cup \{\aleph_\beta~\vert~\beta<\alpha\}.

For your second question, this is not always true. For example:

\{a<\aleph_2 : a \text{ is a cardinal}\}=\mathbb{N}\cup \{\aleph_0,\aleph_1\}.

But this is also countable, so |a<\aleph_2 : a \text{ is a cardinal}|=\aleph_0

Where CH does come into play, is with the set

|a<2^{\aleph_0} : a \text{ is a cardinal}|

If CH is true, then this is \aleph_0. But if CH is not true, then it can be any cardinal. E.g. it is consistent with ZFC that

|a<2^{\aleph_0} : a \text{ is a cardinal}|=\aleph_{666}

and so on if you replace 666 with any ordinal.

I hope this answer was helpful...
 
micromass said:
and so on if you replace 666 with any ordinal.

I don't think you can replace it with any ordinal. What about, e.g., \omega_1? Could that really be consistent?
 
But it is consistent with ZFC that 2^{\aleph_0}=\aleph_{\omega_1}. See Easton's theorem...
 
CRGreathouse said:
I don't think you can replace it with any ordinal. What about, e.g., \omega_1? Could that really be consistent?

I believe there are some very weak limits on possible cardinalities of the continuum - I think the cofinality of the continuum is uncountable, and so, for instance, 2^aleph_0 can't be aleph_omega.
 
Many thanks micromass, you're clarified a few things for me. :smile:
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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