# How many cardinals exist between $\aleph_0$ and $\mathfrak c$? [duplicate]

Assuming the negation of the continuum hypothesis, there is a set $G_1 \subset \mathbb{R}$ such that if $\aleph^\prime_1$ is the cardinality of $G_1$, then $\aleph_0 < \aleph^\prime_1 < \mathfrak c$.

Do there exist models of set theory in which there exist another set, say $G_2$ such that $\aleph_0 < \aleph^\prime_1 < \aleph^\prime _2 < \mathfrak c$ with $\aleph^\prime _2$ the cardinality of $G_2$ and so on (that is, the cardinality of $G_2$ is bigger than that of $G_1$ and less than the cardinal of the set if real numbers)?

Is it possible to repeat this process and find sets with larger and larger cardinalities which still are less than $\mathfrak c$? If the answer is yes, what are the consequences of this assumption in analysis?

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• en.wikipedia.org/wiki/Easton%27s_theorem – Michael Greinecker Jan 4 '18 at 8:48
• It seems that Easton theorem you mentioned is about power sets (as it is well known, power set of $\mathbb N$ has cardinality of continuum). My question is about subsets of $\mathbb R$ which their cardinals are neither one of $\mathbb N$ nor $\mathbb R$. I want to know whether continuum hypothesis fails, how many cardinals exist between them @MichaelGreinecker – Humed Jan 4 '18 at 9:39
• @HamedPourmohammad The cardinality of the continuum is exactly determined by the number of smaller cardinals. – Michael Greinecker Jan 4 '18 at 11:47
• What this means is that the failure of CH tells us basically nothing about what $2^{\aleph_0}$ is - it just tells us that $2^{\aleph_0}\not=\aleph_1$. Essentially, besides the obvious restriction - that $2^{\aleph_0}$ has uncountable cofinality - ZFC can't prove anything about the size of the continuum. – Noah Schweber Jan 4 '18 at 15:20
• As a starting point, you might be happy with the following theorem: Suppose $M$ is a countable transitive model of ZFC, and $\kappa$ is an $M$-cardinal (that is, $\kappa\in M$ and $M\models$"$\kappa$ is a cardinal"). Then there is a forcing extension $N$ of $M$ such that the cardinals of $N$ are exactly the cardinals of $M$ and $N\models$ "$2^{\aleph_0}\ge\kappa$." Proof: add $\kappa$-many Cohen reals to $M$, and use the countable chain condition. – Noah Schweber Jan 4 '18 at 15:30

First, a quick comment: you've phrased your question in terms of $\vert\mathbb{R}\vert$ and sets of reals. However, it's much easier to talk instead in terms of powersets and cardinals $\le 2^{\aleph_0}$, and this gives the same result (since $\vert\mathbb{R}\vert=\vert\mathcal{P}(\mathbb{N})\vert$ and every cardinal $\le 2^{\aleph_0}$ is the cardinality of some set of reals and vice versa). That is, your question is just

How big can $2^{\aleph_0}$ be?

and this is how I treat it below.

It turns out that basically nothing can be said about $2^{\aleph_0}$ in ZFC alone. Specifically, while Konig's theorem tells us that $\mathfrak{c}$ must have uncountable cofinality, that's basically all that can be said. This was proved by Solovay, and later extended dramatically by Easton.

A very small case of Solovay's result is the following:

For any $n>0$, it is consistent that $2^{\aleph_0}=\aleph_n$.

For example, we can have $2^{\aleph_0}=\aleph_{17}$. Even more is possible however: we could have $2^{\aleph_0}=\aleph_{\omega+1}$, or $\aleph_{\omega_1}$, or even be weakly inaccessible.

The whole situation can best be summed up as follows:

Suppose $M$ is a countable (= so that we can actually force over it) model of ZFC, and $\kappa$ is an $M$-cardinal (= $\kappa\in M$ and $M\models$"$\kappa$ is a cardinal"). Then there is a forcing extension $N$ of $M$ such that $(i)$ the cardinals of $M$ are exactly the cardinals of $N$, and $(ii)$ $2^{\aleph_0}\ge\kappa$.

The proof is a straightforward forcing argument: let $G$ be generic over $M$ for the forcing $\mathbb{P}$ which adds $\kappa$-many Cohen reals (specifically, elements of $\mathbb{P}$ are finite partial maps from $\kappa\times\omega$ to $2$, and we order $\mathbb{P}$ by reverse extension). Let $N=M[G]$. We clearly have in $N$ an injection $\kappa\rightarrow\mathbb{R}^N$, and the fact that $Card^M=Card^N$ follows from the fact that $\mathbb{P}$ has the countable chain condition.

So in a precise sense, there is absolutely no ZFC-provable bound on the size of $2^{\aleph_0}$.

Finally, you ask about the possible impacts on analysis of a large continuum. To the best of my knowledge there are very few of these (and fewer still that aren't simply equivalents of $\neg$CH right off the bat). However, there are interesting set-theoretic propositions which imply a large continuum and which have meaningful analytic consequences: e.g. cardinal characteristic inequalities or forcing axioms. But in general, simply knowing that the continuum is large doesn't really tell you very much.

• The continuum being a real measurable cardinal has arguably a big impact on analysis. – Michael Greinecker Jan 4 '18 at 23:41
• @MichaelGreinecker Sure, but I didn't say there weren't any - just that there seem to be few. – Noah Schweber Jan 4 '18 at 23:50
• Point taken.$~$ – Michael Greinecker Jan 4 '18 at 23:50