# Can the real numbers be forced to have arbitrary cardinality?

Let $\mathbb R$ be the real numbers in a given model of set theory.

Given an arbitrary cardinal number $\kappa$, does forcing produce a larger model in which the cardinality of $\mathbb R$ is equal to $\kappa$? In particular, is there always (or ever) a model in which $\mathbb R$ is countable?

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The Lowenheim-Skolem Theorem guarantees the existence of a countable model, where of course $\mathbb{R}$ will be countable. However, $\mathbb{R}$ will not be countable-within-the-model. $\mathbb{R}$ always has cardinality $2^{\aleph_0}$ (within the model), and $2^{\aleph_0}$ can be almost any noncountable cardinal (a few are excluded by Koenig's Theorem, e.g., $\mathbb{R}$ cannot have cardinality $\aleph_{\omega}$). –  Arturo Magidin Apr 15 '12 at 5:04
Arturo: Thank you, but I don't think this answers my question. I am starting with a not-necessarily-countable model, and constructing the real numbers in that model. This gives me a set R. The question is whether there is a larger model in which the set R is countable. (Of course, if so, then in that larger model R is no longer the set of all real numbers.) –  WillO Apr 15 '12 at 5:19
Well, you have the $R$ and $2^R$ are never bijective! –  checkmath Apr 15 '12 at 5:25
@chessmath: Larger models may have "more" functions, so that the set we were calling $R$ and the set we were calling $2^R$ in model $M$ may be bijectable when we are working in a larger model $M'$ (the set we were calling $2^R$ wouldn't be the power set of the set $R$ in $M'$, though). –  Arturo Magidin Apr 15 '12 at 5:32
@WillO You can always make $\mathbb R$ countable via Levy collapse i.e., forcing with finite partial functions from $\omega$ into $\mathbb R$ ordered by reverse inclusion. –  azarel Apr 15 '12 at 5:36

Let us assume ZFC is our set theory.

Given a model of set theory, $M$, we always have that $M$ thinks $\mathbb R^M$ (what $M$ thinks is the real numbers) is of cardinality $2^{\aleph_0}$. It is a theorem of ZF that $\mathbb R\sim{\cal P}(\mathbb N)$, so from the internal point of view $M\models|\mathbb R| = 2^{\aleph_0}$. This, however, does not say much about what $\alpha$ we have $|\mathbb R|=\aleph_\alpha$.

Using forcing we can always ensure there is some $M\subseteq M'$ such that $M'\models|\mathbb R|=\aleph_1$, we may have added real numbers in the process, but we can ensure that in $M'$ this is the situation.

Now given $\kappa$ which is uncountable we only have one requirement from $\kappa$ that $\operatorname{cf}(\kappa)>\aleph_0$, we can now add $\kappa$ many real numbers via forcing to have $M\subseteq M'\subseteq M''$ such that $M''\models |\mathbb R|=\kappa$.

This covers all the cardinals except those with countable cofinality. Suppose now that we started with $M$ such that $M\models CH$, we may collapse $\aleph_1$ to be countable, namely find $M\subseteq M'$ such that $M'\models|\omega_1^M|=\aleph_0$. In the new model we have added real numbers, however externally we have that $M$ is a model in which there are only countably many real numbers.

I cannot give you a concrete answer about singular cardinals of countable cofinality, however I can make a remark about a symmetric extension (i.e. a submodel of a forcing extension):

Start with a model of CH, $M$, and collapse all the $\aleph_n$'s to be countable. We effectively added $\aleph_{\omega+1}$ many real numbers. So we have $M'$ such that $\aleph_{\omega+1}^M=\aleph_1^{M'}$.

Now there is a submodel, $M\subseteq N\subseteq M'$, in which the axiom of choice does not hold, where the continuum is a countable union of countable sets(!!). From $M'$ view (an external point of view) $N$ has $\aleph_\omega$ many real numbers. I believe that this argument can be done with every singular cardinal of countable cofinality, however I did not check that.

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