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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 $\aleph_{\omega}$).2012-04-15
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    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.)2012-04-15
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    Well, you have the $R$ and $2^R$ are never bijective!2012-04-15
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    @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).2012-04-15
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    @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.2012-04-15
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    Azarel: Thanks. That's exactly what I was asking. I'll read up on the details.2012-04-15
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    I added LaTeX, I hope you don't mind I have changed A to $\kappa$.2012-04-15

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