Let M be an uncountable standard model of ZFC, let $\frak c$ be the cardinality of the continuum, and let (just for the sake of argument) $\mathfrak c=\aleph_2$. If one assumes M has 'all' the ordinals, then $L$ is the smallest inner model of ZFC [have I made any mistake here?] and $L \subset M$. Now according to Cohen, $M$ contains all the countable ordinals so if one restricts one's attention to $L$, $M$ contains all the constructible reals, [is this a faulty reinterpretation of Cohen's assumption of $V=L$?]. Also, if my reasoning is correct, the constructible reals form a continuum of size $\aleph_1$, making it (if one restricts oneself to L) a 'subcontinuum'of M's continuum of $\aleph_2$ (if this line of reasoning is correct, of course...). Is this line of reasoning correct as it stands, and if so, can the 'L--people' discover that the 'actual' continuum is of size $\aleph_2$?


(Apparently my previous answer was bad, for some cryptic reason, so I've written it from scratch.)

Let's make some order in the things. We have the universe $V$ and $L$ which is the "real" $L$, and $\sf Ord$ is the class of ordinals. Whenever we add a superscript $M$, like $L^M$ or $\mathsf{Ord}^M$ we mean the interpretation of these in a model $M$. Since $L$ is absolute for transitive models, writing $\omega_1^L$ and $\omega_1^{L^M}$ is the same thing, and we can safely use the former notation.

If $M$ is an uncountable transitive model of $\sf ZFC$, then $M\cap\sf Ord=\alpha$ for some uncountable ordinal $\alpha$. It follows that $L_\alpha\subseteq M$, and moreover $L_\alpha=L^M$. In that case indeed every constructible real is in $M$.

Assume that $|M|=\aleph_1$. Then we have several cases.

  1. $\omega_1^M=\omega_1^L$, or even better $V$ and $L$ agree on all the cardinals (but not on their exponentiation, as you postulated $2^{\aleph_0}=\aleph_2$). The first thing to note that is that $\alpha>\omega_1$. The reason is that $M$ would recognize every countable ordinal as countable (because it is countable in $L$).

    If we assume, for example, that $\sf MA$ holds in $V$, then the argument for the existence of generic filters over $M$ for Cohen forcing follows the same way as it would in the case of countable models. Now we can blow up the continuum inside of $M$, and have $M\models\lnot\sf CH$.

  2. $\omega_1^M>\omega_1^L$, so $M$ knows that $\omega_1^L$ is a countable ordinal in $V$, and in which case it is also aware of the fact that $\Bbb R^L$ is nothing but a countable set.

    So it is possible that $M$, while a model of $\sf CH$, or not, is aware that $L$ only has a countable set of real numbers, and therefore $\Bbb R^L$ is not a subcontinuum (even not when considering only constructible subsets of it) of $\Bbb R^M$ of size $\aleph_1^M$. Because it is not of size $\aleph_1^M$.

So we can have either case along with $M\models\lnot\sf CH$. So it is impossible to conclude that $\Bbb R^L$ is a subcontinuum of size $\aleph_1^M$ of $\Bbb R^M$, simply because it is possible that it is a countable set.

In either case the "L--people" don't know what is beyond their universe. The question itself is meaningless, as the second case shows.

Finally, let me point out there is an inherent ambiguity in the term "subcontinuum" (in the definition given in the comments). You say that a continuum is a Dedekind-complete linear order. But as you note there is a question of what sets one asks for when interpreting the second-order completeness axiom. And from $M$'s point of view if $N\subseteq M$ and $\Bbb R^N\subsetneq\Bbb R^M$, then $\Bbb R^N$ is not a subcontinuum of $\Bbb R^M$, from the point of view of $M$ itself.

  • $\begingroup$ Thomas, \aleph and not \Aleph, also keep a space between \mathbb and R, similarly braces {} on the exponents would solve the problem. $\endgroup$ – Asaf Karagila Dec 5 '13 at 22:55
  • $\begingroup$ @ Asaf: Thanks. $\endgroup$ – Thomas Benjamin Dec 5 '13 at 22:57
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    $\begingroup$ Since I have no idea what's wrong with the answer, a comment accompanying the downvote would be extremely helpful. $\endgroup$ – Asaf Karagila Dec 9 '13 at 0:05
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    $\begingroup$ @Thomas: If $N$ is a class of $M$, then yes. $M$ knows that $N$ is a class, and for every axiom of $\sf ZFC$, $M$ knows that $N$ satisfies that axiom; and so $M$ knows what are the real numbers of $N$. It might be that $N$ is not a class of $M$, so the metatheory knows about it, but not $M$ itself. I don't recall any context where that happened, though, so I can't really give you an example off the top of my head when that happens. In the case where $N\models V=L$, then clearly $N$ is a class of $M$. $\endgroup$ – Asaf Karagila Dec 12 '13 at 10:48
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    $\begingroup$ @Thomas: As Carl said, ordinary mathematics is rarely concerned with its set theoretical assumptions. $\endgroup$ – Asaf Karagila Dec 12 '13 at 19:52

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