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I have a (seeming) contradiction and I can't seem to figure out the (obvious) mistake in my reasoning. The following (related to my previous question):

(1) $\omega$, defined to be the least infinite ordinal, by definition has the same cardinality in every model of set theory (model of ZF or ZFC I'm not sure it's true in every theory), namely it is countably infinite. I think one can prove the countability by showing that its cardinality is less equals the cardinality of every infinite set (give an injection and apply Cantor-Schroeder-Bernstein). Then use the existence of a countably infinite set which follows from the axiom of infinity.

(2) We can force $\omega$ to be finite! To this end use the Levy collapse to collapse $\aleph_0$ to $42$. Let's call this new model $M[G]$.

Now, presumably, $M[G]$ is no longer a model of ZF(C).

Questions:

1.Is this correct? Is $M[G]$ no longer a model of ZFC?

2.And if yes: why not? Because it doesn't have an element representing the natural numbers?

2.a)If yes: how do I know that it doesn't?

3.And more generally: how do I know, after applying forcing to a model $M$ of ZFC, whether the resulting model $M[G]$ is still a model of ZFC or not?

Thanks for your help!

Glorfindel
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Rudy the Reindeer
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    In your previous question, you showed that $\omega$ has the same cardinality in every transitive model of ZF(C). It's worth noting that in ZF, $\omega$ isn't necessarily injectable into every infinite set (look up Dedekind-finite, or D-finite, sets for more on this). To show that $\omega$ is countable depends on how you define "countable". If you define it, for example, as "injectable into $\omega$", this is trivial to show. – Cameron Buie Nov 04 '12 at 10:18
  • Dear @CameronBuie, excellent point. I have been thinking about this, actually. And I was going to post a question about how to define countable. – Rudy the Reindeer Nov 04 '12 at 10:29
  • @Matt: Please don't. There are like 10 questions about the definition of countable on this site as it is. – Asaf Karagila Nov 04 '12 at 14:43
  • @AsafKaragila Can you point me to these 10 questions, I can't seem to find what I'm looking for. That would be very kind of you! – Rudy the Reindeer Nov 04 '12 at 18:52
  • One, two, three, and four. And one you might want to read anyway. All this and I'm not counting the endless repetitions and discussions in other questions about countability of $\mathbb N\times\mathbb N$ and so on. – Asaf Karagila Nov 04 '12 at 19:07
  • @AsafKaragila Thank you but none of these answer my question. – Rudy the Reindeer Nov 05 '12 at 09:28
  • Matt, all of these deal with the definition of countable. If you wish to discuss internal vs. external definitions, that you can find in the last linked post as well in pichael (the OP of that last link) other questions. – Asaf Karagila Nov 05 '12 at 10:14

3 Answers3

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Let us see what the definition of the Levy collapse of $\kappa$ to $\omega$ is:

The conditions are finite functions from $\omega$ to $\kappa$, and we say that $p$ is stronger than $q$ if $q\subseteq p$. Of course when I say a finite function I mean a partial function from a finite subset of $\omega$.

The generic filter, if so, adds a surjection from $\omega$ onto $\kappa$, and therefore it adds an injection from $\kappa$ into $\omega$, and thus $\kappa$ is now countable.

Why doesn't that work for $42$? Well, you can't extend functions from $42$ indefinitely. One of the keys to understanding the idea here is that we can always extend any function to add another element to the range. However for $p\colon 42\to\omega$ you quickly find yourself stuck. If you covered the entire domain then you cannot extend it anymore. The generic set, if so, cannot add any new function which is surjective from $42$ to $\omega$.

For the second question, this is a nontrivial theorem by Cohen from when he developed forcing: If $M$ was a countable transitive model of ZFC, $P$ was a forcing in $M$ and $G$ was $M$-generic over $P$ then $M[G]$ is also a countable transitive model of ZFC.

One can either verify the axioms, one by one, or use Boolean-valued approach and see that the axioms of ZFC all have Boolean value of $1$, and therefore must hold in the generic extension, whereas other sentences might not have Boolean value of $1$ and can be taken to be false.

Rudy the Reindeer
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Asaf Karagila
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I am not sure what you mean by (1). For every model M of ZF, M thinks that "the cardinality of $\omega$ is $\aleph_0$", by definition.

But there are models M such that the set of all m with "M thinks m in omega" is uncountable.

(2) is just false. You cannot collapse $\omega$ to 42. You can only collapse infinite cardinals to infinite cardinals.

Martin
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In the sequel I am assuming that your Levy collapse consists of all partial functions $p : 42 \to \omega$ of size $< 42$.

Note that the important point about the Levy collapse in general is that the correct subsets are dense. When collapsing $\omega_1$ to $\omega$ we use conditions which are finite functions $\omega \to \omega_1$, and then we can prove that for each $\alpha < \omega_1$ the set $$D_\alpha = \{ p \in \mathbb{P} : \alpha \in \mathrm{rng} (p) \}$$ is a dense subset of $\mathbb{P}$. By genericity is then follows that the generic object encodes/is a surjection $\omega^M \to \omega_1^M$.

If you attempt to collapse $\omega$ to $42$, then you will run into the problem that the sets $$D_n = \{ p \in \mathbb{P} : n \in \mathrm{rng} (p) \}$$ are not dense!

(Any function $p$ of size $41$ cannot be extended any further, and so picking any $n \notin \mathrm{rng} (p)$ we see that $p$ cannot be extended to a function in $D_n$.)

Would $M[G]$ be a model of ZFC? Of course, by the Forcing Theorem. But I believe that more is true: your forcing extension will not actually extend anything! Your generic object will become a function $g$ from a subset of $42$ of size $41$ into $\omega^M$, and since the family of these functions is absolute for transitive models of ZFC only the trivial extension will be made.

user642796
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