On Lévy collapsing the reals

197 Views Asked by At

Consider the Lévy forcing notion. Let $M$ be a transitive standard model of $\mathsf{ZFC}$. Let $\aleph_n$ be the cardinality of the real numbers $2^\omega$ in $M$. Now collapse $\aleph_n$ to $\omega$. The resulting model $M[G]$ is again a model of $\mathsf{ZFC}$ but it is provable in $\mathsf{ZF}$ that $2^\omega$ is uncountable.

Could someone help me resolve this seeming contradiction? I'd be most grateful.

And I have a second question: Where can I access the original document containing the forcing notion explained? Or if that is not available: are there any other resources available? (I have a feeling that I might have been able to sort out my confusion on my own with more documentation available but Wiki and Jech are rather too concise for me.)

1

There are 1 best solutions below

7
On BEST ANSWER

Note that for every $\alpha<\aleph_{n+1}$ we add a bijection between $\omega$ and $\alpha$ to $M$. So in $M[G]$ all those ordinals are countable ordinals. So we added subsets of $\omega$ which encode these bijections (or the order type, if you prefer to think about that). So there are $\aleph_{n+1}$ new subsets to $\omega$ in $M[G]$.

Therefore we added $(\aleph_{n+1})^M$ real numbers to $M$, and changed $(\aleph_{n+1})^M$ to be $(\aleph_1)^{M[G]}$, because now whenever $\alpha<(\aleph_{n+1})^M$ we have that $M[G]\models|\alpha|=\aleph_0$. Therefore in $M[G]$ the least ordinal not in bijection with $\omega$ is $(\aleph_{n+1})^M$, which makes it $(\aleph_1)^{M[G]}$.

So we now have that $M[G]\models |2^\omega|=(\aleph_{n+1})^M=\aleph_1$.


As for reference request, I think that Kanamori's The Higher Infinite gives a nice exposition on the Levy collapse, as well one of its famous uses (constructing the Solovay model where all sets of reals are measurable).