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I've read a few different presentations of Cohen's proof. All of them (that I've seen) eventually make a move where a Cartesian product (call it CP) between the (M-form of) $\aleph_2$ and $\aleph_0$ into {1, 0} is imagined. From what I gather, whether $CP \in M$ is what determines whether $\neg$CH holds in M or not such that if $CP \in M$ then $\neg$CH.

Anyway, my question is: Why does this product, CP, play this role? How does it show us that $\aleph_2 \in M$ (the 'relativized' form of $\aleph_2$, not the $\aleph_2 \in V$)? Could not some other set-theoretical object play the same role?

Asaf Karagila
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djkern
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2 Answers2

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I'm not exactly certain what you are asking, but I'll take a stab anyway.

The basic idea of Cohen's proof (or at least modern interpretations of it) are to adjoin to the ground model a family of $\aleph_2^M$ new reals. Since a real is nothing more than a function $\omega \to 2$, a family of $\aleph_2^M$-many reals may be thought of as a function $\omega_2^V \times \omega \to 2$. (Recall that there is a natural bijection between $C^{A \times B}$ and $( C^B )^A$, and so a function $f : \omega_2^M \times \omega \to 2$ may be thought of as an indexing of $\omega_2^M$-many reals $\{ f_\alpha : \alpha < \omega_2^M \}$ where $f_\alpha ( n ) = f ( \alpha , n )$.

If we can generate such a function, we are left with two very important steps:

  1. Show that if $\alpha \neq \beta$, then $f_\alpha \neq f_\beta$, which is a standard genericity argument. Once this is done we know that there are at least $\aleph_2^M$-many reals in the extension $M[G]$.
  2. Show that $\aleph_2^M$ is still the second uncountable cardinal in the generic extension. If it is, then we know that $M[G] \models \text{"}| \mathbb{R} | \geq \aleph_2 \text{"}$, and therefore CH is false in $M[G]$. This step is usually handled by combinatorial properties of the Cohen forcing (i.e., that it has the countable chain condition).

(Of course, that $\aleph_2$ is used is not strictly important, and the methodology of the proof goes through for "almost all" uncountable ordinals.)

user642796
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    Nice answer, though I somehow doubt that the second uncountable ordinal in $M[G]$ is $\omega_2^M$. :-) – Brian M. Scott Dec 03 '12 at 22:41
  • In his original proof, did Cohen assume the consistency of $ZFC$ to obtain $M$? – Rudy the Reindeer Apr 13 '13 at 09:51
  • Alright, after looking at this link pointed out to me by Asaf it looks as if he assumed the consistency of ZF only. – Rudy the Reindeer Apr 13 '13 at 10:18
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    @MattN. While a large part of Cohen's demonstration in his Set Theory and the Continuum Hypothesis involved working under the assumption that "$\mathsf{ZF(C)}\text{ has a transitive standard model}$," Cohen also noted that this can be avoided by using now-standard techniques. (Either working purely syntactically, or by taking countable transitive models of large enough finite fragments of $\mathsf{ZF(C)}$.) – user642796 Apr 13 '13 at 10:49
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In order to prove the continuum hypothesis is independent from the axioms of $ZFC$ what Cohen did was to start with $ZFC+V=L$ (in which the generalized continuum hypothesis holds), and create a new model in which $ZFC$ in which the continuum hypothesis fails.

First we need to understand how to add one real number to the universe, then we can add $\aleph_2$ of them at once. If we are lucky enough then $\aleph_1$ of our original model did not become countable after this addition, and then we have that there are $\aleph_2$ new real numbers, and therefore CH fails.

To add one real number Cohen invented forcing. In this process we "approximate" a new set of natural numbers by finite parts. Some mysterious create known as a "generic filter" then creates a new subset, so if we adjoin the generic filter to the model we can show that there is a new subset of the natural numbers, which is the same thing as saying we add a real number.

We can now use the partial order which adds $\aleph_2$ real numbers at once. This partial order has some good properties which ensure that the ordinals which were initial ordinals (i.e. cardinals) are preserved, and so we have that CH is false in this extension.

(I am really trying to avoid a technical answer here, and if you wish to get the details you will have to sit through some book and learn about forcing. I wrote more about the details in A question regarding the Continuum Hypothesis (Revised))

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Asaf Karagila
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