Intuitive explanation of Cohen's forcing (continuum hypothesis) and how Godel proved CH was consistent within ZFC?

Question

I'm new to set theory and am really struggling to get my head around this. Can anyone give me an intuitive explanation so I can get a general grasp? Cheers

Answer 1

When you want to prove a relative independence result ("If $T$ is consistent, then $T$ doesn't prove $\varphi$") there are essentially two ways to do it:

• Analyze the structure of $T$-proofs as combinatorial objects in their own right. An example of this is the usual (Gödel-Rosser) proof of the second incompleteness theorem, that no recursively axiomatizable consistent theory extending PA can prove its own consistency (and in particular, Rosser's improvement on Gödel's original argument was removing the last vestige of an appeal to models).

• Build a model of $T$ in which $\varphi$ fails (and then apply the soundness theorem). For example, to prove that the theory of groups doesn't prove $\forall x,y(x*y=y*x$), it suffices to exhibit an example of a nonabelian group, together with a proof that it is in fact a nonabelian group.

Generally, the second approach is much easier, and this is what Gödel and Cohen each did - although models of ZFC are extremely complicated objects, so rather than build appropriate models "from scratch" they showed how appropriate models could be constructed given "starting models" of ZFC (which is fine - we're assuming that ZFC is consistent to begin with, and that means we can apply the completeness theorem). Interestingly, they went in opposite directions:

• Gödel showed that for any model $M$ of ZFC, there is a smaller structure $N\subseteq M$ which forms a model of ZFC+CH.

• Cohen showed that for any model $M$ of ZFC, there is a larger structure $M\subseteq N$ which forms a model of ZFC+$\neg$CH.

  • OK, that's not quite true: Cohen only showed that the above statement holds if $M$ is countable. But by the downward Lowenheim-Skolem theorem, if ZFC has a model then it has a countable model, so that's fine.

  • It's also worth pointing out that Cohen's method is much more flexible that Gödel's: Gödel's arguments cannot be used to show the consistency of ZFC+$\neg$CH, whereas Cohen's argument also lets us build models in which CH holds, so Cohen solves both parts of the problem while Godel solves only one. That said, Cohen's method doesn't make Gödel's obsolete, it's just superior for this particular problem.

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That describes what they're doing; now, how did they do it?

Gödel's construction (inner models) is much simpler (and so it's unsurprising that it was discovered earlier). Intuitively, we can define the constructible universe $L$ recursively by $$L_0=\emptyset,\quad L_{\alpha+1}=\mathcal{P}_{def}(L_\alpha),\quad L_\lambda=\bigcup_{\alpha<\lambda}L_\alpha\mbox{ for $\lambda$ limit}$$ where $\mathcal{P}_{def}(X)$ is the set of all subsets of $X$ definable in the structure $(X,\in)$, and let $$L=\bigcup_{\alpha\in Ord}L_\alpha.$$ It turns out that we can run this construction inside an arbitrary model of ZFC (indeed, ZF) and the result satisfies ZFC+CH (indeed, a lot more); roughly speaking, every element of $L$ is "definable" in a certain sense, so we can calculate $2^{\aleph_0}$ by counting the number of definitions of the appropriate type (although it's actually more complicated than that).

• The limitation of Gödel's argument mentioned above comes from the fact that $L$ is "minimal" in a precise sense - there's no hope in general that we can find a nice submodel of a given model in which CH fails, since maybe our starting model is $L$ itself.

Meanwhile, Cohen's method (forcing) is quite involved, and too complicated to explain here. The answers to this MSE question explain it a bit, but there's just too much to say to give a good explanation. This article by Chow may be helpful.

Answer 2

If you had an injection $$ j : \aleph_2 \hookrightarrow \mathcal{P}(\aleph_0),$$ then the continuum hypothesis would be false:

$$ \aleph_1 < \aleph_2 \leq \mathcal(\aleph_0)=\mathfrak c.$$

If you have a model of ZFC, and you smuggle something like $j$ into it, you get a model where you forced the continuum hypothesis to fail. But because ZFC's axioms are complicated, a lot can go wrong.

Here's a synopsis based on Timothy Chow's Beginner's guide to Forcing.

• Find a model $M$ where we can construct something like $j$. Specifically, a countable model.

Because of incompleteness, we can't construct a model of ZFC from scratch, so assume we have one. But because of Löwenheim-Skolem, we can assume it's countable. Cardinality is not absolute: $M$ will think sets are a different cardinality than they "really" are. For example, $M$ will think its copy of $\mathbb{R}$ is uncountable, but it's "really" countable. $M$ just doesn't see any of the bijections between $\mathbb{R}^M$ and $\mathbb N^M$.

Cohen turned this paradox into an advantage. Because $M$ is a countable model, its copies of $\mathcal P(\aleph_0)$ and $\aleph_2$, call them $\mathcal P^M(\aleph_0^M)$ and $\aleph_2^M$, are "really" countable. So an injection

$$j:\aleph_2^M \hookrightarrow \mathcal P^M(\aleph_0^M)$$

exists.

Because the powerset is relative (the model doesn't think all subsets "exist"), think instead about functions from $X\to 2$.

$$ A\to P(B) = \big(P(B)\big)^A = \big(2^B\big)^A = 2^{B \times A}. $$

Instead of our $j$, Chow uses an $$ F : \aleph_2^M \times \aleph_0^M \to 2 $$ with a pairwise-distinctness condition, $$ a\neq b \implies F(a,-)\neq F(b,-), $$ to capture injectivity.

• Create a Boolean-valued model

Instead of smuggling $F$ directly into $M$, Chow uses a Boolean-valued model. Some statements are independent from ZFC's axioms. Why pretend otherwise? Instead of assigning every statement true or false, what if we had some nuance? If we replace true and false with a boolean algebra $B$, we can add nuance while keeping the structure of reasoning. This lets us defer the choice of which independent statements should be true or false in our model. When we're ready to choose, we can quotient out by something that collapses $B$ into $\{\text{True},\text{False}\}$. This makes the choice which independent statements to accept and reject. The algebraic structure of $B$ ensures we do it consistently.

• Smuggling through quotienting

The easiest objects to study in passing from $M_B \to M_B/B$ are subsets of $B$. Instead of worrying about the nuances of ZFC, this is a purely boolean-algebraic problem. Because set membership in $M_B$ is fuzzy, sets in $M_B$ are "really" functions into $B$. The easiest nontrivial B-valued "subset" of $B$ is the identity map. Call it $I$ to emphasize we're thinking of it as a $B$-set.

$$ x\in I \xrightarrow[\text{interpret in }M_B]{} \text{id}(x)=x $$

If we quotient $B$ by some $U$, this becomes $$ x\in q(I) \iff q(X\in I) = \text{true} \iff q(x) = \text{true} \iff x\in U $$ where $q$ stands for the quotient map. But then $$ q(I)= U \in M_B/U$$.

So if we can ensure $M_B$ contains $B$ and $I$, we can force $M_B/U$ to contain $U$.

• Which $B$? Which $U$?

Because we're taking quotients of boolean algebras, there are algebraic constraints on $U$, so we can't make it $F$ on-the-nose. In particular, $U$ has to be an ultrafilter. But if $U$ contains the right ingredients, we can construct $F$ from inside $M_B/U$. Since we're building an $$ F : \aleph_2^M\times \aleph_0^M \to 2, $$ use a boolean algebra that's related to the set of such functions. Specifically, (Chow 13) uses the complete boolean algebra generated by the set of finite partial functions. To ensure we can make $F$ pairwise-distinct, there's an additional condition on $U$: it's a generic ultrafilter. I don't yet understand why that is the correct condition, but I think it's related to how the set of functions that agree at a given point are closed.