1 The Continuum Hypothesis

1398: Gödel showed $\mathrm{Con}(\mathsf{\text{ZFC}})⟹\mathrm{Con}(\mathsf{\text{ZFC}}+\mathsf{\text{CH}})$.

1962: Cohen showed $\mathrm{Con}(\mathsf{\text{ZFC}})⟹\mathrm{Con}(\mathsf{\text{ZFC}}+\neg \mathsf{\text{CH}})$.

We denote this by Hartogs’s aleph of $X$, $\aleph (X)$.

Using Axiom of Choice, well-order $P(X)$ and get ordinal $2^{|X|}$.

We have

Continuum Hypothesis (CH): ${\aleph }_1={\beth }_1$.

Generalised Continuum Hypothesis (GCH): $\forall \alpha {\aleph }_{\alpha }={\beth }_{\alpha }$.

Why “continuum”?

Reminder:

Gödel showed $\mathrm{Con}(\mathsf{\text{ZFC}})⟹\mathrm{Con}(\mathsf{\text{ZFC}}+\mathsf{\text{CH}})$.

Cohen showed $\mathrm{Con}(\mathsf{\text{ZFC}})⟹\mathrm{Con}(\mathsf{\text{ZFC}}+\neg \mathsf{\text{CH}})$.

Relative consistency proofs.

By Completeness Theorem, this means:

If there is $M\models \mathsf{\text{ZFC}}$, then I can construct $N\models \mathsf{\text{ZFC}}+(\neg )\mathsf{\text{CH}}$.

Analogy from algebra:

Axioms of fields: Fields.

Let ${\phi }_{\sqrt{2}}:=\exists x(x\cdot x=1+1)$. Note $\mathbb{Q}\models \neg {\phi }_{\sqrt{2}}$.

Idea: Start with $\mathbb{Q}$ and extend $\mathbb{Q}$ to get $F\models \text{Fields}+{\phi }_{\sqrt{2}}$.

Construct by algebraic closure (not in the usual sense – here we just mean adding in $\sqrt{2}$ and then adding everything else that this requires us to add).

Obtain $\mathbb{Q}(\sqrt{2})\models \text{Fields}+{\phi }_{\sqrt{2}}$.

This is easy because everything that matters (Fields and ${\phi }_{\sqrt{2}}$) is determined by equations; all formulas we need to check are atomic.

WHat if we have models of ZFC? Have

No function symbols nor constant symbols. So: almost nothing is atomic.

$M\subseteq N$ if and only if $M$ is an $L_{\in }$-substructure of $N$.

And: the formulas that we care about are definitely not atomic, but instead very complex.

Try to imagine a proof of:

If $M\models \mathsf{\text{ZFC}}$ then there is $N\supseteq M$ such that $N\models \mathsf{\text{ZFC}}+\mathsf{\text{CH}}$.

Without loss of generality $M\models \neg \mathsf{\text{CH}}$ (else we are done). For simplicity, let’s consider the case ${\beth }_1={\aleph }_2$.

What can we do to “get rid of ${\aleph }_1$”?

Maybe a surjection $f:\mathbb{N}\to {\aleph }_1$. Maybe we can form $M[f]\supseteq M$ to get a smallest model $M[f]\models \mathsf{\text{ZFC}}$.

Clearly, in $M[f]$, ${\aleph }_1^M$ is not a cardinal anymore.

Does that show CH?

All sorts of things can happen

Assuming it is actually possible to form this smallest model $M[f]$, there are lots of ways that this might not end up being enough to deduce CH. For example:

• Maybe ${ℝ}^{M[f]}\ne {ℝ}^M$

• Maybe ${\aleph }_2^M$ is not a cardinal either

A fundamental problem of non-absoluteness

Consider ${\phi }_{\varnothing }(x):=\forall z(z\notin x)$, which means “$x$ is empty”.

Consider $M\models \mathsf{\text{ZFC}}$. Therefore there are $e,e^{\prime }$ such that $M\models {\phi }_{\varnothing }(e)$ and $M\models \forall z(z\in e^{\prime }⟺z=e)$.

Consider $N:=M\setminus \{e\}$.

$N$ is an $L_{\in }$-substructure of $M$. But $N\models {\phi }_{\varnothing }(e^{\prime })$, even though $M\models \neg {\phi }_{\varnothing }(e^{\prime })$.

So ${\phi }_{\varnothing }$ is not absolute between substructures.

Instead of substructures, we will restrict out attention to transitive substructures: in particular, to $M$ transitive ($\forall x,x\in M⟹x\subseteq M$ or equivalently $x\in M\wedge y\in x⟹y\in M$) such that $M\models \mathsf{\text{ZFC}}$.

Next time: theorems about absoluteness for transitive substructures.

Clearly, if $\phi$ is absolute for $M$ and absolute for $N$, then it’s absolute between $M$ and $N$.

Cohen’s proof becomes:

If $M$ is a countable transitive set such that $M\models \mathsf{\text{ZFC}}$, then there is a a countable transitive set $N\supseteq M$ such that $N\models \mathsf{\text{ZFC}}+\neg \mathsf{\text{CH}}$.