# On The Explicit-Ness of $\omega_1$

$\omega_1$ is generally referred to as an ordinal when it is rather more of a formula $\phi$ true for some sets in models of ZFC.

I started to wonder if $\omega_1$ really was an ordinal, unlike $\beth_1$ (which is a cardinal in every model but isn't always the same cardinal; it is more easily described as a formula where $\phi(x)\Leftrightarrow x=\beth_1$). This question was originally hard to describe, but I now have an explicit definition that I feel represents the "Canonicality" of $\omega_1$:

Let a formula $\chi$ be $\phi$-Intact in a theory $T$ when:

$$\exists M(M\models T\land M\not\models\exists x(\phi(x)\land\chi(x)))\land\exists\chi_1(T\models(\exists x(\phi(x)\land\chi(x))\rightarrow\forall x(\chi(x)\Leftrightarrow\chi_1(x))\land\exists x(\phi(x)\land\chi(x))))$$

Broken down into several parts, there is some model $M$ of $T$ such that $M\models\neg\exists x(\phi(x)\land\chi(x))$. In othere words, $T$ cannot prove the existence of some $x$ such that $\phi(x)\land\chi(x)$. However, there is some other formula $\chi_1$ such that $T$ semantically entails that some $x$ with $\chi_1(x)\land\phi(x)$ exists and $T$ semantically entails "if some $x$ exists such that $\phi(x)\land\chi(x)$, then for every $x$ $\chi(x)\Leftrightarrow\chi_1(x)$".

This basically means: There are models $M$ of $T$ which do not contain any $x$ such that both $\chi(x)$ and $\phi(x)$ hold. However, there is some other formula $\chi_1$ such that in every model of $T$ there is some $x$ for which both $\chi_1(x)$ and $\phi(x)$ hold, and in the models of $T$ in which there is some $x$ such that both $\chi(x)$ and $\phi(x)$ both hold, in that model $\chi(x)\Leftrightarrow\chi_1(x)$ for every $x$.

Let $\psi(x)$ be true if and only if $x$ is a countable ordinal. The Canonicality of $\omega_1$ is equivalent to $\chi$ being $\psi$-Intact in ZFC for every $\chi$ such that:

• $\exists M\models\mathrm{ZFC}(M\models\exists x(\chi(x)\land\psi(x)))$
• $\exists M\models\mathrm{ZFC}(M\not\models\exists x(\chi(x)\land\psi(x)))$
• $\forall M\models\mathrm{ZFC}(M\models\exists x(\chi(x)\land\psi(x)))\rightarrow\forall x,y(\chi(x)\land\psi(x)\land\chi(y)\land\psi(y)\rightarrow x=y)$ (In all of the models of ZFC for which some countable ordinal $\alpha$ with $\chi(\alpha)$, there is only one such ordinal.)

If all of the previously mentioned bullet points hold for some $\chi$, $\psi$, and $T$, then one could generalize this by saying $\chi$ is a $\psi$-Undecidability in $T$. The previous bullet points holding for $\chi$ is equivalent to $\chi$ being a $\psi$-Undecidability in ZFC.

My conjecture is that $\omega_1$ is not canonical. In other words, there is a $\psi$-Undecidability in ZFC which is not $\psi$-Intact in ZFC.

This conjecture came about when researching the Proof-Theoretic ordinal of KPI. The Proof-Theoretic ordinal is not only proven to exist from ZFC but proven to be countable, without even assuming the consistency of KPI.

Assuming the consistency of a weakly inaccessible cardinal, there are models of ZFC in which $I_0$, the smallest weakly inaccessible cardinal, exists. In all of these models, the Rathjen collapse $\psi_\Omega(\varepsilon_{I_0+1})$. Consistency-wise this Rathjen collapse (when viewed as a formula) is as strong as a weakly inaccessible cardinal, but whenever it exists it is KPI (which always exists).

With this, I felt that, even though this Rathjen collapse isn't provably existent from ZFC, it is "canonical" in a sort; whenever it does exist it is equal to a Truly Canonical countable ordinal.

Of course, the immediate question is whether or not all such undecidable countable ordinal formulas $\varphi$ are canonical in the same way; and the falsity of that is my conjecture.

Assuming My Conjecture is True, one could "expand" $\omega_1$ by making larger ordinals which are not equivalent to any provably existant and countable ordinal. This would make $\omega_1$ bigger as adding Large Cardinal axioms makes $V$ bigger. It does not increase the cardinality of $\omega_1$ itself, but simply adds more ordinals. This could be a sort of 'antiforcing'; forcing is often used to make large sets smaller (like making an inaccessible $\omega_1$), but this could be used to literally add things to $\omega_1$ and keep it $\omega_1$. This could be generalized to any $\omega_n$ for finite $n$.

My Question is Whether or not there are any immediate proofs or disproofs of my Conjecture.

• $\omega_1$ and $\beth_1$ behave quite similarly. Both are really described by a formula $x=\omega_1$, $x=\beth_1$, and both are "cardinals/ordinals in every model, but not always the 'same' cardinal/ordinal". – Goldstern Oct 15 '17 at 12:03
• As a point of writing style, the use of two versions of $\phi$ ($\phi$ and $\varphi$) is pretty annoying. Is there a reason this is done? – Todd Trimble Oct 15 '17 at 12:04
• No there isn't. If you would like, I could change that, for a better reading time. – Keith Millar Oct 15 '17 at 15:16

CLAIM: A formula $\chi$ is $\phi$-intact in $T$ iff $T$ proves $\exists x: \phi(x)$ and moreover there is a model $M$ of $T$ satisfying $\lnot \exists x: \phi(x)\wedge\chi(x)$ (i.e., if $T$ does not prove the formula $\exists x: \phi(x)\wedge \chi(x)$).

Proof: It is clear that this property is necessary.

If $\phi$ and $\chi$ have this property, we can define $\chi_1(x)$ as $\chi(x) \vee \lnot\exists y:\phi(y)\wedge \chi(y)$.

• In those models where a $y$ satisfying $\phi\wedge \chi$ exists, $\chi_1(x)$ will be equivalent to $\chi(x)$, as demanded; the second clause in your second line will also be true because $\chi$ and $\chi_1$ are equivalent in such models.
• In those models where no such $y$ exists, the first clause is vacuously true, and the second clause $\exists x: \phi(x)\wedge \chi_1(x)$ will be witnessed by any element $x$ satisfying $\phi(x)$.

This ends the proof of my claim.

For your specific $\phi:=\psi$ and $T:=ZFC$, a formula $\chi$ is $\psi$-intact iff ZFC does not prove $\exists \alpha: \psi(\alpha)\wedge \chi(\alpha)$.

Hence every formula satisfying your second bullet point will be $\psi$-intact.

(Note that this argument uses almost nothing specific about ZFC or $\omega_1$.)

• Interesting... with this, if $\omega_1$ contains something which has an interesting property, even if it may be in only a few models, there is always an equivalence to the ordinal with that interesting property. – Keith Millar Oct 15 '17 at 15:25