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In this question, the term “real” implies an infinite binary sequence (a binary sequence of length $\omega$).

Assume that the theory “ZFC + there is a proper class of worldly cardinals” is consistent and use this theory as the background set theory. Additionally, assume that $V \ne L$.

Let $f(\alpha)$ denote the $\alpha$-th worldy cardinal. Let $t(\phi, \alpha)$ denote the truth value (either $0$ or $1$) of a statement $\phi$ (a finite formula of arbitrary complexity) in $V_{f(\alpha)}$.

Assuming that $\beta$ is either $0$ or a limit ordinal, let $r(\phi, \beta)$ denote a real encoded by the following sequence of bits: $$r(\phi, \beta) = t(\phi, \beta), t(\phi, \beta+1), t(\phi, \beta+2), \ldots$$

Let $S$ denote the set of reals such that for any $x \in S$ there exists a statement $\phi$ and an ordinal $\beta$ such that $r(\phi, \beta) = x.$

Question: does there exist a real $y$ which is not an element of $S$? If no, why? If yes, is it possible to prove it and maybe describe a particular example of such real?

lyrically wicked
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  • At the very least, we shouldn’t be able to show all reals can be written as $r(\phi, \alpha)$. If we could, we’d have a definable well-order on the reals. – Mark Saving Jul 09 '22 at 05:16
  • @MarkSaving: I did not claim that all reals can be written as $r(\phi, \alpha)$. I am asking for an example of a real that cannot be encoded this way. – lyrically wicked Jul 09 '22 at 05:36

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Let’s suppose $V = L$, which is relatively consistent with a proper class of worldly cardinals. We know that $L$ is definably well-ordered, so take the smallest bijection $f : 2^{\aleph_0} \to \{f : \omega \to 2\}$ according to this well-ordering.

Then consider the following statement:

There is a set of worldly cardinals. Let $\beta$ be the ordinality of the set of worldly cardinals. Then we can write $\beta$ uniquely as $\beta = 2^{\aleph_0} \cdot \gamma + \delta$ for some $\delta < 2^{\aleph_0}$. Furthermore, we have $\delta < \omega$ and $\gamma < 2^{\aleph_0}$. And the $\delta$th element of $f(\gamma)$ is $1$.

We denote the above statement as $\phi$. Note that $r(\phi, 2^{\aleph_0} \cdot \alpha) = f(\alpha)$. Therefore, all reals can be written as $f(\phi, \beta)$ for some $\beta$. $\square$

The above demonstrates that we cannot prove the existence of a real which is not of the form $f(\phi, \beta)$.

Now suppose we could show that all reals are written in the form $f(\phi, \beta)$. Then we could define a well-order on the reals. But I suspect that even with a proper class of worldly cardinals, we could not produce a definable well-ordering of the reals.

So it appears this question is independent of your axioms.

Mark Saving
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  • Yes, I know that the axiom of constructibility allows to define a well ordering of the reals. But my question is not based on the assumption that $V=L$, so I am interested in the case $V \ne L$. – lyrically wicked Jul 09 '22 at 05:50
  • And I should clarify: if not all reals can be encoded by some $r(\phi, \alpha)$ (assuming $V \ne L$), then I am asking for an abstract description of a particular real that is not encoded by some $r(\phi, \alpha)$ (if it is possible to provide such a description). – lyrically wicked Jul 09 '22 at 06:19
  • @lyricallywicked So you are asking for some $\xi$ such that under the assumptions of ZFC + a proper class of worldly cardinals, we can prove there is a unique real $x$ such that $\xi(x)$, but we cannot prove that $x$ can be encoded as $r(\phi, \alpha)$? – Mark Saving Jul 09 '22 at 15:42
  • What is “$\xi(x)$”? – lyrically wicked Jul 11 '22 at 02:47