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Let $f : \omega \to \{\text{wffs of set theory}\}$ be a function. Let $\leq_f$ be a total order on $\omega$.


Definition: $\langle f,\leq_f \rangle$ is a computable quasi-completion of ZFC if and only if

1. $f$ and $\leq_f$ are both computable
and
2. for all $m,n$ in $\omega$, if $m \leq_f \; n$, then all theorems of $ZFC+f(m)$ are theorems of $ZFC+f(n)$
and
3. for all sentences $s$ in the language of set theory, there exists a member $n$ of $\omega$ such that $ZFC+f(n)$ is consistent and one of $\{s,\lnot s\}$ is a theorem of $ZFC+f(n)$


Is it known that there is no computable quasi-completion of ZFC?


Basically, has the goal of Ultimate L been shown to be impossible?

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By 2., if for some $n$, ZFC+$f(n)$ is incosistent, then for all $m$ with $n\leq_f m$, ZFC+$f(m)$ is inconsistent as well. Hence there is an initial segment $A$ of $(\omega,\leq_f)$ that corresponds to the consistent theories of the form ZFC+$f(n)$. Now, isn't ZFC+$\{f(n):n\in A\}$ just a completion of ZFC? –  Stefan Geschke Feb 19 '11 at 5:05
    
I don't understand what this "goal" is you refer to. –  Andres Caicedo Feb 19 '11 at 5:06
    
I agree, however, that since the initial segment $A$ might not be computable there is no immediate contradiction to the imcompleteness theorems here. So you somehowuse the fact that we don't know when the inconsistency starts to avoid conflicts with the incompleteness theorems. Very clever, I would say. –  Stefan Geschke Feb 19 '11 at 5:11
    
Andres, I think what Ricky wants is a hierarchy of large cardinal axioms (linearly ordered) that might start to become inconsistent from some point of, but the maximal consistent initial segment gives you a complete extension of ZFC. And everything is computable, except for the consistent initial segment, of course. I am not so sure, however, whether this is accurately described as the "goal" of Ultimate L. –  Stefan Geschke Feb 19 '11 at 5:21
    
Although it had me convinced for a while, the argument I posted earlier was flawed and I deleted it. –  François G. Dorais Feb 20 '11 at 15:36
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1 Answer

up vote 4 down vote accepted

Here is something very close which might be adaptable to get a suitable $f$ and ${\leq_f}$.

Outline

Since there is nothing very special about ZFC in the construction, I will instead work with a consistent computably axiomatizable theory $T$ for which Goedel's First Incompleteness Theorem applies. Let $\phi_0,\phi_1,\ldots$ be a computable enumeration of all sentences in the language of $T$. I will assume that $\phi_0$ is your favorite contradiction, which I will also denote $\perp$.

I will define a computable function $f:\omega\to\omega$ and computable linear ordering ${\preceq}$ of $\omega$ by stages. At each stage $s$, I will decide the restrictions of $f$ and ${\preceq}$ on the set $\{0,1,\dots,t_s\}$. I will also keep track of a marker $m_s$ which will mark the ${\preceq}$-first element of $\{0,1,\dots,t_s\}$ for which it is known by stage $s$ that $T + \{\phi_{f(i)} : i \leq t_s \land i \preceq m_s\}$ is inconsistent. By the phrase known by stage $s$ I mean that $\perp$ appears by stage $s$ in the standard computable enumeration of the consequences of $T + \{\phi_{f(i)} : i \leq t_s \land i \preceq m_s\}$. This standard computable enumeration is allowed to enumerate finitely many consequences at each stage. I will also assume that this standard computable enumeration is such that:

  1. If $F$ is a finite set of sentences in the language of $T$, then all elements of $F$ appear at stage $0$ of the enumeration of the consequences of $T + F$.

  2. If $F \subseteq G$ are finite sets of sentences in the language of $T$, then all elements which appear by stage $s$ of the enumeration of the consequences of $T + F$ also appear by stage $s$ of the enumeration of the consequences of $T + G$.

The function $f:\omega\to\omega$ and the linear ordering ${\preceq}$ will have the following properties.

  1. Every $n \in \omega$ has either finitely many ${\preceq}$-predecessors or finitely many ${\preceq}$-successors. The dividing line is whether or not the theory $T + \{\phi_{f(i)} : i \preceq n\}$ is consistent.

  2. If $I$ is the set of all $n \in \omega$ which have finitely many ${\preceq}$-predecessors, then $T_I = T + \{\phi_{f(i)} : i \in I\}$ is a complete theory.

So the only missing requirement is that $i \prec j$ implies that $T \vdash \phi_{f(j)} \rightarrow \phi_{f(i)}$.

Construction

Stage $s = 0$ is trivial: $t_s = 0$, $f(0) = 0$, $0 \preceq 0$. (Note that $m_0 = 0$ since $\phi_0 = \perp$.)

At stage $s+1$, let $\ell_s$ be the ${\preceq}$-first element of the set $\{i \leq t_s : i \preceq m_s\}$ such that $T + \{\phi_{f(i)} : i \leq t_s \land i \preceq \ell_s\}$ is known to be inconsistent by stage $s+1$. Note that $\ell_s$ necessarily exists since $m_s$ is in the set being searched.

Case $\ell_s \neq m_s$: Then simply let $t_{s+1} = t_s$. (Note that $m_{s+1} = \ell_s$.)

Case $\ell_s = m_s$: Then consider the first element $n$ of the set $$\{0,1,\ldots,s+1\}\setminus\{f(i): i \leq t_s \land i \prec m_s\}$$ such that $$T + \phi_n + \{\phi_{f(i)} : i \leq t_s \land f(i) < n \land i \prec m_s\}$$ is not known to be inconsistent by stage $s+1$.

  • If there is no such $n$, then simply let $t_{s+1} = t_s$. (Note that $m_{s+1} = m_s$.)

  • Otherwise, set $t_{s+1} = t_s + 1$ and $f(t_{s+1}) = n$. Place $t_{s+1}$ immediately after the maximum element of $\{i \leq t_s : f(i) < n \land i \prec m_s\}$ in the ${\preceq}$-ordering.

Note that we always have that if $i \prec j \prec m_s$ and $i, j \leq t_s$ then $f(i) < f(j)$.

Verifications

First, observe that $\sup_{s<\omega} t_s = \omega$. Thus $f:\omega\to\omega$ is well-defined and ${\preceq}$ is a linear ordering of $\omega$.

Proof. Suppose, for the sake of contradiction, that $t_s = t_{s_0}$ for all $s \geq s_0$. Note that there are only finitely many stages $s \geq s_0$ such that $\ell_s \neq m_s$. So we can find a stage $s_1 \geq s_0$ such that $\ell_s = m_s$ for all $s \geq s_1$. Let $t = t_{s_0}$ and $m = m_{s_1}$. Note that $T' = T + \{\phi_{f(i)} : i \leq t \land i \prec m\}$ is necessarily consistent. Moreover, for every $n \notin \{f(i) : i \leq t \land i \prec m\}$, we must have that $T' + \phi_n$ is inconsistent and hence $T' \vdash \lnot\phi_n$. Since $\phi_0,\phi_1,\ldots$ enumerates all sentences of the language of $T$, it follows that $T'$ is a complete extension of $T$. However, $T'$ is computably axiomatizable and hence cannot be complete. QED

Next, observe that every $n \in \omega$ has either finitely many ${\preceq}$-predecessors or finitely many ${\preceq}$-successors. The dividing line is whether the theory $T + \{\phi_{f(i)} : i \preceq n\}$ is consistent or not.

Proof. First, suppose that $T + \{\phi_{f(i)} : i \preceq n\}$ is inconsistent. Then, there is some stage $s$ such that $n \leq t_s$ and $T + \{\phi_{f(i)} : i \leq t_s \land i \preceq n\}$ is inconsistent and this is known by stage $s$. It follows that $m_s \preceq n$ and after this stage no new elements will appear after $n$ in the ${\preceq}$-ordering.

Next, suppose that $T + \{\phi_{f(i)} : i \preceq n\}$ is consistent. For any stage $s$ such that $n \leq t_s$, the theory $T + \{\phi_{f(i)} : i \leq t_s \land i \preceq n\}$ is consistent and hence not known to be inconsistent by stage $s$. It follows that $n \prec m_s$ and hence that $f(i) < f(j) < f(n)$ for all $i \prec j \prec n$ such that $i,j \leq t_s$. Therefore, $n$ has no more than $f(n)$ predecessors in the ${\preceq}$-ordering by stage $s$. Since the bound $f(n)$ is independent of $s$, it follows that $n$ has no more than $f(n)$ predecessors in the ${\preceq}$-ordering. QED

Now let $I$ be the set of all $n \in \omega$ which have only finitely many ${\preceq}$-predecessors. I claim that $T_I = T + \{\phi_{f(i)}:i \in I\}$ is complete.

Proof. Let $n_0 \in \omega$ be minimal such that $n_0 \notin \{f(i) : i \in I\}$ and $T_I + \phi_{n_0}$ is consistent. Find $i_0 \in I$ such that $n_0 < f(i_0)$ and let $s_0 \geq n_0$ be such that $\max\{i : i \preceq i_0\} \leq t_{s_0}$. If $s \geq s_0$, then $$\{i : i \prec i_0\} = \{i \leq t_s : f(i) < f(i_0) \land i \prec m_s\}$$ because $i \prec i_0$ iff $f(i) < f(i_0)$ for all $i \prec m_s$.

Since $T_I + \phi_{n_0}$ is consistent, it follows that if $s \geq s_0$ then $$T + \phi_{n_0} + \{\phi_{f(i)} : i \leq t_s \land f(i) < n_0 \land i \prec m_s\}$$ is not known to be inconsistent by stage $s+1$. Indeed, $$\{\phi_{f(i)} : i \leq t_s \land f(i) < n_0 \land i \prec m_s\} \subseteq \{\phi_{f(i)} : i \leq t_s \land f(i) < f(i_0) \land i \prec m_s\}$$ and $$\{i \leq t_s : f(i) < f(i_0) \land i \prec m_s\} = \{i : i \prec i_0\} \subseteq I.$$

If $s \geq s_0$ and $\ell_s = m_s$, then $n_0$ belongs to the set of all $$n \in \{0,1,\ldots,s+1\} \setminus \{f(i) : i \leq t_s \land i \prec m_s\}$$ such that $$T + \phi_n + \{\phi_{f(i)} : i \leq t_s \land f(i) < n \land i \prec m_s\}$$ is not known to be inconsistent by stage $s+1$. Let $n_1 \leq n_0$ be the minimal such $n$. Then, by the construction, we have $f(t_s+1) = n_1$ and $t_s+1$ comes immediately after the ${\preceq}$-maximal element of $$\{i \leq t_s : f(i) < n_1 \land i \prec m_s\}.$$ Since $n_1 \leq n_0 < f(i_0)$ it follows that $t_s+1 \prec i_0$, which contradicts the fact that $\max\{i : i \preceq i_0\} \leq t_{s_0}$.

It follows that $n_0$ does not exist and hence that $n \notin \{f(i) : i \in I\}$ entails that $T_I + \phi_n$ is inconsistent (i.e., $T_I \vdash \lnot\phi_n$). Since $\phi_0,\phi_1,\ldots$ enumerates all sentences in the language of $T$, it follows that $T_I$ is complete. QED

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