Complete extensions of a consistent theory

Question

I understand that I need to use compactness but somehow can't finish it.

Suppose $L$ is a language and $T$ a consistent $L$-theory with only finitely many logically inequivalent complete extensions. Show that there are $L$-sentences $\sigma_1, ..., \sigma_n$ such that every complete extension $T' \supseteq T$ is equivalent to one of $T \cup \{\sigma_i\}$.

My initial idea was to assume the opposite: for any $L$-sentences $\sigma_1, ..., \sigma_n$ there is a complete extension $T' \supseteq T$, such that $T' \neq T \cup \{\sigma_i\}$ for all $i.$

Answer 1

Here's an argument using completeness.

Let $T_1,\ldots,T_k$ be the complete extensions of $T$. Since they're distinct and complete, for each $T_i$ there are formulas $\sigma_{i,1},\ldots,\sigma_{i,_k}$ such that

$T_i\vdash \bigwedge_{j=1}^k\sigma_{i,j}$, but

$T_j\vdash\neg \sigma_{i,j}$ for all $j\neq i$.

Put $\sigma_i=\bigwedge_{j=1}^k\sigma_{i,j}$. Then $\mathcal M\models T\cup \{\sigma_i\}$ only if $\mathcal M\models T_i$. So by the completeness theoreom, $T\cup \{\sigma_i\}\vdash \sigma$ for all $\sigma\in T_i$. So $T_i$ is equivalent to $T\cup\{\sigma_i\}$.

Answer 2

There is also a very simple inductive argument:

Assume $T$ has the complete extensions $T_1,\dots, T_n$ and show by induction on n that there are sentences $\sigma_1,\dots \sigma_n$ such that $T_i$ is equivalent to $T \cup \{\sigma_i\}$

The case n=1 is clear. (For example let $\sigma_1$ be the sentence $\forall x :x=x$)

Now let $n>1$, that means there are two non-equivalent extensions of T and therefore a sentence $\psi$ s.t. both $T\cup \{\psi\}$ and $T\cup \{\neg\psi\}$ are consistent. So there is $1\leq k <n$ such that $T\cup \{\psi\}$ has k-many complete extensions and $T\cup \{\neg\psi\}$ has $(n-k)$-many complete extensions.

In both cases by induction hypothesis there are $\psi_1,\dots,\psi_n$ such that every complete extension of $T\cup \{\psi\}$ is equivalent to some $T\cup \{\psi, \psi_i\}$ with $i\leq k$ and every complete extension of $T\cup \{\neg \psi\}$ is equivalent to some $T\cup \{\neg\psi,\psi_i\}$ with $i>k$.

Now set $\sigma_i=\left\{ \begin{array}{ll}\psi\land\psi_i \qquad i \leq k\\ \neg\psi\land\psi_i \qquad i>k\end{array} \right.$

Answer 3

The following is a more abstract approach - I often find reasoning this way is easier, and helps me come up with the proof faster:

View the set of consistent completions of $T$ as the set of paths through a binary tree. Specifically, order the set of all sentences in the language as $\{\tau_i: i\in\mathbb{N}\}$, and let $T\subseteq 2^{<\omega}$ be the tree of nodes such that $p\in T$ if $T\cup\{\tau_i: p(i)=1\}\cup\{\neg\tau_j: p(j)=0\}$ is consistent. Note that $T$ has no dead ends.

Now saying that $T$ has only finitely many completions is the same as saying that this tree has only finitely many paths. So we may pick some $n$ such that there are no "branching nodes" (elements $p\in T$ such that both $p^\smallfrown\langle 0\rangle$ and $p^\smallfrown\langle 1\rangle$ are on the tree) of length $\ge n$. This is because $T$ has no dead ends: any branching node has to lead to two distinct paths.

Let $p_i$ ($i<m$) be the set of all nodes on $T$ of length $n$ - then each path through $T$ is determined by which $p_i$ it passes through, by assumption on $n$.

To each $p_i$ we may associate the sentence $$\sigma_i=(\bigwedge_{p_i(j)=1}\tau_j)\wedge(\bigwedge_{p_i(j)=0}\neg\tau_j);$$ the finitely many sentences $\sigma_i$ ($i<m$) have the desired property.

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Where did I use "Compactness?" Well, basically the tree picture of consistent extensions is all about compactness - the compactness of first-order logic$^*$ is equivalent to the compactness of Cantor space! (This can even be made rigorous, via Reverse Mathematics.)

$^*$Fine, in a countable language. :P