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Let $L$ be a first order language and $\mathcal T$ be the collection of all complete $L$ theories. For each $L$-sentence (closed $L$-formula) $\phi$, define $B_\phi :=\{T \in \mathcal T : \phi \in T\}$. Let $\mathcal B:=\{ B_\phi : \phi \in T \}$ .

I can see that if $T \in B_{\phi_1} \cap B_{\phi_2}$, then $\phi_1,\phi_2 \in T$, then for every model $M$ of $T$, $M \vDash \phi_1, M \vDash \phi_2$ [since $T$ is complete and $\phi_1, \phi_2$ are $L$-sentences ] , and then $M \vDash \phi_1 \land \phi_2 $ , so $T \vDash \phi_1 \land \phi_2$ , so $T\in B_{\phi_1 \land \phi_2} \subseteq B_{\phi_1} \cap B_{\phi_2} $.

Thus $\mathcal B$ forms a base for a topology on $\mathcal T$ .

How to show that this topology is compact and Hausdorff ?

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Throughout I presume that "complete theory" means "complete consistent theory," and that a theory is complete if it contains (not merely proves) each sentence or its negation.

Hausdorffness and compactness of the topology reflect basic properties of first-order logic: namely, the behavior of negation and the finiteness of proofs, respectively.

EDIT: Before going further, based on your comment below I think it's worth stating explicitly the following: if $T$ is complete (and consistent), then for any sentence $\varphi$ exactly one of $\varphi$ and $\neg\varphi$ is in $T$. In particular, if $T$ is complete and we know $\varphi\not\in T$, then we know $\neg\varphi\in T$.

  • Towards Hausdorffness: Suppose $T_0, T_1$ are distinct complete theories. Then there is some $\varphi$ such that $\varphi\in T_0$ and $\neg\varphi\in T_1$ (why?). Do you see a corresponding pair of basic open sets which appropriately separate $T_0$ and $T_1$?

  • Towards compactness: It will probably be easier, first, to consider basic open covers - that is, covers by basic opens (not just by opens). Suppose $\mathcal{C}=\{B_\varphi:\varphi\in \Gamma\}$ (for some set of sentences $\Gamma$) is a basic open cover. Then every theory $T$ is in some $B_\varphi$. Now think about the set $\check{\Gamma}=\{\neg\varphi: \varphi\in\Gamma\}$. If $\mathcal{C}$ has no finite subcover, then $\check{\Gamma}$ is finitely consistent (why?) and hence actually consistent, since proofs are only finitely long. Consider a complete theory $T\supseteq\check{\Gamma}$; do you see a problem with this theory as a point in our space? HINT: think about the "cover" $\mathcal{C}$ ...


Of course, it might feel a bit odd that we're not using the compactness theorem in proving the compactness of this space. Ultimately this comes down to the precise way we defined the space. "Complete (consistent) theory" is a proof-theoretic property, and doesn't invoke models in any way, so there's no need to bring semantic notions into the picture.

We could, however, have considered the space whose points are theories of structures - that is, sets of sentences of the form $Th(\mathcal{M})$ for $\mathcal{M}$ a structure (equivalently, complete satisfiable theories). Clearly every such set is complete and consistent, but the converse requires the full strength of the completeness theorem. To show that this space is compact, we can either use the completeness theorem to reduce it to the situation above ... or use the (weaker) compactness theorem to (following otherwise the argument above) go from "$\check{\Gamma}$ is finitely satisfiable" to "$\check{\Gamma}$ is satisfiable," and then let $T=Th(\mathcal{M})$ for some $\mathcal{M}\models\check{\Gamma}$.

The space of complete satisfiable theories can also be thought of as the $T_0$-ification](https://en.wikipedia.org/wiki/Kolmogorov_space) of the (proper class sized) space of all structures, where the basic open sets (well, classes) are those of the form $\{\mathcal{M}: \mathcal{M}\models\varphi\}$ for $\varphi$ a sentence. (If you don't like the set/class issue here, by the Lowenheim-Skolem theorem we can safely restrict attention to structures whose underlying set is a subset of $\mathbb{N}$, and there's only set-many of these; note, however, that this fix depends on the specific properties of first-order logic.)

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  • $\begingroup$ how can we find $\phi$ with $\phi \in T_0, \neg \phi \in T_1$ ? $\endgroup$ – user Apr 29 '18 at 16:50
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    $\begingroup$ @users Well, $T_0\not=T_1$; so either there's some $\varphi$ in $T_0\setminus T_1$, or there's some $\varphi\in T_1\setminus T_0$. Let's look at the first possibility: since $\varphi\not\in T_1$ and $T_1$ is complete, what can you say about $\neg\varphi$ and $T_1$? The other possibility is the same but flipped. (See also my edit.) $\endgroup$ – Noah Schweber Apr 29 '18 at 16:52
  • $\begingroup$ Yes thank you ... complete theories are indeed decidable $\endgroup$ – user Apr 29 '18 at 16:54
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    $\begingroup$ @users Are you familiar with Zorn's lemma? If so just think about a maximal consistent theory containing $\check{\Gamma}$ - why does such a thing exist (Zorn) and why must such a thing be complete (if not, you could make it bigger)? If not, I think it's a good idea to first think about the case when the language is countable. We list all the sentences as $\psi_n$ ($n\in\mathbb{N}$), and build an extension of $\check{\Gamma}$ via a "greedy algorithm:" throw in $\psi_0$ if that's consistent, otherwise throw in $\neg\psi_0$; now throw in $\psi_1$ if that's consistent, otherwise $\neg\psi_1$; etc. $\endgroup$ – Noah Schweber Apr 29 '18 at 20:30
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    $\begingroup$ This process clearly produces a complete theory containing $\check{\Gamma}$, and doesn't need any set theory. The fully general case, however, does require (a weak form of) Zorn's lemma, though, so ultimately you're not going to escape the creeping claws of set theory entirely. $\endgroup$ – Noah Schweber Apr 29 '18 at 20:30
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To prove that the space is Hausdorff, consider two $L$-theories $T_1\neq T_2$. Since they are different and complete, there is $\phi$ such that $\phi\in T_1$ but $\phi\not\in T_2$, which implies $\neg\phi\in T_2$. So the open sets $B_{\phi}$ and $B_{\neg\phi}$ witness that the space is Hausdorff.

Suppose that $\mathcal{D}:=\{B_{\phi_i}:i\in I\}$ is an open cover of the space (we can assume that all $B_i$ are basic open sets). If $I$ is finite, we have nothing to do, so suppose $I=\omega$ (there are no other possibilities since the space is second-countable). Suppose for a contradiction that for all $I'$ finite subset of $\omega$ $\{B_{\phi_i}:i\in I'\}$ is not a covering of $\mathcal{T}$. Then, there is an $L$-structure satisfying $\bigwedge_{i\in I'}\neg\phi_i$, which, by compactness (not surprisingly!) means that the theory $T$ containing all the $\neg\phi_i$ is satisfiable. But then, if $T'$ is any completion of $T$, $T'$ cannot be in $\bigcup \mathcal{D}$, a contradiction.

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    $\begingroup$ why is the space second countable ? I don't see it unless $L$ is countable ... $\endgroup$ – user Apr 29 '18 at 17:01
  • $\begingroup$ I was actually assuming $L$ to be countable. But it does not matter. I am not using the fact that $I$ is $\omega$, you may just replace that sentence with "$I$ is infinite". $\endgroup$ – Leo163 Apr 29 '18 at 17:07

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