Ultrafilter Proof of the Compactness theorem

In David Marker's Model Theory: An Introduction, Exercise 2.5.20, he gives the ultrafilter/ultraproduct based proof of the Compactness theorem. I am concerned with a small detail of the proof. Is there a small error in part (a)? Technically, a set $Y$ is in $D$ if and only if $Y$ extends $X_\phi$ for some $\mathcal{L}$-sentence $\phi$. So all $Y$ have some $X_\phi$ as a subset - where $X_\phi$ is the whole set of all finite theories containing $\phi$. By my reasoning, then, every $Y \in D$ must, for some $\phi$, contain all finite theories $\Delta$ that contain $\phi$.

So, choose $A$ and $B$ in $D$; then for some $\phi_1$ and $\phi_2$, \begin{equation} X_{\phi_1} \subset A \subset I \end{equation} \begin{equation} X_{\phi_2} \subset B \subset I \end{equation} So the intersection of $A$ and $B$ must contain $X_\psi$ for some $\psi$, but why is $A \cap B$ guaranteed to contain this, and what would $\psi$ be? I was thinking of $\psi = \phi_1 \land \phi_2$ because $A \cap B$ contains every finite theory that contains both $\phi_1$ and $\phi_2$, but just because a theory contains $\phi_1$ and $\phi_2$ doesn't mean it contains $\phi_1 \land \phi_2$, because there's nothing mentioned about taking logical completion and that doesn't make sense in this context. The proof on ProofWiki is slightly different than this one so I was wondering if this was a bug.

• I think the picture is unrelated to the question. Do you want to correct it? – An old man in the sea. Jul 30 '17 at 19:48
• LOL thanks. That's what happens when you save all your screenshots in a "misc" folder with random alphanumeric strings as titles. – Patrick N. Jul 30 '17 at 19:50
• The proof I know is also a bit different, in that you consider $X_{\{\phi_1,..., \phi_n\}}$ with obvious notations, so that $X_{\{\phi_1,...,\phi_n\}}\cap X_{\{\psi_1,...,\psi_n\}} = X_{\{\phi_1,...,\phi_n, \psi_1,...,\psi_n\}}$, and thus it works fine – Max Jul 30 '17 at 20:04
• Yeah that's what the proof on ProofWiki looks like that I mention at the end. – Patrick N. Jul 30 '17 at 20:29
• Marker's book is not known for its lack of typos. I think you're right that the proof doesn't work as stated and should be amended as suggested by @Max, or by first adding to the theory each sentence which is a consequence of a finite subset. – Alex Kruckman Jul 30 '17 at 23:10

You're right. Interestingly, Marker has in his the errata that in the rest of the exercise you're supposed to work with the filter generated by $D$. So, as noted by Alex Kruckman, you just need to show that $D$ has the finite intersection property, which is not very difficult: in the case of the op, $A \cap B$ is not empty, since it contains $\{\phi_1, \phi_2\}$. As noted by Alex in his comments, this clearly generalizes for every finite intersection.
• I don't think $D$ is a filter basis. The intersection of $X_{\phi_1}$ and $X_{\phi_2}$ does not contain anything of the form $X_\phi$ since every element of it have at least two elements. – Idéophage Jul 30 '17 at 23:39
• @Idéophage - Yes, yes, I read the op's a bit too hastily; when he said "the intersection of $A$ and $B$ must contain...", I thought that was a conclusion! I'm going to edit the answer soon. – Nagase Jul 30 '17 at 23:41
• Marker's errata is correct: $D$ is not a filter basis, but Marker doesn't claim it is. He says "work with the filter generated by $D$", meaning that we have to close $D$ under finite intersections. Any set with the finite intersection property (any intersection of finitely many elements is nonempty) generates a proper filter in this way. And if $Y_1,\dots,Y_n\in D$, witnessed by $X_{\varphi_i}\subseteq Y_i$ for all $1\leq i\leq n$, then $Y_1\cap \dots\cap Y_n$ is nonempty, since it contains $\{\varphi_1,\dots,\varphi_n\}$. – Alex Kruckman Jul 31 '17 at 0:02
• @AlexKruckman - So another patch would be to first throw $Y_1 \cap \dots \cap Y_n$ into $D$ to get $D'$ and then generating a filter? – Nagase Jul 31 '17 at 0:09