Problem on finitely axiomatizable theory

Question

Here is a problem on finitely axiomatizable theory that I need help.

1. If $T_1$ and $T_2$ are two theories such that for all $\mathfrak{A}$, $\mathfrak{A}$ is a model of $T_1$ iff $\mathfrak{A}$ is not a model of $T_2$, $T_1$ and $T_2$ are finitely axiomatizable.
2. Let $\,T_1\subset T_2\subset \cdots\,$ be a strictly increasing sequence of closed theories in $\mathscr{L}$. Show that their union $\,T = \underset{n<\omega}\bigcup T_n\,$ is a consistent closed theory in $\mathscr{L}$ and it is not finitely axiomatizable.

Part 1 means that $\,T_1\cup T_2\,$ is inconsistent. Part 2 means that $\,T = \underset{n<\omega}\bigcup T_n\,$ has an infinite model. However, it is not clear how they are related to prove $T$ is not finitely axiomatizable.

Answer 1

For 1 you already noted that $T_1 \cup T_2$ must be inconsistent. Following Chris Eagle's hint from the comments we can apply compactness to find $\phi \in T_1$ and $\psi \in T_2$ such that $\{ \phi, \psi \}$ is inconsistent (this assumes that $T_1$ and $T_2$ are closed under conjunctions, but we may as well assume that). We claim that $\phi$ axiomatises $T_1$ and $\psi$ axiomatises $T_2$. Every model of $T_1$ is a model of $\phi$. Now let $M \models \phi$, and suppose that $M \not \models T_1$. Then by assumption $M \models T_2$, and so $M \models \psi$. This contradicts $\{ \phi, \psi \}$ being inconsistent, so we must have $M \models T_1$. We see that $\phi$ axiomatises $T_1$. Analogously we find that $\psi$ axiomatises $T_2$.

For 2 I assume that "closed" means "closed under consequence". We first prove that $T$ is closed. Let $\phi$ be a consequence of $T$. Then $\{ \neg \phi \} \cup T$ is inconsistent, so by compactness there are $\psi_1, \ldots, \psi_n \in T$ such that $\{\neg \phi, \psi_1, \ldots, \psi_n\}$ is inconsistent. There must then be some $k < \omega$ such that $\psi_1, \ldots, \psi_n \in T_k$. It follows that $\phi$ is a consequence of $T_k$ and so $\phi \in T_k$ as $T_k$ is closed. We conclude that $\phi \in T$, as required.

Finally, to see that $T$ is not finitely axiomatisable we use TomKern's hint from the comments. Suppose for a contradiction that $T$ is finitely axiomatisable. Let $\{\phi_1, \ldots, \phi_n\}$ be a finite axiomatisation of $T$. Write $\phi$ for $\phi_1 \wedge \ldots \wedge \phi_n$, so $\phi$ axiomatises $T$ and hence $\phi$ is a consequence of $T$. We saw before that $T$ is closed, so $\phi \in T$. That means that $\phi \in T_k$ for some $k < \omega$. We then get $T_k = T$, because $\phi$ implies every sentence in $T$ and $T_k$ is closed. We thus have that $T = T_k \subsetneq T_{k+1} \subseteq T$, which is a contradiction. So $T$ cannot be finitely axiomatisable.