Why would this be a $T_1$-space?

Question

Let $L$ be a first order language, and let $F$ be the set of the sentences in $L$. Let $\simeq$ denote semantic equivalence between sentences (of course, according to the completeness theorem, this is the same as syntactic equivalence, but in the context I'm working in, semantics are used rather thab syntax); and let $T := F/\simeq$.

For $H \subset F$, let $H^*$ denote the union of all equivalence classes that intersect $H$ (letting $\pi : F\to T$ denote the canonical mapping, we have $H^* =\bigcup\pi[H]$). Set $\Phi^* := \{\Phi\}^*$ for $\Phi\in F$.

For $H\subset T$, let $\overline{H} := \displaystyle\bigcap_{\Phi \in F, H\subset \{\Psi^* \mid \Phi\models \Psi\}} \{\Psi^* \mid \Phi \models \Psi\}$ (where $\models$ is the semantic implication relation)

I am asked in an exercise to show that this closure operator makes $T$ into a topological space, which is $T_1$.

But this seems false, as in a topological space, $\overline{\emptyset} = \emptyset$, whereas here, if we let $p := (\exists x, x=x)^*$, then for all $\Phi\in F$, $\Phi\models \exists x, x=x$, and so $p\in \overline{\emptyset}$.

Moreover, if we decide to artificially add $\emptyset$ to this set, turning it into a family of closed sets, the resulting space cannot be $T_1$, because for any $q\in T$, $p\in \overline{\{q\}}$.

What do you make of this ? Did the exercise forget to mention something, or am I wrong for some reason ?

Answer 1

You're right on all counts: This definition of the closure operator does not define a topological space, but if you add $\overline{\emptyset} = \emptyset$, you get a topological space (and almost certainly this was the intended space), but not a $T_1$ space.

To give some context on what might seem like a bizarre definition, the set $F$ of sentences comes with a preorder $\models$. Two sentences $\varphi$ and $\psi$ are logically equivalent if and only if $\varphi \models \psi$ and $\psi\models \varphi$. Thus $T$ is the poset associated to the preordered set $F$.

Now the definition specifies that the closed sets in the topology on $T$ are exactly those which are closed upwards in the induced partial order on $T$. This kind of topology has a name, an Alexandrov Topology (usually you take the upwards closed sets to be open in the Alexandrov topology, so we're really looking at the Alexandrov topology on $T^\text{op}$, the dual partial order).

The Alexandrov topology on a partial order is never $T_1$, unless the partial order is discrete (which $T$ clearly is not). However, it is always $T_0$ (you can also define the Alexandrov topology on any preordered set, and it will be $T_0$ iff the preorder is a partial order).

Maybe whoever wrote the exercise confused $T_1$ for $T_0$?

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Added: I've looked at Grätzer's book. The situation is very strange: it's a standard exercise in logic to show that given a first-order language $L$, you can associate a topological space which is $T_1$, totally disconnected, and compact (a.k.a. a Stone space), and the compactness of these spaces is equivalent to the compactness theorem. But Grätzer just describes the wrong space!

Here's the real space you should use when you try to solve exercises 85 and 86:

Let $S$ be the set of all complete consistent $L$-theories. For every sentence $\varphi$, let $U_\varphi = \{T\in S\mid \varphi\in T\}$. Then a basis for the topology on $S$ is given by the family of (clopen) sets $\{U_\varphi\mid \varphi\text{ a sentence}\}$.