5

I've heard the result before that the theory of topologies cannot be expressed as a first-order theory, but I can come up with a simple multisorted theory that seems to capture the open set axiomatization of a topology.

I'm familiar with the statement that the theory of topologies isn't first order mostly as a quick way to demonstrate that first-order logic doesn't capture every theory we might be interested in (similar to referencing torsion groups).

I'm curious whether this attempt at a first-order theory for a single topology succeeds or fails, and what extra subtlety there is in the statement "topologies are not a first-order theory".

This axiomatization uses variations of the axioms and axiom schemas in the simple theory of types and the set theory NFU.

Multiple sorts can be paraphrased away into predicates giving us a single-sorted theory at the cost of some verbosity.

First, we have three sorts. $X$ is the sort of the underlying topological space, $S$ for "set" is the sort corresponding to $2^X$ informally and $F$ for "family" is the sort corresponding to $2^{2^X}$ informally.

I will now attempt to construct a theory of topology. The sorts are as above and there are three predicate symbols. $\in$ is a predicate symbol corresponding to the elementhood relation. $O$ is a unary predicate on $S$ that identifies open sets and $O^F$ is a unary predicte on $F$ that identifies families of open sets.

First, we have two axiom schemas of comprehension.

Comprehension from $X$ to $S$.

$$ \exists s : S \mathop. \forall x : X \mathop. x \in s \leftrightarrow \varphi(x) $$

Comprehension from $S$ to $F$.

$$ \exists f : F \mathop. \forall s : S \mathop. s \in f \leftrightarrow \varphi(x) $$

And we have two axioms of extensionality, one for each level.

$$ \forall a : S \mathop. \forall b : S \mathop. (\forall x : X \mathop. x \in a \leftrightarrow x \in b) \to a = b $$

$$ \forall a : F \mathop. \forall b : F \mathop. (\forall s : F \mathop. s \in a \leftrightarrow s \in b) \to a = b $$

For the remaining cases I'll omit sort annotations for the sake of brevity. These axioms should be understood to apply at both the $S$ and $F$ levels.

Definition of the complement

$$ \forall x \mathop. \forall a \mathop. a \in x^c \leftrightarrow a \not\in x $$

Definition of binary intersection

$$ \forall a \mathop. \forall b \mathop. \forall x \mathop. (x \in a \cap b) \leftrightarrow (x \in a \land x \in b) $$

Definition of binary union

$$ \forall a \mathop. \forall b \mathop. \forall x \mathop. (x \in a \cup b) \leftrightarrow (x \in a \lor x \in b) $$

The definition of arbitrary union applies only to $F$.

$$ \forall f \in F \mathop. \forall x \in X \mathop. (x \in \cup f) \leftrightarrow (\exists b : S \mathop. x \in b \land b \in f) $$

$O^F$ is defined as follows. A family of open sets is a family of sets where every set is open.

$$ \forall f : F \mathop. O^F(f) \leftrightarrow (\forall s : S \mathop. s \in f \to O(s)) $$

And, finally the axioms of topology.

The empty set is open. $\varnothing$ is a constant in sort $S$.

$$ O(\varnothing) $$

The whole space is open. $X$ is a constant in sort $S$.

$$ O(X) $$

Open sets are closed under binary intersection.

$$ \forall a : S \mathop. \forall b : S\mathop. O(a) \land O(b) \to O(a \cap b) $$

Open sets are closed under arbitrary union.

$$ \forall f : F \mathop. O^F(f) \to O(\cup f) $$

This completes the multisorted axiomatization of topological space.

Greg Nisbet
  • 11,657
  • 2
    Models of this theory aren't all topologies - they include every topology-like thing in which the collection of "open sets" is closed under definable unions. – Noah Schweber Jun 09 '21 at 02:17
  • But, even in valid topologies (valid in view of a "specific" structures criterion), one can find properties that are not found in other topologies. It does not mean much. It's almost cyclic reasoning. Any one can pick a stricter criterion and then say that another criterion (these axioms) is not good, because it admits too much. I do not mean at all that the stricter class (the standard topologies) is not interesting. Still, it does not say that the axioms are not practically correct. – Dominic108 Nov 29 '21 at 13:24
  • This should be linked to https://math.stackexchange.com/questions/46656/why-is-topology-nonfirstorderizable where it is written "Virtually all of mathematics can be directly formalized in first-order set theory, and in particular this is how topology is usually formalized." – Dominic108 Nov 29 '21 at 16:49

1 Answers1

8

Every topological space can be naturally construed as a model of your theory, but models of your theory aren't all topologies. For example, by downward Lowenheim-Skolem there is a model of your theory in which $X$ is infinite but $S$ is countable. Basically, you're only getting "definable approximations" to the powerset/iterated powerset.

This is the sense in which topology is not first-order: any attempt to "first-orderize" topology will result in things which look like topological spaces "up to first-order facts" but are not in fact topological spaces.

EDIT: That said, there are definitely ways we can bring first-order logic into play to analyze topological spaces. For example, we could look at first-order properties of associated structures such as the lattice of closed sets. See this article of Bankston for some results along these lines.

Noah Schweber
  • 245,398
  • Thank you, I have a follow-up question if that's okay. Does this mean that I also could not add an "is a topology" predicate $T(\cdot)$ to $\mathrm{ZFC}$ and give it its intended semantics without swapping out FOL for a more powerful logic? (The intent behind my original question was to try to "pull in" fragments of a weaker set theory into my theory so now I'm wondering if we can do the opposite) – Greg Nisbet Jun 09 '21 at 16:19
  • 2
    @GregoryNisbet Well, here we hit a subtlety. We can certainly write down the axioms of topologies in the first-order language of set theory (in particular, the arbitrary unions axiom is "For every object which is a set of open sets, ..."). However, the limitations of first-order logic mean that what a given model of ZFC thinks is a topological space may not be a topological space. Specifically, if $M\models\mathsf{ZFC}$ then we may have some $x,y\in M$ such that $M\models$ "$x$ is a topology on $y$" but in reality $x$ is not a topology on $y$ since it's not closed under arbitrary unions. – Noah Schweber Jun 09 '21 at 16:30
  • 2
    So when you say "give it its intended semantics," there's an ambiguity here, and in order to untangle it it's best to talk about structures explicitly: what properties do you want models of your theory to have? Depending on those properties, first-order logic may or may not be enough. – Noah Schweber Jun 09 '21 at 16:31
  • I down voted it, because it does not say anything about the most important. Can we use these axioms to prove every thing that is true and useful in all topologies? I know the useful part makes it philosophical, but it also makes it practical. It will be nice to remain practical and give an important theorem about topology that cannot be deduced from these axioms. – Dominic108 Nov 29 '21 at 07:36