Why some say virtually all of mathematics can be done with FOL when even the class of topologies is not axiomatizable in FOL?

Question

The question can be seen as: please explain the apparent contradiction (the paradox) between the answers in Can we express the theory of a single topology as a multi-sorted theory? and in (Why) is topology nonfirstorderizable?. The first answer states "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." The second answer states "Virtually all of mathematics can be directly formalized in first-order set theory, and in particular this is how topology is usually formalized". The similar statement "FOL is all the logic we have and all the logic we need." by John Alan Robinson in his paper in computational logic mentioned in this other answer is also relevant.

For example, using arithmetic for simplicity, in Manzano Extension of First Order Logic (p.116), Peano's axiom of induction for arithmetic is expressed in second order logic (with $c$ a constant intended to be $0$ and $\sigma$ a unary function intended to be the addition of $1$) as $$\forall X\,(Xc \wedge \forall z(Xz \rightarrow X\sigma z) \rightarrow \forall x\; Xx$$ In this formulation, $Xc$ has the fixed interpretation $c \in X$. Manzano does not use the symbol $\in$ because its interpretation is fixed in the second order semantic. But there is a problem with this formulation, because it quantifies over all possible subsets of numbers $X$. No calculus can keep up with that much expressiveness. In any specific context where this axiom is to be used to prove things, this extra expressiveness is useless. When we think about proof systems, which we must in practice, we are led to first-order semantics because those semantics satisfy the completeness and compactness theorems. Any sentence that is provable in first-order semantics is also valid in second-order semantics. But not every sentence valid in second-order semantics is provable, regardless which effective proof system we choose. This is saying that in practice, in the context of a practical calculus, we are essentially always using the first order version, which can be stated with the same syntax, but using the Henkin semantic, which restrict the meaning of $\forall$ on the higher order variables. So, all those who insist that we are using the standard higher order formulation (without Henkin's semantic) seem to be ignoring the practical side of mathematics. On the other hand, with the Henkin's semantic, the Löwenheim-Skolem theorem comes back as a true theorem and the non standard arithmetics as valid interpretations. Using this argument, many argue that we need the standard higher order logic. I ask the question, because it's hard not to agree with that other side of the story. The examples provided in the answers below are very convincing. Yet, again, we could also say with Manzano that "second order logic with standard semantics has great expressive power (too much, we might say)". In this analysis, am I overlooking some folk's knowledge regarding logic that would explain this paradox?

Why the answers below are unsatisfactory, but useful: They do not reconciliate at all the answers in the two linked question. On the contrary, HallaSurvivor emphasizes that "there are plenty of properties we're interested in topologically that are not first order". This means that FOL cannot be practically used to axiomatize the class of topologies. On the other hand, the answer of Carl Mummert says "virtually all of mathematics" can be done in FOL. The natural meaning of "virtually all of mathematics" must clearly include topology. So, we need to understand what Carl Mummert means by "virtually all of mathematics can be done with FOL" and how it includes topologies, even though their class is not FOL axiomatizable. The answers do not do that at all. They are very useful answers with examples and I asked these examples as part of the original question. I said that these examples will shed some light and they did, but yet they simply do not answer the question.

There is also no answer to my question in Why is the class of topological spaces not axiomatizable?, which mentions the same paradox and then simply proves and discusses the non axiomatizability of the class of topologies, without really explaining the paradox.

I think I know what is meant by "virtually all of mathematics" so that there is no contradiction and I am just checking that. The idea is that when we translate a statement into a FOL version, we might lose important properties, but the meaning of the statement corresponds to what can be proven with it in the available calculus. It's also that at this proof level, we do not care about the whole class, but about the specific property used in the proof system. This applies also for set theory, number theory and many other theories. Take number theory for example. One can argue that FOL is not sufficient to capture every thing about number theory. Yet, in many applications, it's fine to use number theory defined by FOL. If you need a number theory that is more special, perhaps you can also define it with FOL. In this manner, FOL would be sufficient to do virtually all mathematics and that would not contradict the fact that the FOL statement does not capture the full meaning. This would unify the answers of the two linked questions. This appears to be the simple meaning of "virtually all of mathematics can be done in FOL". Yet, this is not satisfying, because we still need to talk of topologies, etc. in terms of specific higher order standard structures. In that sense, one should refuse to say that essentially all of mathematics is done with first order logic. So, the question is clear. How, do we explain this paradox?

Answer 1

You want a "well known result in topology" that doesn't work in this setting, and since this setting preserves first order results we'll need to move to higher order questions. Thankfully, there are plenty of properties we're interested in topologically that are not first order! You might be interested in Goldblatt's Lectures on the Hyperreals, which discusses models of the first order theory of $\mathbb{R}$ (with extra sorts for definable subsets, and definable families of definable subsets, and ...). In particular, some time is spent discussing topological properties of ${}^*\mathbb{R}$, and reading the book will give you a better sense of how these "nonstandard" models can act bizarrely when you try to inspect them with questions that aren't first order.

Let's look at compactness as a case study. That is, let's say that every cover of $X$ admits a finite subcover. Recall your linked question includes a sort $X$ for points, $S$ for (definable) subsets, and $F$ for (definable) families of (definable) subsets. Expressed formally in this notation, compactness says:

$$ \forall f \in F . O^F(f) \land \bigcup f = X \longrightarrow \bigvee_{n \in \mathbb{N}} \left ( \exists U_1, \ldots, U_n \in S . \left ( \bigwedge_{k = 1}^n O(U_k) \right ) \land U_1 \cup U_2 \ldots \cup U_n = X \right ) $$

Importantly, since we don't know how big the subcover should be, we must have this countable disjunction (or something of similar strength, like a natural number quantifier). Of course, countable disjunctions aren't first order syntax, so we should expect there to be spaces which are first-order equivalent to a compact space, but which are not themselves compact.

Interestingly, there is an alternative definition of compactness which is first order expressible: "Every open cover totally ordered by subset inclusion contains $X$" can be expressed as

$$ \forall f \in F . O^F(f) \land \bigcup f = X \land \forall U_1, U_2 \in f . \big ( U_1 \subseteq U_2 \lor U_2 \subseteq U_1 \big ) \longrightarrow X \in f $$

technically we don't have $\subseteq$ in our language, but I'll leave it to you to show that it's an abbreviation for something that is expressible in our language.

Now for the contradiction: we'll build a model of this theory which is not compact in the sense of finite subcovers, but which is compact in the sense of this linear ordering definition. That will show that the "classical" topological fact that these two definitions are equivalent is no longer true. Moreover, it's the usual definition of compactness that is the problem here! So while I guess you could just redefine compactness to be this linear-ordering condition, you would need to basically redo all of topology to see what you can still prove using this (now weaker) notion of compactness. I wouldn't hold my breath, though, because lots of other properties to do with compactness (every sequence having a convergent subsequence, for instance) are also inherently non-first order.

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Ok, so, the proof:

Let $(X,\tau)$ be an infinite compact space, and let $\mathbf{X} = (X, S, F, \in, O, O^F)$ be its associated model.

Let's add a new constant symbol $f^*$ of sort $F$ to our language, as well as constant symbols for every $x \in X$, $s \in S$, and $f \in F$. Next, let's add as axioms to our theory

• everything we can write down that's true of $\mathbf{X}$ (the elementary diagram of $\mathbf{X}$)
• $O^F(f^*)$
• $\bigcup f^* = X$
• $\forall U_1 \in f^* . U_1 \neq X$
• $\forall U_1, U_2 \in f^* . U_1 \cup U_2 \neq X$
• $\forall U_1, U_2, U_3 \in f^* . U_1 \cup U_2 \cup U_3 \neq X$
• ...

Now, this new theory is finitely satisfiable (do you see why? I'll include a description under the fold), thus satisfiable by the compactness theorem. Fix a model, say, $\mathbf{X}^*$, and notice

1. $\mathbf{X}^*$ models the elementary diagram of $\mathbf{X}$, so they agree on all first order formulas. In particular, $\mathbf{X}^*$ thinks that the linear-order definition of compactness is true.
2. $f^*$ is an open cover of $\mathbf{X}^*$ with no finite subcover

But this is a problem! The two formerly equivalent notions of compactness are now distinct!

Any finite collection of these axioms contains only finitely many axioms of the form "there is no finite subcover of size $n$". So pick an $N$ large enough to be bigger than all of the $n$ forbidden by our finitely many axioms, and look at $\mathbf{X}$ where we interpret $f^*$ to be an open cover of size $N$, with no smaller subcover (this is why we require $\mathbf{X}$ to be infinite).

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So the notion of finite-subcover compactness is not captured by this definition of "topological space". Again, this is not an isolated example. Intuitively, any theorem about sequences isn't first order (how can you possibly write down a function $\mathbb{N} \to X$ in a first order way?), any question about separability (how can you write down "I have a countable dense subset"?), etc.

Of course, you could try to write down a higher-order axomatization of topological spaces. But this is basically what we do in set theory! We wouldn't be gaining anything, since (as the famous quote goes): "higher order logic is set theory in sheep’s clothing" (see here, say).

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I hope this helps ^_^

Answer 2

Let me at least try address your last sentence:

If there is a well known theorem that cannot be proven using the FOL axiomatic, it will be a clear setback to the belief that this axiomatic can practically represent topologies.

Consider instead of topological spaces (for simplicity) the theory of real numbers, which is also nonfirstorderizable.

Suppose you define real numbers by a first order theory, then by the Lowenheim-Skölem theorem there is a countable model of the real numbers. This means that the property "The set of real numbers is uncountable" cannot be deduced from a first order axiomatization of real numbers.

For concreteness, this model can be constructed by letting $\mathbb R^*$ be the set of all constructible Cauchy sequences of real numbers modulo an equivalence relation. Then $\mathbb R^*$ is countable because every real number can be expressed by a finite sequence of symbols.

You can do a similar construction for the topology on $\mathbb R$, by defining the set of open sets $\mathcal O^*$ to consist of only the open sets (in the usual sense) which are constructible. Although this will satisfy the first order axioms for topological spaces, I claim this does not form a topology because one can (nonconstructively) specify a collection of opens in $\mathcal O^*$, and their union will not lie in $\mathcal O^*$ (this is my intuition, please correct me if I am wrong here).

See also Skolem's paradox.