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This is a follow-up question to this one, where in the answer it is explained how topological spaces may very well be described in a purely first-order manner. Furthermore, the set of first-order sentences that hold in every topological space seems to include equivalents of the axioms of topology. On the other hand it is stated - as a key fact - that the class of topological spaces is not (first-order) axiomatizable. At first sight, this seems paradoxical.

Is there a simple proof of the nonaxiomatizability of topology?

References are welcome!

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Are you asking why the class is not elementary? –  Andres Caicedo Jun 22 '11 at 21:05
    
Yes, elementary and axiomatizable seem to be equivalent names. –  Hans Stricker Jun 23 '11 at 6:24

2 Answers 2

up vote 9 down vote accepted

Shelah proved (in "Number of open sets for a topology with a countable basis", Israel J. Math. 83 (1993), no. 3, 369-374) that if a topology admits a countable base, then the number of open sets is either countable or equal to $2^{\aleph_0}$. (The argument is short and elegant, you may enjoy reading it.)

Using the Löwenheim-Skolem theorem, the result you ask for follows if CH fails: Start with a space with a countable basis and continuum many open sets (for example, the reals), modeled as a first order (multi-sorted) structure as explained in previous answers, and take an elementary substructure of size $\omega_1$ where all the elements of the basis are included. It follows that this is not a topological space, as there are at most $\omega_1$ open sets, but there should be continuum many.

Of course, the failure of CH is not needed here: Start with the the topological space of the rationals, seen as a first-order structure. Note this is a structure of size continuum, and take a countable elementary substructure.


I mentioned Shelah's result because I think that there are many interesting questions about topology that can be approached using the techniques of elementary substructures (in part, precisely because the class of topological spaces is not elementary). It is actually a common move in set theory: You study second (or higher) order objects using first-order approximations. In the context of topology, I strongly recommend that you take a look at the paper "More set-theory for topologists" by Alan Dow, in Topology and its Applications 64 (1995) 243-300, especially section 5.

The fact that the class of topological spaces is not elementary leads to interesting mathematics, as you can start with a space, take an elementary substructure, and study the properties that the new structure has, either on its own, or as an approximation to the original space. This has been developed in some detail by Tall and his collaborators, starting with the paper by Lucia R. Junqueira and Franklin D. Tall, "The topology of elementary submodels", Topology and its Applications 82 (1998) 239-266.

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Andres, is there some descriptive set-theoretic analysis of the complexity of the topologies with a countable base? –  JDH Jun 22 '11 at 21:49

Although the discreteness of the space is expressible in the set-up proposed by Carl's answer to your other question (where the open sets are treated as first-order objects with their own sort), nevertheless the class of infinite discrete topological spaces will not form an elementary class. The reason is that it has no countable models. Once there are infinitely many points, then by discreteness we know that the open sets will consist of all subsets of those points, and this forces the model to have at least continuum many open sets. This would violate the downward Löwenheim-Skolem theorem if it were an elementary class.

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Maybe a dumb question: Can't there be infinite models - countable from the outside - which "believe" to be uncountable and thus nevertheless are uncountable topological spaces? (I guess not, since if this was the case you couldn't prove a lot with the downward Löwenheim-Skolem theorem.) –  Hans Stricker Jun 23 '11 at 14:14

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