Idle remarks on the Furstenburg topology

There are numerous proofs of the infinitude of primes in the literature, but for my money the “cutest” is the “topological proof” due to Hillel Furstenburg.

Make $\mathbb{Z}$ into a topological space by taking as a neighborhood basis the set of arithmetic progressions $A(a,d):=\{a+nd:n\in\mathbb{Z}\}$ (where d>0, of course).  It’s not hard to check that this really is a topology, and that it has the following interesting properties.

  1. The union of finitely many arithmetic progressions is both closed and open.  (Look modulo the gcd of the various differences in the progressions; this set and its complement are each unions of residue classes.)
  2. Any open set which is not empty is infinite.  (Obvious, since the neighborhoods are all infinite.)
  3. The set S=\bigcup_{p\mbox{ prime}} (p\mathbb{Z}) is open; by 1, it is also closed if there are only finitely many primes.
  4. But \mathbb{Z}\setminus S=\{\pm 1\} is finite, so \mathbb{Z}\setminus S is not open, and S is not closed.

That’s the whole proof.  Once you introduce the topology, the theorem practically proves itself!

(If someone has written about this topology and given it an official name, then I don’t know about it; “Furstenburg topology” seems as good a name as any for now.)

It is surprising to me that we don’t seen this topology (or its natural generalization to general number rings) mentioned more often.   Sure, Furstenburg’s proof isn’t so much a new idea about primes as a convenient distillation of some key bits of elementary number theory, but there are lots of places in number theory where the discussion can be phrased in terms of this topology.  Perhaps there is something to gain from this.

For example, if S is a set of integers, then the Furstenburg closure of S is the set of all integers t such that, for every modulus M, there is some x\in S such that x\equiv t \pmod{M}.

Exponential Diophantine equations. For example, in recent threads on MathOverflow (beginning here, then moving here), Kevin Buzzard asks whether, if you want to know whether 3^n-2^m=t has solutions in natural numbers n,m, it is always possible to either give a solution or give some modulus where the equation is “obviously” impossible.  This is a good question, because it is a routine thing to enumerate the possible residues of 3^n-2^m, or any similar exponential Diophantine expression in any given modulus; it would be good to know whether the process of successively trying moduli and see what’s impossible would eventually rule out any given t for which there is no solution.  If the answer to his question is no, then one wonders what kind of proof methods would be required to prove that 3^n-2^m=t has no solutions in a case where it has solutions modulo M for every M.  In other words, is \{3^n-2^m:n,m\in \mathbb{N}\}.

(As noted in one of the comments on the second linked-to MathOverflow threads, it is a nice theorem that for any integer a>1, the set \{a^n:n=0,1,2,3,\ldots\} is Furstenburg closed.)

In some very fluffy sense, a basic form of the Hasse principle is the assertion that the sets of numbers expressible by certain types of Diophantine expressions are Furstenburg closed.

Apollonian circle packings. A primary open problem in integer Apollonian circle packings is about the set of curvatures that occur in a given packing.  Fix a primitive packing P, and let S be the set of positive curvatures appearing in P.  Then it’s well-known that the elements of S miss many residue classes modulo 2^n3^m for some n,m depending on P.  (From a modular arithmetic point of view, this is in some sense the only obstruction, since it is also known that S intersects every residue class modulo M if M is coprime to 6.)  The big question is, does S contain all but finitely many of the numbers not ruled out by this modular arithmetic obstruction?

In the context of the Furstenburg topology, the question has a nice summary.  Is S Furstenburg-relatively-closed in some set \{N,N+1,N+2,N+3,\ldots\}?


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