A bit of topology

This proof is due to Hillel Fürstenberg, and is topological.

Let’s define a topology on \mathbb Z. We’ll declare any infinite arithmetic sequence to be open. That is, each set

U_{a,b}=\{\dots,-2a+b,-a+b,b,a+b,2a+b,\dots\}

is open. These are the basic open sets, so any open set can be obtained by taking unions and finite intersections of these. It’s not hard to see that a finite intersection of these is empty, or infinite. That is, there are no finite open sets other than \varnothing.

Another fun fact about this topology is that each U_{a,b} is also closed. Indeed

U_{a,b}=\mathbb Z\setminus\displaystyle\bigcup_{i\ne b}U_{a,i}.

This is probably poor notation, but I mean that i runs from 0 to a-1, but skips b. So U_{a,b} is the complement of an open set and hence closed. All I will really need is that U_{a,0} is closed

Now, if there were only finitely many primes, then \bigcup_{p}U_{p,0} would be a closed set. This set consists of any number which is a multiple of some prime. It’s complement is \{-1,1\} which has to be open. This is a contradiction, because finite sets cannot be open. Cool, huh?

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2 Responses to A bit of topology

  1. fusionball says:

    There was a huge thread on mathoverflow a while back arguing about whether this is really just Euclid’s proof couched in the language of topology. The argument could also be phrased this way. Say that a set S of integers is periodic if for some positive integer n, we have x \in S iff x + n \in S. Then the union of a period p set with a period q set is a set of period pq (here periods do not have to be minimal). Complements of periodic sets are also periodic (if the period is p then the complement will be the union of the “missing” congruence classes mod p). If there were only finitely many primes, then \{1,-1\} is periodic, contradiction. But, if we unravel this argument, really we are just looking at p_1 p_2 \ldots p_n \pm 1 so it’s Euclid’s argument all over again.

  2. Pingback: And one more similar one… « Andy Soffer

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