6
$\begingroup$

The theorem of Furstenberg showing there exists infinitely many primes (and variants, including those stripping away the topological side of things) has been discussed several times on MSE, e.g. in this question.

For clarity let me follow the paper of Mercer in the Monthly which sets it up as :

  • Claim 1. A finite intersection of Arithmetic Progressions is either empty or infinite.
  • Claim 2. If S is any collection of sets, then a finite intersection of finite unions of sets in S is also a finite union of finite intersections of sets in S.
  • Claim 3. If $p_1$,...,$p_n$ were all primes, then since the finite intersection of non-multiples is a two-element set, we'd get a contradiction (i.e. it is a fact that $\{-1;1\} =$ $NM(p_1) \cap\dots\cap NM(p_n)$, where $NM(p_i):=$ $(1+p_i\mathbb{Z})\cup\dots\cup ((p_i-1)+p_i\mathbb{Z})$).

Now, surely the following is wrong (otherwise how on earth has it not been found earlier), but let's try to apply this to show the existence of infinitely many twin primes.

Let us assume

  • that there exists infinitely many primes $p_1,p_2,\dots$
  • that there exists only finitely many pairs of twin primes, say $(q_1;q_1+2),\dots,(q_m;q_m+2)$, where for each $i$ there exists some $j=j(i)$ such that $q_i=p_j$.

But then, we have that $\{-1;1\} =$ $NM(q_1+2) \cap\dots\cap NM(q_m+2)$ since all the $q_i+2$ are prime. A contradiction just as well, i.e. primes of the form $q_i+2$ must occur infinitely many times, too (indeed of any form $q_i+2k$, by the same token).

Question: could someone please point out where is the flaw in this reasoning? (It feels like there's gap in logic right at the end, but I can't articulate it.)

Thank you!

$\endgroup$
3
  • 1
    $\begingroup$ The statement $\{-1;1\} =NM(q_1+2) \cap\dots\cap NM(q_m+2)$ has no basis in the argument since we aren't covering the entire set of primes by referencing $q_1,\dots,q_m$. This same problem occurs when attempting to use an approach analogous to that used by Euclid. $\endgroup$
    – abiessu
    Jun 26 '15 at 20:07
  • 1
    $\begingroup$ Basically, the Furstenberg argument works because we are covering all the primes in existence, and the only numbers that are not represented in arithmetic progressions from any prime are exactly $-1,1$. $\endgroup$
    – abiessu
    Jun 26 '15 at 20:10
  • $\begingroup$ Thanks, I think I understand now! Feel free to post this as an answer and I'll accept it. $\endgroup$
    – Butini
    Jun 26 '15 at 20:28
1
$\begingroup$

The problem arises when we first state $$\{-1;1\} =NM(q_1+2) \cap\dots\cap NM(q_m+2)$$

The equation is not true before we attempt to use it for contradiction purposes since we haven't eliminated all the other primes as sources for arriving at the set $\{-1,1\}$. In the Furstenberg proof, we must use all the possible primes in order to arrive at this set, which does not allow us (so far) to distinguish which primes are part of twin prime pairs and which are not.

$\endgroup$
1
  • $\begingroup$ I accept this answer, and to state it in a way I understand better : if I were to intersect the finite intersection $NM(q_1+2)\cap\dots\cap NM(q_m+2)$ with all the other $NM(p_i)$ the result would still be $\{-1;1\}$ and because this is an infinite intersection we can't apply claim 1. Thanks! $\endgroup$
    – Butini
    Jun 26 '15 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.