It is known that the set of non trivial zeros is an infinite set. But is it known if it is a countable, or uncountable infinite set?

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    $\begingroup$ @Lucian I'm quite confused at your comment, what about $\Bbb Z\subseteq\Bbb C$, surely this is a countable, but non-dense subset. $\endgroup$ Oct 22 '15 at 3:44
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    $\begingroup$ @AdamHughes: I meant uncountable; sorry. $\endgroup$
    – Lucian
    Oct 22 '15 at 3:50
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    $\begingroup$ @Lucian Ah, that makes more sense, though I would note $[0,1]\subseteq\Bbb C$ is uncountable and not dense. $\endgroup$ Oct 22 '15 at 3:51
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    $\begingroup$ @AdamHughes: It contains a a dense subset. Replace not dense with nowhere dense. $\endgroup$
    – Lucian
    Oct 22 '15 at 3:56
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    $\begingroup$ @Lucian Cantor's set is uncountable and nowhere dense. $\endgroup$
    – Wojowu
    Oct 22 '15 at 12:46

If the set $Z$ of zeroes of $\zeta(s)$ were uncountable, then it would have an accumulation point. Now, by certain version of identity theorem, this implies that $\zeta(s)$ is identically zero on its domain, which is absurd.


The comment has left me a bit less willing to commit to the statement of my original answer, though the spirit remains, in essence, true since the two are intimately related, but as I know of no other proof of the explicit formula without zero density arguments, I think it best to avoid that as an a priori reason.

The reasons given in the other answers are my other standard go-to explanations, so I'll try and make this accepted one more developed to do justice to the status of "accepted," as the lazy approach (simple deletion) is not available to me.

The problem with uncountably many zeros is that we may cover $\Bbb C\setminus\{1\}$ by a countable number of compact sets, eg. by $A_n, K_n$ which are closed annuli centered at $1$ of inner radius $n$ and outer radius $n+1$ in the cast $A_N$. The $K_n$ are a cover of the disc

$$D=\{z\in\Bbb C | 0<|z-1|<1\}.$$

which can be taken to be

$$K_n =\left\{z\in\Bbb C : {1\over n+1}\le |z-1|\le {1\over n}\right\}.$$

But the set of zeroes, $S_0$ having no limit points would indicate that $S_0\cap K_n$ is finite (it is discrete and compact), hence $S_0$ may be written as a countable union of finite sets

$$S_0=\bigcup_{m}A_m\cap S_0\sqcup\bigcup_n K_n\cap S_0$$

and so is itself finite. This fact is not special about the $\zeta$ function, it holds for any holomorphic function on any relatively nice set, eg. $\Bbb C\setminus\{1\}$.

The original, accepted answer

The set must be countable by a simple argument. If you look at the so-called "explicit formula" it involves a sum over non-trivial zeroes of the $\zeta$ function.

$$\Psi(x)=x-\sum_{\rho}{x^\rho\over\rho}-{\zeta'(0)\over\zeta(0)}-{1\over 2}\log\left(1-{1\over x^2}\right).$$

Here the sum, $\rho$ is taken over non-trivial zeros. But we know that any infinite sum for which more than countably infinitely many terms are non-zero yields a divergent series. Since we know the explicity formula converges, it must be that the number of zeroes is countable.

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    $\begingroup$ This is a circular argument. The explicit formula requires a countable number of zeroes to begin with; its derivation involves the fact that $\zeta$ has a memorphic extension across $\mathbb{C}$ and a somewhat delicate treatment of its growth. $\endgroup$
    – anomaly
    Oct 21 '15 at 23:37
  • $\begingroup$ @AdamHughes Edit the answer to reflect this fact. $\endgroup$
    – isaacg
    Oct 22 '15 at 3:05
  • $\begingroup$ @isaacg I thought of the same solution. :-) $\endgroup$ Oct 22 '15 at 3:14
  • $\begingroup$ @anomaly I retract my earlier comment: I was able to give a modicum of improvement to help address the systematic difficulties while preserving the fundamental understanding of the course of events. Thanks again for your comment. $\endgroup$ Oct 22 '15 at 3:15
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    $\begingroup$ The argument based on the explicit formula may be circular, but it still shows that this fact follows quickly and elegantly from a 'known' fact. If I were the OP, I would be immediately convinced of that the answer to my question was 'yes'. $\endgroup$ Oct 22 '15 at 10:58

In short, the number of nontrivial zeroes of $\zeta(s)$ with $\lvert \text{Im}(s) \rvert < T$ is on the order of $T \log T$. So there are countably many nontrivial zeroes.


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