# Prove the integers in the arithmetic progression topology is not compact

I've been studying for my final exam in a general topology course, and I came upon this problem about compactness that I'm have a really tough time solving.

Let $a$ and $b$ be integers, with $b\neq 0$, and let $A_{a,b} = \{a+nb\,|\,n\in\mathbb{Z}\}$ be an arithmetic progression. The arithmetic progression topology on $\mathbb{Z}$ has a basis $\mathcal{A}=\{A_{a,b}\,|\,a,b\in\mathbb{Z} \text{ and } b\neq 0\}$. The Dirichlet Prime Number Theorem asserts that the arithmetic progression $A_{a,b}$ contains infinitely many prime numbers if $a$ and $b$ are relatively prime. Use this result to prove that $\mathbb{Z}$ is not compact in this topology.

I'm unsure how to use the Dirichlet Prime Number Theorem, as I'm inclined to think that I should be using the infinite primes in an arithmetic progression $A_{a,b}$ satisfying the theorem to create an infinite open cover, but I'm having trouble ensuring that I can create a cover of $\mathbb{Z}$. Elsewhere, I found online that $\bigcup_pA_{-1,p}$ over all prime numbers $p$ should cover the integers and in fact indicates $\mathbb{Z}_+$ is not compact in the arithmetic progression topology, but I can't see how we know it actually covers all integers. If this does hold, though, then wouldn't $A_{0,q}\cup(\bigcup_pA_{-1,p})$, where $q$ is prime, cover the integers and be an infinite cover with no finite subcover, as without $A_{0,q}$, $0$ is not covered, and if any $A_{-1,p}$ is removed, then $p-1$ is uncovered. If $p = q$, removing one basis element from this infinite open cover doesn't make it a finite subcover. However, I feel that this approach is wrong since it doesn't using the Dirichlet theorem at all.

Completely revised. I don’t see how to use Dirichlet’s theorem, but I also don’t see any need for it. Let $X$ be the space, and let $D=\{2^n:n\in\Bbb Z^+\}$.

• Show that for each $n\in\Bbb Z^+$, $$D\cap A_{2^n,2^{n+1}}=\{2^n\}\;,$$ so $D$ is discrete.

I claim that $D$ is also closed; if so, then $X$ is not even countably compact, let alone compact.

• If $n$ has an odd prime divisor $p$, then $A_{0,p}$ is an open nbhd of $n$ disjoint from $D$. Note that this includes the case $n=0$.
• If $n=-2^k$ for some $k\in\Bbb Z^+$, show that $n\in A_{2^n,2^{n+1}}$, and then use the fact that $X$ is $T_1$ to show that $n$ has an open nbhd disjoint from $D$.
• I’ll leave the case $n=\pm 1$ to you; it’s not hard to find an open nbhd of both that is disjoint from $D$.
• If we let $z$ be the limit point of a subsequence of $\pi$, then if we show that there exists a neighborhood $U$ of $z$ that does not contain infinitely many primes, is that sufficient to say $U$ also does not contain infinitely many points of any subsequence? – hallowedelegy May 14 '16 at 17:34
• @hallowedelegy: I have to apologize: the argument that I had in mind doesn't quite work, and at the moment I don’t see what was intended. You can either let me think about it for a while to see if I can fix the answer, or you can uncheck it for the time being, and I’ll delete it until I can fix it. – Brian M. Scott May 14 '16 at 17:54
• If it helps, the only theorems about compactness introduced in the section this question appears are: 1. Continuous images of compact spaces are compact. 2. Finite unions of compact sets are compact. 3. Arbitrary intersections of compact sets in a Hausdorff space are compact. 4. A closed set contained in a compact set is also compact. 5. If a set is compact in a Hausdorff space, it is closed. – hallowedelegy May 14 '16 at 18:06
• @hallowedelegy: Basically I’m using your fourth result in the revised answer: the set $D$ is closed but clearly not compact. – Brian M. Scott May 14 '16 at 18:11
• I believe there is a typo in the second-to-last bullet. $n$ is negative, so $A_{2^n,2^{n+1}}$ isn't well-defined. However, the set $A_{-2^k,2^{k+2}}$ would give the desired conclusion: $n \in A_{-2^k,2^{k+2}}$ and $A_{-2^k,2^{k+2}} \cap D = \emptyset$. – spenceryue Oct 12 '18 at 3:48