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I know $\mathbb{Z}\subset \mathbb{R}$ is not compact for the standard topology on $\mathbb{R}$. [In $\mathbb{R}$, compact=closed and bounded, and $\mathbb{Z}$ is not bounded]

How about the topology for the infinitude of primes? i.e., topology generated by $s(a,b) = \{a+nb|n\in \mathbb{Z}\}$.

First I know $B= \{ s(a,b)|a,b \in \mathbb{Z}\}$ is a basis for $\mathbb{Z}$ and furthere $s(a,b)$ is both open and closed [$\because s(a,b) = \mathbb{Z} \setminus \cup_{j=1}^{a-1} s(a,b+j)$].

Note that $s(0,1) = \mathbb{Z}$, so if we consider $s(a,b)$ as an open set of $\mathbb{Z}$, then it seems for me $\mathbb{Z}$ is compact[because it can cover by one open set $s(0,1)$] in this topology, but I am not sure whether my approach is valid or not.

Is my approach correct? or is there any other approach to see whether $\mathbb{Z}$ is compact in this topology?

phy_math
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1 Answers1

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The approach is not valid. All sets, compact or not, can be covered by a single open set: the space itself. You need to consider all possible covers of $\Bbb{Z}$ by other open sets, and show that you can reduce these to a finite subset.

That said, $\Bbb{Z}$ is not compact. Indeed, $\mathcal{U}_p = \{pn : n \in \Bbb{Z}\}$ is open for all primes $p$, and so $$\{5n + 1 : n \in \Bbb{Z}\} \cup \{5n - 1 : n \in \Bbb{Z}\} \cup \bigcup_{p \text{ prime}} \mathcal{U}_p$$ is an open cover of $\Bbb{Z}$ by the fundamental theorem of arithmetic. But, since there are infinitely many primes of the form $5n + 2$ or $5n + 3$, and each of these primes appear exactly one open set in this union, this union has no proper finite subcover.

Theo Bendit
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