I'm trying to solve this exercise from Dugundji's Topology textbook:
Given $a,b\in\mathbb{Z}^+$, define $U(a,b)=\{an+b:n\in\mathbb{Z}\}\cap\mathbb{Z}^+$. The set $\{U(a,b) : \text{g.c.d.}(a,b)=1\}$ is a basis for some topology $\tau$ in $\mathbb{Z}^+$. Prove that the countable connected Hausdorff space $(\mathbb{Z}^+,\tau)$ is not compact [Hint: Consider the open covering by the sets $\{U(p,1) : \text{all primes p}\}$].
Well, I found the hint useless cause the open covering suggested doesn't cover $\mathbb{Z}^+$ ($2\notin U(p,1)$ for all $p$ prime, considering that the elements $m$ in $U(p,1)$ satisfy that $m\equiv1\mod{p}$). I looked in the answer of the following question:
Prove the integers in the arithmetic progression topology is not compact ,
but the reasoning that they used works because they were working in all $\mathbb{Z}$ and $a,b$ don't have to satisfy that $\text{g.c.d.}(a,b)=1$. I don't know how to solve this. Any help, hint or suggestion will be really appreciated.
Consider the cover $\{U(p,1)\;|\; p:\textrm{ prime}\} \cup \{U(5,2)\}$. This fixes the problem with two not being in the original cover. If this had a finite subcover, then there would have to exist finitely many primes $p_1,...,p_n$ so that for all $N\in\mathbb{Z}_+$ either $N\equiv 2\mod 5$ or for some $i$ we have $N \equiv 1\mod p_i$. But, by the Chinese Remainder Theorem, we can find a number $N$ that is congruent to $3$ mod $5$ and is congruent to $2$ mod each $p_i$ not equal to $5$. This $N$ would fail each of these equations.