Determine Connectedness for a Topology on $\mathbb{Z}$

Question

The question is

Consider the topology on $\mathbb{Z}$ that has basis of the form $B(a,k) = \{a + p^k m \mid m \in \mathbb{Z}\}$. Where $a$, $k$ are integers and $k$ is nonnegative, $p$ is a arbitrary prime number. Is $\mathbb{Z}$ connected in this topology?

Here is my attempt.

I want to give a counterexample. I observed that when $p = 2$, $B(1,1) = \{1+2m \mid m \in \mathbb{Z}\}$ and $B(1,0) = \{2m \mid m\in \mathbb{Z}\}$ is a separation, hence disconnected. Then, I concluded that when $p$ is arbitrary, there always exists $B(1,1) \cup \cdots \cup B(p-1,1)$ which are disjoint and seem to be a separation. Am I in the right direction? Any help is greatly appreciated!!

Answer 1

For any $p$, the basic open sets are equivalence classes $\pmod {p^k}$, and those partition $\Bbb Z$, so this topology is disconnected for any $p$.

Answer 2

Thanks guys. I figured out eventually. $\mathbb{Z}$ is indeed not connected. To show that we have to create a separation: Take $U = B(1,1)$ and $V = B(2,1) \cup \cdots \cup B(p-1,1)$. We have to check whether $U, V$ are disjoint, nonempty, open and their union is $\mathbb{Z}$. Nonempty and open is trivial.

To show disjoint, by contradiction we suppose $x \in U $ and $x \in V$. We can get $ 1 + pm = a + pm'$ for $m, m' \in \mathbb{Z}$. Arrange this equation to get $p(m - m') = a-1$. Since $a - 1 < p$ and $p(m - m') \ge p$, we formed the contradiction. Hence, $U \cap V = \emptyset$.

To show $U \cup V = \mathbb{Z}$, we take any integer $x$ and divide it by $p$. Then it must have a remainder $a$ such that $1 \le a \le p-1$. And $x$ can be expressed as $x = a + pm$.

Hence it's disconnected (i.e., not connected).