Showing that all sets in $\mathcal{B}:=\{a+b\mathbb{Z}:a\in \mathbb{Z}, b\in \mathbb{N}\}$ are closed wrt the topology $\mathcal B$ is the basis of

Question

$\mathcal{B}:=\{a+b\mathbb{Z}:a\in \mathbb{Z}, b\in \mathbb{N}\}$

I already showed that $\mathcal B$ is a basis of a topology on $\mathbb Z$. How can I show that every set in $\mathcal B$ is closed w.r.t. that topology?

For this I want to show that their complements are open which means they can be written as unions of elements of $\mathcal B$.

Buw how do I do this? What's $(a+b \mathbb Z)^c$ and how can I write it as such a union of elements of $\mathcal B$?

(I know that $a + b\mathbb Z=x +b\mathbb Z \ \forall x \in a+b \mathbb Z$ if that helps)

Answer 1

Just to make clear what $a+b\mathbb{Z}$ means: \begin{align} 3 + 5\mathbb{Z} &= \{\ldots,3-5\times 2,3-5\times 1,3,3+5\times 1,3+5\times 2,\ldots\} \\ &= \{\ldots,-7,-2,3,8,13,\ldots\}. \end{align} Therefore the complement of $a+b\mathbb{Z}$ is equal to \begin{align} (a+b\mathbb{Z})^c = (a+1)+b\mathbb{Z} \cup (a+2)+b\mathbb{Z} \cup \cdots \cup(a+b-1)+b\mathbb{Z}. \end{align}

Answer 2

The statement that all sets are closed is actually the same as the statements that all sets are open (see my comment on the question).

So if indeed all sets are closed then the set $\{0\}$ must be an open set.

Then some $B\in\mathcal B$ must exist with $0\in B\subseteq\{0\}$.

There is only one candidate: $B=\{0\}$ so we conclude that $\{0\}\in\mathcal B$.

However if here $0\notin\mathbb N$ then all elements of $\mathcal B$ are infinite sets so that in that case $\{0\}\notin\mathcal B$.

This indicates that here $0\in\mathbb N$ and based on that we find that every singleton is an element of $\mathcal B$ because:$$\{a\}=a+0\mathbb Z\in\mathcal B$$

Then all singletons are open and consequently all sets are open.

Then also all sets are closed.