Show that the digital line topology on the integers is connected.

The digital line topology on the integers maps $\{n\}$ to $\{n\}$ when $n$ is odd, and $\{n\}$ to $\{n-1,n,n+1\}$ when $n$ is even.

My hint is to start with finite subsets $[-n,n]$.

This statement feels very intuitively true, but I cannot figure out how to formalize it. Specifically, if we picture this topology on $\mathbb{Z}$, there is no way to express $\mathbb{Z}$ as the union of two disjoint subsets $A$ and $B$. I think the problem boils down to the fact that if we want to make sure $A$ and $B$ contain the even integers, must include all subsets of the form $\{n1,n,n+1\}$, but then these subsets all overlap at the odd integers, and there is no way to make these subsets disjoint.

I would appreciate any guidance on how to formalize this reasoning!

The sets of the form $\{n\}$ for $n$ odd and $\{n-1,n, n+1\}$ for $n$ even are a basis for the digital line topology, which means that every open set in this topology is an arbitrary union of this kind of set.
If you try to express $\mathbb{Z}$ as union of two disjoint nonempty open subsets, at some point you'll have some $n$ odd separated from $n+1$, which is even. But that's not possible, since every nonempty open set containing $n+1$ contains also $n$.
If you start with $n$ even, the argument is analogue.
More rigorously, let's call those open sets $U$ and $V$. Now, there exists $n\in U$ such that $n+1$ or $n-1$ is not in $U$ (otherwise $U$ would be the whole $\mathbb{Z}$ by induction, thus $V=\emptyset$). Assume $n$ is odd and $n+1\in V$. Then, since $V$ is open and contains $n+1$ (which is even), $V$ must contain a subset of the form $\{n,n+1,n+2\}$, but then $n\in V$. Contradiction. Analogously for $n-1\in V$ and for $n$ even.
• A good hint, but those with less proof experience might benefit from more detail on how to formalize the argument "at some point you'll have some $n$ odd separated from $n+1$, which is even". – rosterherik Apr 25 '18 at 8:53