The digital line topology is
B(n) = {n} when n is odd
B(n) = {n-1, n, n+1} when n is even.
I've figured out how to prove that the topology is not Hausdorff, but was wondering if there was a way to prove that there is a singleton set in the digital line topology that is not open.
Right now I have that if n is even then the set {n} is not closed since it does not have a complement in T that is open. Is this reasoning sufficient, or am I missing something?
Assume shown that the collection $B(n)$ as described is a basis, and further assume $n$ is odd. We show that $\lbrace n \rbrace$ is not closed, by showing that its complement is not open.
If its complement $U$ were open, then $n-1$ would be in $U$ and there would be a basis element $B$ containing $n-1$ in $U$. But since $n$ is odd, we have that $n-1$ is even and thus every basis element $B$ containing $n-1$ contains the elements $n-2$ and $n$. In particular we have that $n \in B$. But $B \subset U = \lbrace n \rbrace^c$ was assumed to be disjoint from $\lbrace n \rbrace$, a contradiction.