Example of a $T_0$ and normal that is not $T_4$

Question

I'm pretty sure that the search for this example resumes to finding a topological space that is $T_0$ and normal that is not $T_1$, But I can't think of any that possesses those characteristics. Please Help.

Answer 1

Define a topology on the natural numbers by saying a subset is open if it contains 31 or is empty. You are right that once a space has points as closed sets and is normal it is a Hausdorff space.

Answer 2

Normality of a space $X$ says that if $F$ and $G$ are disjoint closed sets, there are disjoint open sets $U$ and $V$ such that $F\subseteq U$ and $G\subseteq V$. This is always possible if one of $F$ and $G$ is empty. (Say $F=\varnothing$; then we take $U=\varnothing$ and $V=X$.) Suppose that $X$ does not contain two disjoint, non-empty, closed sets: then $X$ is automatically normal. Thus, all we need to do is find a $T_0$ space that does not contain two disjoint, non-empty, closed sets. (As long as $X$ has at least two points, that guarantees that $X$ is not $T_1$, since singletons are closed in a $T_1$ space.)

One of the standard examples of a $T_0$ space that is not $T_1$ is $\Bbb N$ with the increasing nest topology: for $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let

$$\tau=\{\Bbb N\}\cup\{U_n:n\in\Bbb N\}\;.$$

You can easily check that $\langle\Bbb N,\tau\rangle$ is $T_0$. The closed sets are $\varnothing$ and the ‘tails’

$$T_n=\Bbb N\setminus U_n=\{k\in\Bbb N:k\ge n\}\;,$$

and clearly no two of the tails are disjoint.