TITLE: Why is $T_1$ required for a topological space to be $T_4$?
QUESTION [15 upvotes]: Let's say we have some topological space.
Axiom $T_1$ states that for any two points $y \neq x$, there is an open neighborhood $U_y$ of $y$ such that $x \notin U_y$.
Then we say that a topological space is $T_4$ if it is $T_1$ and also satisfies that for any two closed, non-intersecting sets $A,B$, there are open neighborhoods $U_A,U_B$ respectively, such that $U_A\cap U_B = \emptyset$.
Could anyone give an example of a topological space which satisfies the second condition of $T_4$, but which is not $T_1$?

REPLY [3 votes]: As an addition: the separating closed disjoint sets part is often called normality (a space is normal if it satisfies this), so $T_4$ is normal plus $T_1$, and similarly for regular: $T_3$ is regular and $T_1$. Spaces that have no disjoint non-empty closed sets (besides the trivial topology, we have examples like $\mathbf{N}$ with the topology generated by the sets of the form $U(n) = \{ k : k \ge n \}$ e.g.) trivially satisfy normality. $T_4$ is to avoid these pathologies: the extra $T_1$ ensures that at least all finite sets are closed, so we have some "relevant" closed sets to apply normality to...