Let $X$ be a topological space and $\{X^n\}_{n\in\mathbb{Z}}$ be a filtration of $X$ (i.e. $X^n\subset X^{n+1}$) whose union is $X$.
Assume that $X^{-2}=\emptyset$ and $X^0\setminus X^{-1}$ is discrete. Assume also that $X$ has the weak topology determined by $\{X^n\}_{n\in \mathbb{Z}}$. Define $A:=X^{-1}$ and $B:=\bigcup_{n\in\mathbb{N}} X^n\setminus A$.
Now the question: If $A,B$ are both Hausdorff, then is $X$ Hausdorff?
Moreover, if $A,B$ are $T_4$, then is $X$ $T_4$?
Thank you in advance!
No. For instance, $X=X^0$ could be any space with two points, with $X^{-1}$ containing only one of the points. Then $A$ and $B$ both have only one point and so are $T_4$, but $X$ does not have to be Hausdorff.