Is Right ray topology is normal space?

Question

Let $\mathbb{R}$ be the set all real numbers with right ray topology $\Im$={$(a,\infty):a\in \mathbb{R}$}$\cup${$\mathbb{R},\emptyset$}.

Is $(\mathbb{R},\Im)$ a normal space?

Each non-empty closed set in $\mathbb{R}$ is $(-\infty,a]$ for some $a\in\mathbb{R}$.

Thanks.

Answer 1

That depends on the definition of "normal":

1. If "normal" includes Hausdorff then the answer is no, because (as you've correctly noted) closed sets are of the form $(-\infty,a]$, in particular points are not closed.
2. But if "normal" doesn't include Hausdorff, i.e. it only means that two disjoint closed subsets can be separated by disjoint neighbourhoods, then it is normal. Simply because no two closed subsets are disjoint.

Note that I've been taught that "normal" means 2., while $T_4$ means "normal + Hausdorff". Some authors however use "normal" and $T_4$ as synonomys that includes Hausdorff.

Answer 2

No, because it's not Hausdorff. Take $0,1\in\mathbb{R}$, then every open set containing $0$, also contains $1$. Also, it's not even a $T_1$ space, because $\mathbb{R}-\{0\}=(-\infty,0)\cup(0,+\infty)$ is not open, then $\{0\}$ is not closed with this topology.