Is $(\mathbb{R}, \tau)$ normal, irrespective of the nature of $\tau$?

Question

Is a topological space $(\mathbb{R}, \tau)$ normal, irrespective of the nature of $\tau$?

For convenience I post the definition of normality:

Definition A topological space $(X, \tau)$ is said to be normal, if for closed sets $C$, $D \subseteq X$, if $C \cap D = \emptyset$, then there exist open sets $U$ and $V$ such that $C \subseteq U$, $D \subseteq V$ and $U \cap V = \emptyset$

Using the definition above it seems to be the case for me that if disjoint subsets $C$ and $D$ of $\mathbb{R}$ can be found, a proof similar to this one can be given. I quote from the chosen answer of that post:

Every metric space is normal, in particular $\mathbb R^n$. The proof goes as follows:

For each $a\in A$, let $r_a=\frac{1}{3}d(a,B)$, and for each $b\in B$, let $s_b=\frac{1}{3}d(b,A)$. Now define $U=\bigcup_{a\in A}B(a,r_a)$ and $V=\bigcup_{b\in B}B(b,s_b)$. It is not hard to show that $A\subseteq U,$ $B\subseteq V,$ and $U\cap V=\varnothing$.

If no two such sets can be found, the statement would be vacuously true like in this case.

Am I missing anything crucial?

Answer 1

Normality is a topological property because its very definition relies on open and closed sets. It is not a property of bare sets. The topology of choice must matter.

Give $\mathbb{R}$ the topology with open sets $\varnothing$ and any subset $A$ of $\mathbb{R}$ with $1 \in A$. Check that this is indeed a topology on $\mathbb{R}$ and that a set $C$ is closed in this topology if and only if $1 \notin C$ or $C=\mathbb{R}$.

Now, consider the disjoint closed sets $C=\{4\}$ and $D=\{5\}$. There cannot exist open sets $U$ and $V$ surrounding $C$ and $D$ respectively with $U \cap V = \varnothing$ because both $U$ and $V$ must contain $1$ by definition.

Answer 2

There are lots of non-normal topologies on $\mathbb{R}$, or indeed (almost) any set whatsoever. For example, the cofinite topology on an infinite set is never normal.

It's a good exercise to exactly determine which sets cannot be equipped with a non-normal topology. You can check by hand that all topologies on a set with at most two elements are normal; what if there are more than two elements in the set?

Suppose $X$ is a set with more than two elements. Let $a,b$ be distinct elements of $X$ and consider the topology $\tau$ generated by $\{X\setminus\{a\},X\setminus\{b\}\}$. Do you see why this is not normal?