Co-finite and Co-countable topologies are not normal

Question

I’ve been investigating these two topologies on $\mathbb R$ and believe that they are both not normal as not every 2 disjoint closed sets have disjoint open neighbourhoods but I’m struggling to find an example that shows this (a counter example to the definition of normal).

I believe the same example will work for both cofinite and cocountable topology.

Can anyone help please?

Thanks

Answer 1

Take for example the closed subsets $A=\{0\}$ and $B=\{1\}$. Suppose you have neighborhoods $U$ of $0$ and $V$ of $1$ such that $U\cap V=\varnothing$. That would mean that $U^c\cup V^c=\mathbb{R}$. But $U^c$ and $V^c$ are finite (countable) in the co-finite (co-countable) topology, leading to a contradiction. So that would be a counterexample.

Answer 2

Every two non-empty open sets have non-empty intersection, therefore any two disjoint non-empty closed sets are instances that disprove normality.