We only consider $\mathbb R$. In the finite complement topology, only finite sets are closed whereas in the standard topology, finite sets are closed but not the only closed sets. For example $[0,1]$ is closed in the standard topology, but is infinite. So, the standard topology is "larger".
This is my intuition for saying the standard topology is finer but how can I go about showing this formally? That is, by showing that a basis element for the finite complement topology contains a basis element from the standard topology.
What is a general basis element for the finite complement topology?
Best Answer
There’s no reason to look at bases for either topology: you know exactly what every open set is in the finite complement topology. Let $\tau$ be the finite complement topology on $\Bbb R$. If $U\in\tau$, then either $U=\varnothing$, which is certainly open in the usual topology, or $\Bbb R\setminus U$ is finite. Let $F=\Bbb R\setminus U$. The usual topology is $T_1$, so $F$ is closed in the usual topology, and therefore $U=\Bbb R\setminus F$ is open in the usual topology. This shows that $\tau$ is at least as coarse as the usual topology. And $(0,1)$ is open in the usual topology but not in $\tau$, so in fact $\tau$ is strictly coarser than the usual topology on $\Bbb R$.