# Proving the cofinite topology is coarser then the Euclidean topology.

Let $$\tau_1$$ and $$\tau_2$$ be two topological spaces on a set $$X$$. Then $$\tau_1$$ is said to be a finer topology than $$\tau_2$$ if $$\tau_1\supset\tau_2$$.

Prove that the Euclidean topology in $$\mathbb{R}$$ is finer than the finite-closed topology.

Let's define the cofinite topology$$\tau_c=\{A:\mathbb{R}\setminus A\: \text{is finite} \}$$

Therefore $$A$$ cannot be an open interval of the form $$(a,b)$$ once the compliment is not finite. $$A$$ can be an interval of the form $$(-\infty,a]\cup[b,+\infty)$$ or the form $$(-\infty,x)\cup (a,+\infty)$$. So both of them are contained in a open set in the usual topology on $$\mathbb{R}$$, generated by the open sets, which proves the assertion.

If $$F$$ is a finite subset of $$\mathbb{R}$$, enumerate it as $$x_1 < x_2 < x_3 < \ldots x_n$$. Then $$X\setminus F = (-\infty, x_1) \cup (x_1, x_2) \cup (x_2, x_3) \cup \ldots \cup (x_n, +\infty)$$
So all sets in $$\tau_c$$ (these are of the form $$X\setminus F$$ with $$F$$ finite, or equal to $$\emptyset$$) are "usual open", and this is what you had to show.
Being contained in an open set isn't enough. What must be proved is that every element of $$\tau_c$$ is an open subset of $$\mathbb R$$,with respect to the usual topology. Let $$A\in\tau_c$$. Then either $$A=\emptyset$$ or $$A^\complement$$ is finite. If $$A=\emptyset$$, it's trivial. And if $$A^\complement$$ is finite with respect to the usual topology, then $$A^\complement$$ is closed with respect to the usual topology. But then $$A$$ is open with respect to the usual topology.