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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.

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If $F$ is a finite subset of $\mathbb{R}$, enumerate it as $x_1 < x_2 < x_3 < \ldots x_n$. Then $$X\setminus F = (-\infty, x_1) \cup (x_1, x_2) \cup (x_2, x_3) \cup \ldots \cup (x_n, +\infty)$$

which is a union of open intervals/segments so usual open.

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.

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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.

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