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Let $(X, \tau)$ be a topological space where $\tau$ is the closed-finite(co-finite) topology. Consider $A \subset X$, is the topology$\tau_{A}$ induced on $A$ by $(X, \tau)$ going to be closed-finite?

A closed finite topology is a topology where the closed sets are either the underlying set or any finite subset.

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Yes, suppose $A$ is a subspace of $X$, let $U$ be open in $X$, then since $X\setminus U$ is finite, $(X\setminus U) \cap A$ is finite so $U\cap A$ is open in the cofinite topology on $A$. Now if $U$ is open in the cofinite topology on $A$, then let $V=(X\setminus A)\cup U$. Then $V\cap A=U$, so all we need to do is show that $V$ is open in the cofinite topology on $X$ since then $U$ will be open in the relative topology on $A$. But $X\setminus V = X\setminus (X\setminus A) \setminus U = A\setminus U$, which is finite since $U$ is open in the cofinite topology on $A$.

Then the open sets in the relative topology and cofinite topology on $A$ are the same, so the topologies are the same.

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