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Let $(X,\tau)$ be a topological space. Prove that $\tau$ is the finite-closed topology on $X$ if and only if (i)$(X,\tau)$ is a $T_1$-space, and (ii) every infinite subset of $X$ is dense in $X$.

I already proved the forward direction but I'm stuck on the backward direction. We know that everyone singleton is closed because of i) and from ii) every open set intersects any infinite set non trivially. Now I need to figure out how to show that every open set are just infinite sets with countably finite points removed, thus the topology is finite-closed.

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Let $U$ be a non-empty open set. If $X \setminus U$ has an infinite number of points then the infinite set $X \setminus U$ cannot be dense. This is because it does not intersect $U$. This contradiction proves that $X \setminus U$ is finite whenever $U$ is a non-empty open set.

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Let $O$ be an open set in $(X,\tau)$, so $O \in \tau$.

  • If $O=\emptyset$, $O \in \tau_{fc}$, as required.
  • If $O=X$, likewise, $O \in \tau_{cf}$.
  • In the final case $O$ is non-empty and is disjoint from $O^\complement$ by definition, and by $(ii)$, $O$ must intersect every infinite set. So $O^\complement$ is not infinite, hence $O^\complement$ is finite, i.e. $O \in \tau_{cf}$.

QED

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You have to prove that, under hypothesis $(i),(ii)$, a non-trivial set (i.e. non-empnty and not $X$) is closed if and only if it is finite.

Any closed set different from $X$ is not dense (because it coincides with its closure). Then by $(ii)$ it is not infinite, hence it is finite. Conversely, any finite set is finite union of points, that are closed because of $(i)$, so it is closed.

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