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Let $X_1$ and $X_2$ be infinite sets and $T_1$ and $T_2$ be the finite-closed topology on $X_1$ and $X_2$, respectively. Show that the product topology, $T$, on $X_2 \times X_2$ is not the finite closed topology.

I know that I have to show that there exist a set $X \backslash F$ such that it's not open in $T$. So: $$(X_1 \times X_2) \backslash (F_1 \times F_2) = (X_1 \backslash F_1) \times X_2 \bigcup X_1 \times (X_2\backslash F_2). $$ But here I'm stuck.

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A space $X$ has the "finite-closed" topology, if the only closed sets are $X$ and the finite subsets of $X$.

If $X_1,X_2$ are two infinite spaces, then $X_1 \setminus \{x_1\}$ is open in $X_1$ if $X_1$ has the finite-closed topology. But then $(X_1 \setminus \{x_1\}) \times X_2$ is also open, so its complement in the product, the set $\{x_1\} \times X_2$, is closed in $X_1 \times X_2$. But it is infinite, as $X_2$ is. So $X_1 \times X_2$ does not have the finite closed topology, even regardless of the topology on $X_2$, we just need it is infinite.

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