Consider the full subcategory of Top consisting of all spaces $X$ such that a subset $A$ of $X$ is closed if and only if $A \cap K$ is closed in $K$ for all subspaces $K$ of $X$ which are countably compact, that is every open covering has a countable subcovering. It would be convenient for a proof I'm trying to write if this were known to be cartesian closed; is it known whether this is the case?
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$\begingroup$ I doubt it -- the usual way to show cartesian closure is to use that the category is generated under colimits by exponentiable spaces, which won't be the case here. $\endgroup$– Tim CampionCommented May 4, 2020 at 17:53
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$\begingroup$ In fact -- doesn't the usual counterexample to cartesian closure of $Top$ just use the quotient of $\mathbb R$ by some countable subset? If I understand correctly, this construction should be perform-able in your category. $\endgroup$– Tim CampionCommented May 4, 2020 at 17:55
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