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Let $X$ be a topological space. For any $A \subseteq X$, consider two possible conditions on $A$:

1) $A$ is open in $X$;

2) $A \cap K$ is open in $K$, for each compact set $K \subseteq X$.

Then $(1)$ implies $(2)$. but, necessarily $(2)$ doesn't imply $(1)$. Give an example of $A$ such that $A$ satisfies in $(2)$ but $A$ is not open in $X$.

Spenser
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user302646
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    (I think in fairness to Goblin, you should reference his/her post: http://math.stackexchange.com/questions/1241768/spaces-in-which-a-cap-k-is-closed-for-all-compact-k-implies-a-is-close?rq=1, whose wording is very similar to yours) – Jason DeVito - on hiatus Jul 01 '16 at 13:21

2 Answers2

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Take $X$ to be an uncountable set with the cocountable topology (a set is open if and only if its complement is countable). You can verify that the compact subsets of $X$ are finite sets with the discrete topology.

Hence, take any $A\subseteq X$ that is not open. Then, for any compact $K\subseteq X$, $A\cap K$ is open in $K$.

Spenser
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A space $X$ is anticompact if and only if the only compact subsets of $X$ are the finite subsets. Let $X$ be $T_1$, anticompact, and not discrete. Since $X$ is not discrete, it has a non-empty subset $A$ that is not open. $X$ is $T_1$, so the relative topology on each finite subset of $X$ is discrete, and therefore $A\cap K$ is open in $K$ for each compact $K\subseteq X$. Thus, any such $X$ provides an example.

Such spaces can even be Hausdorff: this answer by bof contains three of them, and I gave another in this answer. Three of those four examples are even zero-dimensional and normal.

Community
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Brian M. Scott
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