$\mathcal J = \{ U \subset X \mid X \setminus U \ \ \text {is compact} \} \cup \{\emptyset\}$ defines a topology on Hausdorff space $X$? Suppose $(X,\mathcal I)$ is a topological space that is Hausdorff.
Consider the set $$\mathcal J = \{ U \subset X \mid X \setminus U \ \ \text {is compact} \} \cup \{\emptyset\}$$
How do I see that $\mathcal J$ is a topology on $X$ ?
First I've tried to verify whether $X \in \mathcal J$. However, I cannot seem to utilize that $X$ is Hausdorff ?
 A: $X$ is in $\cal T$ because $\emptyset$ is compact.
Union of arbitrary number of sets in $\cal T$ is again in $\cal T$, because taking complements, we get some intersection of compact sets, which is again compact. The Hausdorff condition is required to show that intersection of compact sets are compact. We use the fact that closed subsets of Hausdoff spaces.
Intersection of finitely many sets in $\cal T$ is again in $\cal T$, because taking complements, we get some union of finitely many compact sets, which is again compact. This is clear from a direct application of the definition of compactness by open covers.
Therefore $\cal T$ is a topology.
A: I'm don't think that the Hausdorffness condition is required, but some extra condition is.
Consider the set $X = \mathbb Z \cup \{ - \infty , + \infty \}$ with the topology generated by base consisting of the following sets:


*

*$\{ n \}$ for $n \in \mathbb Z$;

*$\{ - \infty \} \cup \mathbb Z$;

*$\{ + \infty \} \cup \mathbb Z$.


This topology is not Hausdorff since $-\infty , + \infty$ cannot be separated by disjoint open sets.
Note that both $K_1 = \{ -\infty \} \cup \mathbb Z$ and $K_2 = \{ +\infty \} \cup \mathbb Z$ are compact subsets of $X$, and so $\{ + \infty \} = X \setminus K_1$ and $\{ - \infty \} = X \setminus K_2$ are in the family $\mathcal{J}$. However $K_1 \cap K_2 = \mathbb Z$ is not compact, meaning $\{ -\infty , + \infty \}$ is not in the collection $\mathcal{J}$.  As $\mathcal J$ is not even closed under finite unions, it cannot be a topology on $X$.
What the Hausdorffness condition gives us is that every compact subset of the space is closed, and as closed subsets of compact sets are always compact, it follows that in Hausdorff spaces intersections of compact sets are compact. Translating this into the family $\mathcal J$, for Hausdorff $X$ the collection $\mathcal J$ is closed under arbitrary unions (which is necessary for it to be a topology).
I'm pretty sure you can show that $\mathcal J$ is a topology just assuming that all compact subsets of $X$ are closed, which is weaker than Hausdorffness.
