A topological space $X$ is called Artinian if the descending chain condition holds for open subsets of $X$. If the descending chain condition holds for open basis subsets of a Hausdorff space $X$ with the property that the intersection of two open basis is an open basis, can we prove that $X$ is Artinian?
Also, is there any proof for the fact that every Artinian Hausdorff space is finite?
If $X$ is Artinian, then for every $x \in X$ there is a minimal neighbourhood $U_x$ such that if $O$ is open with $x \in O$, then $U_x \subseteq O$.
This is a simple application of Zorn's lemma on the poset of open neighbourhoods of $x$, ordered by reverse inclusion. The Artianness of $X$ implies that all chains in this poset are finite and then the intersection is an upperbound. So a maximal element exists. If $O$ is then open and contains $x$, we have $O \cap U_x \ge U_x$ so that by minimality $O \cap U_x = U_x$ or $U_x \subseteq O$.
If $X$ is moreover Hausdorff or even $T_1$, for $x \in X$ we have that $U_x =\{x\}$ or otherwise $y \neq x$ with $y \in U_x$ would exist. But $x$ must have a neighbourhood missing $y$ and this contradicts $y \in U_x$. So $X$ is discrete. And a discrete space is Artinian iff it is finite.
