In any quasi-ordered set X, we say that a subset A is upward-closed iff whenever $x\in A \ $and$\ x \leqslant y$, then $y\in A$.
Consider the collection $\mathcal{O}$ of all upward closed subsets of $X$. This is a topology. (In Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.P57), I must admit it's obvious.(1. $X,\emptyset$ is upward closed; 2. for every $x\in \bigcup_{\alpha \in I}U_\alpha$, $x \leqslant y$ $\exists \alpha _0\in I$ s.t. $x\in U_\alpha$ ($U_\alpha$ is upward closed subsets of $X$), $y\in U_\alpha \subset \bigcup_{\alpha \in I}U_\alpha$; 3. for every $x\in \bigcap_{i=1}^{n}U_i$, $x\in U_i$ ($\forall 1\leqslant i \leqslant n$), $y\in U_i$ ($\forall 1\leqslant i \leqslant n$), $y\in \bigcap_{i=1}^{n}U_i$).
My question is whether there is a situation where any intersection of upward-closed subset is not upward-closed(can be infinite)? If so, how should it be constructed?
There is not. Characterization by a pre/quasi-order is the Alexandrov property, equivalent to the arbitrary intersection of open being open.