Throughout this post, let $(P,\leq)$ be a poset. The interval topology $\tau_i(P)$ on $P$ is generated by $$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$ where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.

Now we define the order convergence topology, denoted by $\tau_o(P)$. By a set filter $\mathcal{F}$ on $P$ we mean a collection of subsets of $P$ such that:

  • $\emptyset \notin \mathcal{F}$;
  • $A, B\in \mathcal{F}$ implies $A\cap B\in \mathcal{F}$;
  • $U\in \mathcal{F}$, $U'\subseteq P$ and $U'\supseteq U$ implies $U'\in \mathcal{F}$.

If $S\subseteq P$ we define $S^u= \{x\in P: x\geq s\text{ for all } s\in S\}$, and $S^l= \{x\in P: x\leq s\text{ for all } s\in S\}$. If $\cal{F}$ is a set filter, then we set ${\cal F}^u = \bigcup\{F^u: F\in \cal{F}\}$ and define ${\cal F}^l$ similarly. For $x\in P$ and ${\cal F}$ a set filter on $P$ we write $${\cal F}\to x \textrm{ iff } \bigwedge\cal{F}^u = x = \bigvee \cal{F}^l.$$

Then we set $\tau_o(P)=\{U\subseteq P: \textrm{ for any } x\in U \text{ and any filter }\mathcal{F} \text{ with } \mathcal{F}\to x \text{ we have } U\in \mathcal{F}\}$. It is not hard to verify that this defines a topology.

Question: Given any poset $(P,\leq)$, does $\tau_i(P)\subseteq \tau_o(P)$ hold?


1 Answer 1


The answer is Yes, and it suffices to show that for $p\in P$ we have $(P\setminus \downarrow p) \in \tau_o(P)$ for all $p\in P$. (A similar argument then also shows that $(P\setminus \downarrow p) \in \tau_o(P)$, which proves that a subbasis of $\tau_i(P)$ is contained in $\tau_o(P)$.)

Let $x\in P\setminus \downarrow p$. This is equivalent to saying that $x\not\leq p$. Let $\mathcal{F}$ be a filter such that $\mathcal{F}\to x$, so we have $x = \bigvee \mathcal{F}^l = \bigwedge\mathcal{F}^u$.

We assume that $(P\setminus\downarrow p)\not \in \mathcal{F}$ and derive a contradiction.

Now if $(P\setminus\downarrow p)\not \in \mathcal{F}$, then we have $F\cap \downarrow p \neq \emptyset$ for every $F\in\mathcal{F}$. We look at 2 cases concerning $\mathcal{F}^l$:

Case 1: $\mathcal{F}^l = \emptyset$. This implies that $x$ is the smallest element of $P$ because $x = \bigvee \emptyset$, but this means that $x\leq p$, contradicting our assumption.

Case 2: $\mathcal{F}^l \neq \emptyset$. So since every member $F\in\mathcal{F}$ intersects $\downarrow p$, we have $\mathcal{F}^l \subseteq \downarrow p$, which in turn implies $\bigvee \mathcal{F}^l \leq p$. But we have $\bigvee\mathcal{F}^l = x$, and $x\neq p$, contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy