# Is $\mathcal{P}(\omega)/(fin)$ with the interval topology path-connected?

Given a poset $$(P,\leq)$$ the interval topology 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 $$\uparrow x = \{y\in P: y\geq x\}$$.

Is $$\mathcal{P}(\omega)/(fin)$$ with the interval topology path-connected?

• According to your definition of "interval topology", it looks as though every subbasic open set is also closed, in any poset ... – Nik Weaver Mar 30 '15 at 14:36
• No, otherwise $\mathcal{P}(\omega)/fin$ would be disconnected, but it is connected (see mathoverflow.net/questions/200784/… ). My question is if it is even path-connected – Dominic van der Zypen Mar 30 '15 at 15:36
• Oh, I misread the definition, sorry. – Nik Weaver Mar 30 '15 at 17:57
• If we use $J=\mathbb{Q}\cap[0,1]$ in place of $\omega$, is the map from the unit interval $x\in[0,1]$ to the rational cut $\{q\in J\mid q<x\}$ a path from $\emptyset$ to $J$ in $P(J)/\text{Fin}$? If so, then we could translate this all around and path-connect any point to the top, deducing path-connectedness of $P(\omega)/\text{Fin}$. That is, we basically map reals to their cut in the rationals. Is it a path? – Joel David Hamkins Nov 8 '17 at 17:02
• It seems to me that it is; I'll write up something later when I have a chance. – Joel David Hamkins Nov 8 '17 at 18:45