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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?

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  • $\begingroup$ According to your definition of "interval topology", it looks as though every subbasic open set is also closed, in any poset ... $\endgroup$
    – Nik Weaver
    Commented Mar 30, 2015 at 14:36
  • $\begingroup$ 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 $\endgroup$
    – Dominic van der Zypen
    Commented Mar 30, 2015 at 15:36
  • $\begingroup$ Oh, I misread the definition, sorry. $\endgroup$
    – Nik Weaver
    Commented Mar 30, 2015 at 17:57
  • $\begingroup$ 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? $\endgroup$
    – Joel David Hamkins
    Commented Nov 8, 2017 at 17:02
  • $\begingroup$ It seems to me that it is; I'll write up something later when I have a chance. $\endgroup$
    – Joel David Hamkins
    Commented Nov 8, 2017 at 18:45

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