2
$\begingroup$

Is $\mathcal{P}(\omega)/fin$ with the interval topology a connected space? (You find the definition of $\mathcal{P}(\omega)/fin$ here.)


Remark: According to this, the interval topology of $\mathcal{P}(\omega)/fin$ is not Hausdorff, but I haven't found out whether it contains non-trivial clopen sets.

$\endgroup$
7
  • 1
    $\begingroup$ Can you show us two disjoint non-empty open subsets? $\endgroup$ – Ramiro de la Vega Mar 23 '15 at 17:51
  • $\begingroup$ I can't... But this doesn't mean there aren't such open sets. My intuition for the interval topology on this particular poset is quite poor. $\endgroup$ – Dominic van der Zypen Mar 23 '15 at 18:32
  • $\begingroup$ There aren't, so the space is connected for trivial reasons. $\endgroup$ – Ramiro de la Vega Mar 23 '15 at 21:31
  • $\begingroup$ @RamirodelaVega, are you saying that it is trivial to see that any two nonempty open sets intersect? This is true, as I argue in my answer, but I found it to be a subtle issue. $\endgroup$ – Joel David Hamkins Mar 23 '15 at 23:59
  • $\begingroup$ @JoelDavidHamkins, I'm saying that there are no proper clopen subsets just because there are no disjoint non-empty open subsets (and this is a trivial reason). The fact that non-empty subbasic open sets intersect is not trivial but it is something that anyone interested in this space should be able to prove by himself. $\endgroup$ – Ramiro de la Vega Mar 24 '15 at 0:35
5
$\begingroup$

The answer is yes, because any two nonempty open sets have points in common. And this also shows directly that the space is not Hausdorff.

From what you describe in the other question, the topology is generated by the complements of the upper cones $\uparrow f=[f,1]=\{h\mid h\geq^* f\}$ and the lower cones $\downarrow g=[0,g]=\{h\mid h\leq^* g\}$, where I am regarding the points as equivalence classes of functions $f:\omega\to 2$ (but I suppress the equivalence classes). Thus, a set is open if it is a union of finite intersections of such cone complements. Let us take those finite intersections as basic open sets.

What I claim is that any two nonempty basic open sets have nonempty intersection.

Suppose that $U$ is the intersection of the complements of the intervals $[f_i,1]$ and $[0,g_j]$ for finitely many $i,j$. So a function $h:\omega\to 2$ is in $U$ just in case it is not almost-above any $f_i$ and not almost-below any $g_j$.

Similarly, suppose $V$ is the intersection of the complements of the intervals $[f_i',1]$ and $[0,g_j']$. So $V$ consists of the functions $h$ that are not almost-above any $f_i'$ and not almost-below any $g_j'$.

Assume that $U$ and $V$ are not empty. Thus, we may assume that the functions $f_i$ are not almost-always $0$ and the $g_j$ are not almost-always $1$. It follows that there are infinite sets $A_i$ and $B_j$ such that $f_i(n)=1$ for $n\in A_i$ and $g_j(n)=0$ for $n\in B_j$. And similarly $A_i'$ and $B_j'$ for $f_i'$ and $g_j'$. By shrinking these sets, we may assume that all $A_i, A_i', B_j, B_j'$ are pairwise disjoint.

Let $h$ be the function that is $0$ on every $A_i$ and $A_i'$ and $1$ on every $B_j$ and $B_j'$. Thus, $h$ is not above any $f_i$ nor any $f_i'$ and not below any $g_j$ nor $g_j'$. So $h\in U\cap V$, as desired.

$\endgroup$
3
  • $\begingroup$ I fixed an issue with my earlier answer. It seems to me that one must consider finite intersections of cone complements, which I do in the update. $\endgroup$ – Joel David Hamkins Mar 23 '15 at 21:07
  • 1
    $\begingroup$ Nice! In Steen-Seebach's "Counterexamples in topology" such spaces where any two nonempty open sets intersect are called hyperconnected. $\endgroup$ – Goldstern Mar 23 '15 at 22:41
  • $\begingroup$ Very nice answer, thanks Joel! Is this space also path-connected? $\endgroup$ – Dominic van der Zypen Mar 24 '15 at 5:53

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.