Is there a connected $T_2$-space $(X,\tau)$ with more than 1 point and with the following property?

Whenever $D\subseteq X$ is dense, $X\setminus D$ is not dense.

  • 2
    The one-point space is an example, so probably you want to insist that there is more than one point. Or perhaps replace "connected" with "no isolated points"? – Joel David Hamkins Jul 14 '17 at 13:25
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    Some vague, trivial thoughts: the property is equivalent to the fact that any two dense sets must intersect. This looks like a local property: consider all the open sets, that can be decomposed as the disjoint union of their dense subsets. A countable union of such sets also has this property. Perhaps this works for ANY union as well. If our space is 2nd countable and also locally connected, then the reminder of the closure of the union will contain an open connected set $Y$ with the property that $int \overline{A}\subset \overline{int A}$, if $A\subset Y$. The latter looks very suspicious. – erz Jul 14 '17 at 13:44
  • If all open sets have the same cardinality, and this is equal to or larger than the size of a base for the topology, then we can build disjoint dense sets by transfinite recursion, by promising some points in one and other points in the other, in such a way that both are dense. – Joel David Hamkins Jul 14 '17 at 13:47
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    If there is a base for the topology of size less than the smallest open set (a common but not universal situation), then pick a point from each set in the base, and since this will not contain any open set, the complement will be dense. One can improve this to: a base for the topology of size at most the size of the smallest open set, by combining with the transfinite recursion to build the dense set and its complement. – Joel David Hamkins Jul 14 '17 at 13:55
up vote 12 down vote accepted

A topological space is called irresolvable if it is not the disjoint union of two dense subsets. So you are asking whether there is a connected, $T_2$, irresolvable space with more than one point. The answer is yes! You can find a proof, along with a few references to relevant literature, in

D. Anderson, "On connected irresolvable Hausdorff spaces," Proceedings of the AMS, 1965. link

There is such a thing as a submaximal topology, in which every dense subset is open. These obviously satisfy your condition.

Take any connected Hausdorff space $(X,\tau)$. Let $\mathscr F$ be an ultrafilter of $\tau$-dense sets. Let $\tau'$ be the topology generated by $\tau\cup \mathscr F$. Then $(X,\tau')$ is submaximal Hausdorff connected.

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