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Let $(X,\tau)$ be a connected Hausdorff space.

Suppose $S\subseteq X$ is such that for every $U\in\tau$, $$U\cap S\neq\varnothing \implies U\cap \overline S\setminus S\neq\varnothing.$$

Is it true that the topology generated by $\tau\cup \{S\}$ is also connected?

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    $\begingroup$ I think it is false because what if $X=\mathbb R$ in the usual topology and $S=\mathbb Q \cap (-\infty,0]$. $\endgroup$
    – aposyndetic
    Commented Dec 11, 2018 at 18:35
  • $\begingroup$ But isn't $(-\infty,0)\cup S$ open, with open complement? $\endgroup$
    – aposyndetic
    Commented Dec 11, 2018 at 22:39
  • $\begingroup$ but $S$ contains $0$ $\endgroup$
    – aposyndetic
    Commented Dec 11, 2018 at 23:10
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    $\begingroup$ Yes, that is right, this counterexample seems totally fine. $\endgroup$
    – Joel David Hamkins
    Commented Dec 11, 2018 at 23:17

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