# Slightly finer topology on a connected space

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?

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