# connected and quasi-connected separators of a space

Does there exist a connected topological space $$X$$ and a subset $$A\subseteq X$$ such that no connected component of $$A$$ separates $$X$$, but some quasi-component of $$A$$ separates $$X$$?

Meaning $$X\setminus C$$ is connected for every connected component $$C$$ of $$A$$, but there is a quasi-component (intersection of all relatively clopen sets of $$A$$ containing a particular point) $$Q$$ of $$A$$ such that $$X\setminus Q$$ is disconnected?

Here is my idea.

Let $$S$$ be the dyadic solenoid. Let $$X_0$$ and $$X_1$$ be two disjoint composants of $$S$$. Let $$\alpha\simeq [0,1]$$ be an arc in $$X_0$$. Let $$\delta_0$$ and $$\delta_1$$ be disjoint countable dense subsets of $$\alpha$$, like disjoint rationals. Define $$X=(X_0\setminus \delta_0)\cup (X_1\cup \delta_1)$$. The topology on $$X$$ will be generated by the Solenoid subspace topology together with the sets $$X_0\setminus \delta_0$$ and $$X_1\cup \delta_1$$.

1. $$X$$ is connected.

2. Let $$A$$ be a be a closed Solenoid neighborhood of $$\alpha$$ such that $$A\neq X$$. Then $$\alpha\setminus \delta_0$$ is contained in a quasi-component $$Q$$ of $$A$$. To see this, note that an irrational set in $$\alpha\setminus (\delta_0\cup \delta_1)$$ is retained by $$X$$, and the basic neighborhoods at these points are precisely the Solenoid neighborhoods. A sequence of arcs in $$X$$ uniformly limits to this set, so $$\alpha\setminus (\delta_0\cup \delta_1)\subseteq Q.$$ Since $$Q$$ is closed, $$\alpha\setminus \delta_0\subseteq Q$$.

3. Each connected component of $$\alpha\setminus \delta_0$$ is a singleton which does not separate $$X$$.

4. $$X\setminus \delta_1$$ is the union of two separated sets $$X_0\setminus \delta_1$$ and $$X_1\setminus \delta_1$$.

Does this sound right? The topology will only be Hausdorff, and of course ultimately I'd like regular or metrizable.

Here is a simple example. Let $$X$$ be $$({\mathbb{R}}^2 \setminus (\{0\} \times \mathbb{R})) \cup \{(0,0),(0,1)\})$$, that is, $$X$$ is the plane with all points on the $$y$$-axis removed except for the points $$(0,0)$$ and $$(0,1)$$. Let $$A = (\{ \frac{1}{n} : n = 1, 2, \cdots \} \times [0,1]) \cup \{(0,0), (0,1)\}$$. Then the components of $$A$$ are the line segments $$\{\frac{1}{n}\} \times [0,1]$$ along with the singletons $$\{(0,0)\}$$ and $$\{(0,1)\}$$. None of these components disconnect $$X$$ but the quasi-component $$\{(0,0),(0,1)\}$$ does disconnect $$X$$.