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.


1 Answer 1


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 $X$.


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.