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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.

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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 $X$.

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