Let $(X,\tau)$ be a topological space, and let $Q$ be a quasi-component of $X$. Let $S$ be a subset of $X\setminus Q$. Then is $Q$ necessarily a quasi-component of $X$ in the topology generated by $\tau\cup\{S\}$?

  • $\begingroup$ For reference, the quasi-component of a point is the intersection of all clopen sets containing that point. en.wikipedia.org/wiki/Connected_space#Connected_components It is usually best to explain such background concepts when posting on MathOverflow. $\endgroup$ – Joel David Hamkins Dec 12 '18 at 20:30
  • $\begingroup$ @JoelDavidHamkins yes that is correct. thank you for this background. I thought of this problem and found it very difficult to solve. It it helps, assume spaces are separable metric. This will mean each quasi-component is the intersection of a decreasing countable sequence of clopen sets. $\endgroup$ – aposyndetic Dec 14 '18 at 0:04
  • $\begingroup$ You need to add an assumption which will avoid triviality: let $\ (X\ \tau)\ $ be disconnected. $\endgroup$ – Wlod AA Dec 15 '18 at 3:19

Let $X$ be the subspace of the plane given by $X = \{ (\frac{1}{n},y) : n = 1, 2, \cdots,\ 0 \leq y \leq 1 \} \cup \{(0,0),(0,1)\}$, and let $S = \{ \frac{1}{n} : n = 1, 2, \cdots\} \times \{\frac{1}{2}\}$. Then the quasi-component of $(0,0)$ in $X$ is $\{(0,0),(0,1)\}$ but in the topology generated by $X$ and $S$ the quasi-component of $(0,0)$ is $\{(0,0)\}$.

In fact, if $X$ is a normal space and $Q$ is a disconnected quasi-component, then there is a subset $S$ of $X \setminus Q$ such that $Q$ is not a quasi-component of the topology generated by adding $S$ to the topology of $X$. For this let $(H,K)$ be a disconnection of $Q$. Then $H$ and $K$ are disjoint closed subsets of $X$. Let $U$ and $V$ be disjoint open subsets of $X$ such that $H \subseteq U$ and $K \subseteq V$. Let $S = X \setminus (U \cup V)$.


First I'll present a special class of counter-examples. Then I'll show that this class is not empty (far from it).

Let $\ (X\ \tau)\ $ be a topological space such that the entire $\ X\ $ is a quasi-component of $\ (X\ \tau)\ $ while $\ G\ H\in\tau\setminus\{\emptyset\}\ $ are disjoint. Then $\ X\ $ is not a quasi-component with respect to topology generated by $\ \tau\cup F,\ $ for $\ F:=X\setminus(G\cup H).$

There are non-trivial (meaning that $\ (X\ \tau)\ $ is disconnected) separable metric spaces $\ (X\ \tau)\ $ and $\ (X\ \ \tau\cup F)\ $, e.g. let $\ E\ $ be the Erdos space of all points of $\ \ell^2\ $ which have all coordinates rational. (It's easy to fill up this with the necessary obvious details).


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