# Slightly finer topology vs a quasi-component

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 $$\tau\cup\{S\}$$?

• 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. – Joel David Hamkins Dec 12 '18 at 20:30
• @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. – aposyndetic Dec 14 '18 at 0:04
• You need to add an assumption which will avoid triviality: let $\ (X\ \tau)\$ be disconnected. – 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 $$S = X \setminus (U \cup V)$$.
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).