Let $X$ be a topological space. I have already shown that
(i) If $Q$ is a quasi-component of $X$ and $Q$ is open, then $Q$ is connected
and
(ii) If $X$ has only finitely many quasi-components, then each quasi-component is connected.
Now I want to show that if $X$ is a locally connected, then each quasi-component of $X$ is connected.
NOTE: The definition of quasi-component I am using: The quasi-component of $x$ is the intersection of all clopen subsets of $X$ containing $x$. Please use this definition.
I have three apparent options. Assuming $X$ is locally connected, I could show
(a) $X$ has finitely many quasi-components
(b) Any given quasi-component of $X$ is open
or, directly,
(c) Any given quasi-component is connected.
Option $(a)$ doesn't seem viable and so I haven't given it much thought. I have spent quite a while trying to show that a given quasi-component is open and trying to show that a given quasi-component is connected but to no avail.
Since $X$ is locally connected, given $x \in X$ and $U$ open in $X$ with $x \in U$, there exists a connected open subset $V \subset U$ with $x \in V$. I was thinking that from here I could show that $V$ is contained in the quasi-component of $x$, call it $Q_x$, so that $Q_x$ would be open, but that didn't work. Then I tried to run into a contradiction by supposing that $V$ intersects two distinct quasi-components of $X$ and nothing happened.
Any suggestions? Thanks 🙂
Best Answer
Suppose $C$ is a component of $X$, where $X$ is locally connected. If $x \in C$, then $x$ has a connected neighbourhood $U_x$. So $C \cup U_x$ is connected, as the union of two intersecting (in $x$) connected sets and $C \subseteq C \cup U_x$. So by maximality $U_x \subseteq C$. So $x$ is an interior point of $C$, and as $x \in C$ was arbitrary, $C$ is open. A connected component is always closed (or $\overline{C}$ would be a strictly larger connected subset). So $C$ is clopen.
This means that if $C$ is a component of $x$ in $X$, $C$ is one of the clopen subsets we intersect in computing the pseudocomponent. So the pseudocomponent $P_x$ of $x$ is a subset of $C$. By general theory, which I hope you have covered, the component of $x$ is a subset of the pseudocomponent $P_x$ of $x$ and so we have equality.