# $X$ locally connected. Show quasi-components are connected.

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 :)

• Option d) Every connected component in a locally connected space is ... what? – Daniel Fischer Feb 12 '15 at 21:06

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.

• Thanks! Do you know of a good textbook that covers quasi-components? I took one semester of undergrad topology last year, and now I am taking the 2nd sem w/ a diff prof. We're using Munkres but he constantly lectures on topics not covered in the text, such as an in depth analysis of the Cantor set, projective systems of topological spaces, and profinite spaces. – Sam Feb 12 '15 at 21:32
• They're pretty obscure. They get a mention in Engelking, and in some other textbooks. They prove a few theorems like the one above, and give some examples of spaces where they're different from components. And then move on... Maybe books on continuum theory would have more. Why are you interested in them? – Henno Brandsma Feb 12 '15 at 21:35
• I don't know. My professor keeps teaching us about them, and I just want a textbook that at least talks about them. – Sam Feb 12 '15 at 21:38
• $C$ is the component of $x$ which means by definition that it is the largest by inclusion (maximal) connected subset of $X$ that contains $x$. $U_x$ is a connected subset of $X$ that contains $x$ and because $C$ is the maximal one it must be a subset of $C$. @ensbana – Henno Brandsma Mar 1 '18 at 10:41
• @ensbana if $U_x$ were not a subset of $C$, then $U_x \cup C$ would be connected (as both sets are connected as intersect in $x$) and strictly larger than $C$. This cannot happen as $C$ is a component. – Henno Brandsma Mar 1 '18 at 11:53