Assume all spaces are metric.
Question. Does there exist a space $X$ which is totally disconnected (the components of $X$ are singletons), yet some quasicomponent of $X$ has nonempty interior?
I know the deleted Knaster–Kuratowski fan (Cantor's leaky tent) is totally disconnected and has no singleton quasicomponents. But the quasicomponents of this space have empty interior, right? The quasicomponents are the lines of rationals or irrationals starting at the base (the Cantor set) and ending at the top of the tent, so I think this would follow from the fact that the Cantor set has no isolated points.
If the answer to my question is yes, are there extra conditions which guarantee all of the quasicomponents in a totally disconnected space have empty interior? If a space is
- totally disconnected,
- has no isolated points, and
- is locally compact (Hausdorff of course),
then the quasicomponents would be equal to the components - singletons - none of which have interior. But local compactness seems a bit extreme! Based on the K-K fan example I want to guess:
Conjecture. If a totally disconnected space has no isolated points, then all of its quasicomponents have empty interior.
I believe the following is a counterexample to your conjecture.
In order to construct the Knaster-Kuratowski fan, we let $C\subseteq[0,1]$ be the standard Cantor set and, for each $c\in C$, let $L(c)$ be the segment from $(c,0)$ to $(\frac12,\frac12)$. We define $$X_c =\{(x,y)\in L(c)\mid y\in\mathbb Q\}$$ if $c\in C$ is an endpoint of one of the intervals at some finite stage of the construction of $C$ and $$X_c =\{(x,y)\in L(c)\mid y\notin\mathbb Q\}$$ for all other $c\in C$. Then, the Knaster-Kuratowski fan is defined as $$X=\bigcup_{c\in C}X_c.$$ Now, define a linear map $f:\mathbb R^2\to\mathbb R^2$ by $f(x,y)=(\frac{x-y}2,\frac{x+y}2)$ and let $$Y=f(X).$$ This is just a smaller (and rotated) copy of the Knaster-Kuratowski fan, based on the segment from $(0,0)$ to $(\frac12,\frac12)$ and with apex $(0,\frac12)$.
To construct our counterexample, let $X'=X\setminus\{(\frac12,\frac12)\}$ and $Y'=Y\setminus\{(\frac12,\frac12),(0,\frac12)\}$. We shall also use the notations $X_c'=X_c\setminus\{(\frac12,\frac12)\}$, $Y_c=f(X_c)$ and $Y_c'=Y_c\setminus\{(\frac12,\frac12),(0,\frac12)\}$. Our counterexample is provided by $$Z=X'\cup Y'.$$
Claim. The space $Z$ is totally disconnected.
Proof. Suppose $Q$ is a connected component of $Z$. If $Q$ intersects $X_c$ for some $c>0$, we have $Q\subseteq X_c$ by the usual argument (construct a separation using a line through $(\frac12,\frac12)$) and $Q$ must be a singleton. Otherwise, we have $Q\subseteq Y'\cup X_0'$. In this case, observe that $Q$ must be contained in a segment $S(x)$ with endpoints $(0,\frac12)$ and $(x,x)$ for some $x\in[0,1]$: $C\cup(\mathbb Q\cap[0,1])$ is totally disconnected, which allows us to separate two points in $Y'\cup X_0'$ lying in distinct segments $S(x_1)$ and $S(x_2)$ by a line through $(0,\frac12)$. Again, this implies that $Q$ is a singleton. $\square$
Next, we need a lemma.
Lemma. Suppose $X\subseteq Y$, and $Q\subseteq X$ and $R\subseteq Y$ are quasicomponents such that $Q\cap R\neq\emptyset$. Then $Q\subseteq R$.
Proof. Let $q\in Q\cap R$. Then $Q$ is the intersection $\cap_{\lambda}U_\lambda$ of all clopen sets in $X$ containing $q$. If $V$ is a clopen set in $Y$ containing $q$, $V\cap X$ is clopen in $X$ and therefore $V\cap X\supseteq U_\lambda$ for some $\lambda$. Hence, the interesection of all such sets $V$ contains $\cap_{\lambda}U_\lambda = Q$. $\square$
Claim. The set $Y'\cup X_0'$ is a quasicomponent of $Z$.
Proof. First of all, $Y'\cup X_0'$ is the intersection of clopen sets $Y'\cup\bigcup\limits_{c\in C_n}X_c'$, where $C_n=C\cap[0,\frac1{3^n}]$. It remains to show that if a clopen set $U$ intersects $Y'\cup X_0'$, we have $Y'\cup X_0'\subseteq U$. Observe that if $U$ intersects $Y_c'$, we have $Y_c'\subseteq U$ by the lemma, since $Y_c'$ is a quasicomponent in $Y'$. Similarly, if $U$ intersects $X_0'$, we have $X_0'\subseteq U$ by the lemma, since $X_0'$ is a quasicomponent in $X'$.
Now, suppose $U$ intersects $(Y'\cup X_0')\setminus Y_1'$. Then $U$ contains a segment of the form $Y_c'$ or $X_0'$, so it contains a point of the form $(x,x)$ for some $x\in[0,1]$. Since $U$ is open, it also contains a neighborhood of $(x,x)$ and therefore intersects $X_0'$. So $X_0'\subseteq U$. But $U$ is also closed, so it contains the closure of $X_0'$, which is $(X_0'\cup f(C))\setminus\{(\frac12,\frac12)\}$. This in turn intersects each $Y_c'$ for $c<1$ so $Y_c'\subseteq U$ for all $c<1$. So, $(Y'\cup X_0')\setminus Y_1'\subseteq U$. But since $U$ is closed, is also contains the closure of this set, i.e. $Y'\cup X_0'\subseteq U$.
Finally, if $U$ intersects $Y_1'$, it must also contain some element of $(Y'\cup X_0')\setminus Y_1'$ since it is open. We may again conclude that $Y'\cup X_0'\subseteq U$. This completes the proof. $\square$
Therefore $Z$ is a totally disconnected space without isolated points whose quasicomponent $Y'\cup X_0'$ has nonempty interior.