Components vs Quasicomponents

Question

Forgive me if this question has been asked before, but I did a quick search and nothing came up. My book (Geometry and Topology by Bredon) Defines components of a topological space $X$ as

The collection of equivalence classes of the equivalence relation "$p$ and $q$ belong to a connected subset of $X$"

while it defines quasicomponents as

The collection of equivalence classes of the equivalence relation "$d(p)=d(q)$ for every discrete valued map $d$ on $X$"

While I understand these definitions, I do have one misunderstanding: I don't understand how these concepts are different. The theorem

A topological space is connected iff every discrete valued map is constant

seems to show that they are one and the same:

Let $[\,\cdot\,]_C$ be the equivalence classes defined in the former definition and let $[\,\cdot\,]_Q$ be the equivalence classes defined in the latter definition.

$(\Rightarrow)$ (this direction is apparently true) If $[x]_C$ is the component in $X$ with $x$ as its representative, then for any $y\in [x]_C$, $x,y$ both belong to a connected subset of $X$. By the above we must then have $d(x)=d(y)$ for all discrete valued maps $d$. Thus $y\in [x]_Q$ and $[x]_C\subseteq [x]_Q$.

$(\Leftarrow)$ If $z,w\in [x]_Q$ are chosen arbitrarily, then $d(z)=d(x)$ and $d(w)=d(x)$ for all discrete valued maps $d$. But then $d(z)=d(w)$ for all discrete valued maps, so the above shows that $[x]_Q$ is connected, and hence $[x]_Q\subseteq [x]_C$.

The $(\Leftarrow)$ direction is evidently wrong. But why? A few months ago I encountered this same issue and apparently overcame it, but I forgot how. Thanks in advance.

Answer 1

The problem is that $[x]_Q$ is not necessarily connected; there just isn’t a function from $X$ to $\{0,1\}$ that takes both values on $[x]_Q$. The sketch below shows an example of such a situation.

The space is a subspace of $\Bbb R^2$ consisting of the two infinite horizontal lines at top and bottom and infinitely many rectangles expanding so that the $x$-coordinates of their left and right sides approach $-\infty$ and $\infty$, respectively, and their top and bottom sides approach the upper and lower horizontal lines, respectively. The quasicomponent of $x$ consists of the top and bottom horizontal lines: if $f:X\to\{0,1\}$ is continuous, and $f(x)=0$, then $f(y)=0$ for each point $y$ not only on the top line, but also on the bottom line. (To prove this, try to prove that $f$ must be $0$ on all sufficiently large rectangles.) Clearly, though, this set is not connected: it has two components, the top line and the bottom line.