Characterization of Connected Components for Compact Topological Spaces

Question

I'm trying to prove the following statement from James Dugundji's Topology (XI - section 2 - problem 6).

Let $X$ be compact (Hausdorff). Define an equivalence relation $R$ as follows: $xRy$ if for every continuous $f\colon X\to E^1$ such that $f(x)=0$, $f(y)=1$, there is a $u\in X$ with $f(u)=\frac{1}{2}$. Show that the equivalence classes are the components of $X$.

(I'll denote the equivalence class of $x$ as $[x]$ and it's component as $C(x)$).
I want to show that $[x]=C(x)$ for each $x\in X$. Showing $C(x)\subseteq [x]$ is easy: let $y\in C(x)$ and let $f\colon X\to E^1$ be continuous with $f(x)=0$ and $f(y)=1$. Since $C(x)$ is connected, $f\big(C(x)\big)$ is connected and thus $[0,1]=[f(x),f(y)]\subseteq f(C(x))$. It is then clear that $C(x)\subseteq [x]$.

In order to prove $[x]\subseteq C(x)$, I want prove that $[x]$ is connected. I have no idea how to prove this. We haven't used compactness yet so clearly this is the hard part of the problem. I'm looking for some hints on how to proceed

What are some hints in order to prove $[x]$ is connected?

Update: I'm now looking for hints, partial solutions, complete solutions, or anything that might allow me to eventually read a proof (either written by me or someone else) of the property.

Definitions. A space $X$ is Connected if it can't be decomposed into two disjoint open sets (i.e. no continuous $f\colon X\to \{0,1\}$ is surjective). A connected component for a point $x$ (written $C(x)$) is defined as the union of all connected sets containing $x$. $E^1$ denotes the first Euclidean space: $(\mathbb{R},\mathscr{T})$ where $\mathscr{T}$ is the usual topology.
$[x]=\{y\in X\mid \forall f\in C(X), [(f(x)=0\land f(y)=1)\implies (\exists u\in X,\; f(u)=1/2)]\}$.

Answer 1

As Eric Wofsey comments, we can invoke the fact that in compact Hausdorff spaces components agree with quasicomponents. See Any two points in a Stone space can be disconnected by clopen sets and Quasi-components and components coincide for compact Hausdorff spaces.

The quasicomponent $Q(x)$ of a point $x$ of a space $X$ is the intersection of all clopen subsets of $X$ containing $x$. This is a closed subset of $X$ which always contains the component $C(x)$ of $x$ in $X$. To see this, let $C \subset X$ be clopen with $x \in C$. Then $U = C \cap C(x)$ and $V = (X \setminus C) \cap C(x)$ are disjoint open subsets of $C(x)$ such that $U \cup V = C(x)$. Since $C(x)$ is connected and $x \in U$, we conclude $V =\emptyset$ and thus $C(x) \subset C$.

The quasicomponents of $X$ form a partition of $X$ into pairwise disjoint closed subsets of $X$. To see this, let us first show $(1)$ $Q(x) \subset Q(z)$ for all $z \in Q(x)$. Let $C \subset X$ be clopen with $z \in C$. Assume $x \notin C$. Then $C' = X \setminus C$ is clopen with $x \in C'$, thus $Q(x) \subset C'$. Hence $z \in C' = X \setminus C$ which contradicts $z \in C$. Next let us show $(2)$ $Q(x) = Q(z)$ for all $z \in Q(x)$. In fact, since $x \in Q(x) \subset Q(z)$ we get $Q(z) \subset Q(x)$ by $(1)$. But $(2)$ proves that $Q(x) \cap Q(y) \ne \emptyset$ implies $Q(x) = Q(y)$.

Lemma. In an arbitrary space $X$ the equivalence classes with respect to $R$ are the quasicomponents of $X$.

Proof. 1. $[x] \subset Q(x)$.

Assume there exists $y \in [x]$ such that $y \notin Q(x)$. Then there exists a clopen $C \subset X$ such that $x \in C$ and $y \notin C$. Define $f : X \to E^1, f(z) = 0$ for $z \in C$ and $f(z) = 1$ for $z \notin C$. This map is continuous wiuth $f(x)= 0$ and $f(y) =1$, but there is no $u \in X$ with $f(u)=1/2$. Hence $y \notin [x]$, a contradiction.

2. $Q(x) \subset [x]$.

Assume there exists $y \in Q(x)$ such that $y \notin [x]$. Then there exists a continuous $f : X \to E^1$ with $f(x) = 0$ and $f(y)= 1$ such that $1/2 \notin f(X)$. The sets $U_1= (-\infty,1/2)$ and $U_2 = (1/2,\infty)$ are disjoint open subsets of $E^1$, thus the $V_i = f^{-1}(U_i)$ are disjoint open subsets of $X$. Since $1/2 \notin f(X)$, we have $V_1 \cup V_2 = X$. Hence $V_1$ is clopen in $X$. Since $x \in V_1$, we have $Q(x) \subset V_1$. But $y \in V_2 = X \setminus V_1$, thus $y \notin Q(x)$, a contradiction.

Answer 2

If $[x]$ is not connected, then it is disconnected. As you say, there is a continuous surjective function $f: [x] \to \{ 0 ; 1 \}$. But this is a continuous function $f : [x] \to E^1$ with at least two points $a,b \in [x]$ such that $f(a)=0$ and $f(b)=1$, and with no $u$ with $f(u)= 1/2$. This is exactly the negation of the hypothesis $a R b$.