# Component and quasi-component

Let $X$ be a topological space and $x\in X$. Then the quasi-component of the point $x$, denoted by $C_x$, is the intersection of all clopen (closed-and-open) subsets of $X$ which contain the point $x$. Clearly the quasi-components of two distinct points of a topological space $X$ either coincide or are disjoint, so the set of quasi-components constitutes a decomposition of the space $X$ into pairwise disjoint closed subsets. And clearly the connected component $C$ of a point $x$ in a topological space $X$ is contained in the quasi-component $C_x$ of the point $x$, and so every quasi-component is the union of the connected components of its points.

Question: Let $C_1$ and $C_2$ be two connected components of $X$ such that there exists $x\in X$ with $C_1\cup C_2\subseteq C_x$ and let $f$ be a real-valued, continuous function on $X$ such that $f(C_1)=\{0\}$. How can we show that $f(C_2)=\{0\}$?

Let $X = ((\omega_1 + 1) \times [-1,1]) \setminus (\omega_1,0)$, where $\omega_1 + 1$ has the order topology and $[-1,1]$ has its usual topology. Then $C_x = \{ \omega_1\} \times ([-1,0) \cup (0,1])$ is a quasi-component containing the components $C_1 = \{\omega_1\} \times [-1,0)$ and $C_2 = \{\omega_1\} \times (0,1]$. However, any continuous real-valued function on $X$ which is $0$ on $C_1$ and $1$ on $C_2$ would be $0$ on $\{\alpha \} \times [-1,0)$ and $1$ on $\{\alpha \} \times (0,1]$ for some $\alpha < \omega_1$, which cannot happen.
The question seems to have changed since the above answer was given. It is not in general the case that $f(C_1) = \{0\}$ implies $f(C_2) =\{0\}$. For instance, in the above example, let $f$ be the function which is given by $f(x,y) = 0$ if $y \leq 0$ and $f(x,y) = y$ if $y > 0$. Then $f(C_1) = \{0\}$ but $f(C_2) = (0,1]$