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 $f(C_2) = (0,1]$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.