1
$\begingroup$

Let $X$ be a topological space and $\{C_i\}_{i\in I}$ be the family of its connected components. Now let $A\subseteq I$, define $f:X\rightarrow \mathbb{R}$ by $f(\cup_{i\in A}C_i)=1$ and $f(X\setminus\cup_{i\in A}C_i)=0$. How can we prove that $f$ is continuous

$\endgroup$
2
  • $\begingroup$ that's just writing down the definition. Have you tried anything? $\endgroup$ Aug 24, 2018 at 1:43
  • $\begingroup$ Maybe try it for just two components first $\endgroup$
    –  mheldman
    Aug 24, 2018 at 1:50

1 Answer 1

2
$\begingroup$

Actually, it’s not true.

Take $\Bbb Z_p$, the $p$-adic integers, which is totally disconnected, so that the only connected components are the singleton sets $\{p\}$. In this case, your $I$ is in natural correspondence with the space $\Bbb Z_p$ itself. Now let $A=I\setminus\{0\}$, whose union is, of course, the set of nonzero elements of $\Bbb Z_p$. Defining $\,f$ as you have done, we get a function that’s $1$ everywhere but at zero, where $f(0)=0$. Discontinuous, since zero is in the closure of the other set.

(Lots of other totally disconnected spaces would work as well. Like $\Bbb Q$, for instance.)

$\endgroup$

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.