prove characteristic function of an open set is continuous.

Question

The question is asking : prove $\chi_G$ is continuous at each point of $G$, where $G$ is an open set in $R$.
it may seem silly! but I ended up proving $\chi_G$ is NOT continuous.
$\chi^{-1}(0)=G^c$
$\chi^{-1}(1)=G$
G is open but singletons in R are closed, it is not like the preimage of "EVERY" open set is open then. so I assume we must conclude the characteristic function above is not continuous. I know I am wrong of course. I need help with understanding where did I go wrong?

Answer 1

Well, you're almost right. The characteristic function of $A$, where $A$ is a subset of a topological space $X$, is continuous if and only if $A$ is open and closed.

Your argument essentially shows this. More precisely, $\{0\}$ and $\{1\}$ are open in $\{0,1\}$, so $\chi_A^{-1}(\{1\})=A$ and $\chi_A^{-1}(\{0\})=X\setminus A$ must be open in order that $\chi_A$ is continuous. Therefore $A$ is open and closed.

The converse is easy.

However, if the problem is whether $\chi_G$ is continuous at every point of $G$, then it's completely obvious: every point of $G$ has a neighborhood where the function is constant.

Answer 2

The "pre-images of open sets are open" definition of continuity is only valid when you are saying the function is continuous as a whole, i.e. continuous everywhere in its domain. Your question is talking about a function being continuous at a point.

While there is a definition of "continuous at a point" that only uses open sets (see RobArthan's comment below), since your question refers to subsets of $\mathbb{R}$, you can fall back on the good old $\epsilon$-$\delta$ definition.