I have been working on the following example for sometime now and wanted to ensure my reasoning is correct. This example is from a chapter in my textbook on The Continuity of Maps between Metric spaces.
Is there a continuous characteristic function, $X_{A}$, on $\mathbb{R}$? Also, if $A \subset \mathbb{R}$ show that $X_A$ is continuous at each point of the interior of $A^{o}$ (the interior or $A$).
Note that the characteristic function $X_A$ I am referring to (usually denoted with chi, but MathJax doesn't take subscripts to this greek letter well) between two metric spaces $(M,d_{1})$ and $(N,d_{2})$ is defined on a subset $A \subset M$ such that $X_{A}(x) := 1$ if $x \in A$ and $X_{A}(x) := 0$ if $x \notin A$.
$\bullet$ For: Is there a continuous characteristic function, $X_{A}$, on $\mathbb{R}$?
Yes, take $\mathbb{R}$ as a subset of itself. So $X_{\mathbb{R}}(x) = 1$ $\forall x \in \mathbb{R}$. Therefore, the preimage, $X_{\mathbb{R}}^{-1}(\{1\}) = \mathbb{R}$ is closed and since the preimage of a closed set is closed, $X_{\mathbb{R}}$ is continuous. As well, if we take $\emptyset \subset \mathbb{R}$: $X_{\emptyset}(x) = 0$. So $X_{\emptyset}^{-1}(\{0\}) = \emptyset$ is closed. By similar reasoning, the preimage of a closed set is closed so we can conclude $X_{\emptyset}$ is continuous.
$\bullet$ For: Also, if $A \subset \mathbb{R}$ show that $X_A$ is continuous at each point of the interior of $A^{o}$ (the interior or $A$).
We know that $A^{o} \subset A$, so let $x \in A^{o}$. So we know $X_A(\{x\}) =\{1\}$ since $x$ is also in $A$. Therefore, $X_A^{-1}(\{1\}) = \{x\}$ where $x$ is any element in $A$, and since $A^{o} \subset A$, it can be any element in the interior. So since the preimage of a closed set is closed, $X_A$ is continuos at each point of the interior of $A$, and also continuous on all of $A$.
I feel like this reasoning is correct, but I have my doubts. Also, are there any other points of continuity such as in $A^c$? Any criticism is welcome.
Best Answer
It is not true that $\chi_A^{-1}\big(\{1\}\big)=\{x\}$ for $x\in\operatorname{int}A$: $\chi_A^{-1}\big(\{1\}\big)=\{x\}$ iff $A=\{x\}$, in which case $x\notin\operatorname{int}A$. Remember, by definition $$\chi_A^{-1}\big(\{1\}\big)=\{y\in\Bbb R:\chi_A(y)=1\}=A\,.$$
To show that $\chi_A$ is continuous at $x$, let $\epsilon>0$. Since $x\in\operatorname{int}A$, there is a $\delta>0$ such that $(x-\delta,x+\delta)\subseteq A$. Thus, if $|y-x|<\delta$, then $\chi_A(y)=1$, so $|\chi_A(y)-\chi_A(x)|=0<\epsilon$, and $\chi_A$ is therefore continuous at $x$.