Discontinuity of the characteristic function Let $A \subseteq \mathbb{R}^n$. Let $f(x) = \chi_A $ be the characteristic function, and put $D = \{ x : f(x) \; \; \text{is discontinuous} \} $. Then $\partial A = D $.
MY try:
Let $y \in D $. Then, $f(x)$ is discontinuous at $y$. Therefore, we can find $\epsilon > 0$ such that for all $\delta > 0$ there exists $x \in A $ such that $||x - y|| < \delta $ implies $||f(x) - f(y) || > \epsilon $. Our goal is to show that for any $r > 0$, $B(y, r)$ intersects both $A$ and $\mathbb{R}^n \setminus A$. Suppose not. Then either $B(y,r) \subseteq A $ or $B(y, r) \subseteq ( \mathbb{R}^n \setminus A ) $ for some $r>0$. I don't see how to obtain a contradiction from here. Any help would be appreciated.
The other direction. Suppose $y \in \partial A$. We want to show that $f(x)$ is discontinuous at $y$. Suppose it is continuous, then for any $\epsilon > 0$, there exists $\delta > 0$ such that for all $x \in A $, we have 
$$ ||x-y|| < \delta \implies ||f(x) - f(y) || < \epsilon $$
In particular, we see that $x \in B(y, \delta) \subseteq A $. Hence $B(y, \delta) \cap ( \mathbb{R}^n \setminus A ) = \varnothing $. Hence, $y$ cannot be a boundary point. Contradiction. Is this direction correct?
thanks in advanced.
 A: Hint: More generally, characteristic function of a set $A$ in a given topological space $(\Omega,\tau)$ is continuous exactly on $\Omega\setminus \partial A$. We first note that  $\partial A$, $\mathrm{int}(A)$ (interior of $A$) and $\mathrm{int}(\Omega \setminus A)$ (interior of complement of $A$) form a partition of $\Omega$.
If $\omega \notin \partial A$, then $\omega \in \mathrm{int}(A)$ or $\mathrm{int}(\Omega \setminus A)$. So there is a neighborhood of $\omega$ fully included in $A$ (or $\Omega \setminus A$) on which $\chi_A$ is constant, $1$ (or $0$). Its image through $\chi_A$ is included in any given neighborhood of $\chi_A(\omega)$.
If $\omega \in \partial A$, then any neighborhood of $\omega$ has elements in $A$ that map to $1$ and elements in $\Omega\setminus A$ that map to $0$. So, there are neighborhoods of $\chi_A(\omega)$ (those that include $1$, but not $0$, or vice versa) that cannot include any image through $\chi_A$ of any neighborhood of $\omega$. 
A: The boundary $\partial A$ of $A \subset \mathbb{R}^n$ is the set of points $x\in \mathbb{R}^n$ satisfying one of the following:
1) $x$ is a point of $A$ and a limit point of the complement of $A$.
2) $x$ is a point of the complement of $A$ and a limit point of $A$.  
Therefore it is clear from the limit definition of continuity that the set of discontinuities of the characteristic function of $A$ is $\partial A$.  
