Show that $\chi_A$ is continuous on $\operatorname{int}{A}$ and $A'$ but not $\partial_A = \overline{A} \cap \overline{A'}$ Let $A$ be a subset of $\mathbb{R}^n$. Show that the characteristic function $\chi_A$ is continuous on the interior of $A$ and of its complement $A'$, but is discontinuous on the boundary $\partial_A = \overline{A} \cap \overline{A'}$

Thoughts:
$\chi_A(x) = \left\{ 
  \begin{array}{l l}
    1 & \quad \text{if $x\in{A}$}\\
    0 & \quad \text{if $x\notin{A}$}
  \end{array} \right.$
After plotting the graph,
Obviously,
$\operatorname{int}{A}$ is the continuous horizontal line for all segments with $y=1$
$A'$ is the continuous horizontal line for all segments with $y=0$
for $\operatorname{int}$ and ${A'}$ must cover all possible x, how come intersection is empty????
 A: To begin with, here is an example in $\mathbb{R}$. Take $A=[0,1)$. Then the interior of $A$ is $(0,1)$, the interior of $A'$ is $(-\infty,0)\cup(1,+\infty)$, and the boundary is $\{0,1\}$. Note that this gives a partition of $\mathbb{R}$. If you take $A$ to be a line in the plane, then the interior of $A$ is empty, the interior of $A'$ is $A'$, and the boundary is the line $A$. Note this is again a partition of the plane.
As pointed out by AlexBecker, we need to be careful with the wording. I suppose you regard $\chi_A$ as a function from $\mathbb{R}$ to $\mathbb{R}$, equipped with the usual topology inherited from the norm/absolute value.
If a function is continuous at every point of $S\subseteq X$, then it is continuous on $S$ (good exercise on the induced topology). The converse is not true in general. For instance, $\chi_\mathbb{Z}$ is continuous on $\mathbb{Z}$, but it is discontinuous at every integer as a function on $\mathbb{R}$. 
If a function is constant on $S\subseteq X$, it is easily seen to be continuous on $S$. So $\chi_A$ is continuous on the interior of $A$, and on the interior of $A'$, because it is constant there.
If a function is continuous on an open set $S\subseteq X$, then it is continuous at every point of $S$ (good exercise again). So $\chi_A$ is also continuous at every point of the interiors of $A$ and $A'$.
Now what about the boundary?
Assume that $\chi_A$ is continuous at some point $x$ of the boundary. You can use sequences. By assumption, there exist $(x_n)$ in $A$ and $(x'_n)$ in $A'$ which both converge to $x$. What is the value of $\chi_A$ on these sequences? What are these values supposed to converge to? Look for the contradiction. 
Now you've shown that $\chi_A$ is discontinuous at every point of the boundary. But this does not mean it can't be continuous on the boundary.
