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????

• This is false. For example, let $A=\{0\}$. Then $\chi_A(x)$ is continuous on its boundary, which is simply $A$. But perhaps you mean that $\chi_A$ has no points of continuity (as a function on $\mathbb R^n$) in $\partial_A$, which is different than not being continuous on $\partial_A$. – Alex Becker Mar 13 '13 at 3:52
• @AlexBecker Yes, I think Paul is like me and is not careful enough with his wording. He most likely meant the latter. – Julien Mar 13 '13 at 3:56
• Restrict the function to the interior, that is, look into $\chi_A:\operatorname{int}A\to \{0,1\}$. I assume you endow $\{0,1\}$ with the discrete topology. Then $f^{-1}(X)$ is open for any choice of $X\subseteq \{0,1\}$: it is either $\operatorname{int}(A)$ or $\varnothing$. Do the same for $\Bbb R^n\setminus \operatorname{int}(A)=\operatorname{cl}(\Bbb R^n\setminus A)$ which is closed. In any case the preimage of any open/closed set is open/closed, so you're done. – Pedro Tamaroff Mar 13 '13 at 4:03
• @julien Do you agree with the above? – Pedro Tamaroff Mar 13 '13 at 4:12
• @AlexBecker What do you think? – Pedro Tamaroff Mar 13 '13 at 4:12

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 $A'$.

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.