Suppose $A = \{ (x,f(x)) : x \in \mathbb{R} \} $. $f$ is a continuous function. I want to find the boundary, closure and interior of $A$. We know that $A$ is closed since $f$ is continuous, then we must have that $\overline{A} = A $. However, I'm having hard time trying to find $\operatorname{Int} A$ and $\partial A $. Can someone help me? thanks
[Math] interior, closure, and boundary of a graph of a continuous function
general-topologyreal-analysis
Best Answer
Hint: One result you may want to prove (if you don't already know it) is that if $A^\circ$ denotes the interior of $A,$ then $\partial A=\overline A\setminus A^\circ.$ This will make finding the boundary simple, once you've found the interior.
Now, consider the sets $$A_\epsilon:=\bigl\{(x,y)\in\Bbb R^2:f(x)-\epsilon<y<f(x)+\epsilon\bigr\}$$ for $\epsilon>0.$ You should be able to show that $A\subseteq A_\epsilon$ for all $\epsilon>0,$ and that $A=\bigcap_{\epsilon>0}A_\epsilon.$ Now, use this to show that $A$ cannot contain any open ball, and so $A^\circ=$ . . . what?
Added: Take any $(x,y)\in A$--meaning $y=f(x)$--and take any $r>0.$ Note that $(x,f(x)+\frac r2)$ lies in the open ball about $(x,y)$ of radius $r,$ but fails to lie in $A_{\frac r2}.$ Hence, since $A\subseteq A_{\frac r2},$ then $(x,f(x)+\frac r2)$ fails to lie in $A.$ Hence, the open ball about $(x,y)$ of radius $r$ is not contained by $A.$ Thus, $A$ does not contain any open ball (why?), and so $A^\circ=\emptyset.$