I need to find the interior and boundary of this set:
$$A=\{(x,y,z)\in R^3 : 0\leq x\leq1 ,0 \leq y\leq2, 0\leq z\lt3\} \setminus\{(0,0,0)\}. $$
We defined the interior as the set of all interior points, where we defined an interior point as:
point $a\in R^n$ is interior for $A \subseteq R^n$ if $\exists r>0 $ so that $K(a,r) \subseteq A$. (K being an open ball with a centre in a and a radius of r).
I understand the definitions in a logical sense but don't know how to apply the "ball condition" to a real example. I also don't understand how the different boundaries ($<, \leq$) impact it.
Best Answer
When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have $$ \begin{align} A &=\{x \in \mathbb{R} : 0\leq x\leq1 \} \setminus\{0\}\\ &=(0,1] \end{align}$$
In order to show that $\mathrm{int}(A)=(0,1)$ ... take $a \in (0,1)$ and define $r:=\mathrm{min}\{\frac{a}{2},\frac{1-a}{2}\}$. Then (perhaps with the help of a sketch) you can show that $K(a,r) \subset (0,1)$.
Can you complete this proof, then do similar in $\mathbb{R}^2$ and $\mathbb{R}^3$?