Find the interior and boundary of a set.

Question

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.

Answer 1

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 $\mathbb{R}^3$?

Answer 2

If $a=(x,y,z)\in A$ and $0\in \{x,y,z\}$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) \not \subset A.$ Because $(x-r/2, y-r/2, z-r/2)\in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,\;y-r/2,\;z-r/2$ is negative.

Similarly if $a\in A$ and $(x=1\lor y=2\lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)\in K(a,r)$ but $(x+r/2>1\lor y+r/2>2 \lor z+r/2>3)$ so $K(a,r)\not \subset A.$

Now $[\;a=(x,y,z)\in A$ $ \land 0\not \in \{x,y,z)\}\land (x\ne 1\land y\ne 2 \land z\ne 3)\;]$ if and only if $a\in B=(0,1) \times (0,2)\times (0,3).$.. (where $(u,v)$ denotes an open interval in $\Bbb R$.) .... If $a\in B$ let $r=\min \{x,\;1-x,\;y,\;2-y,\;z,\;3-z\}$. Then $r>0.$ And $(x',y',z')\in K(a,r)\implies \max (|x'-x|, |y'-y|,|z'-z|)<r$ $ \implies (x',y',z')\in B.$.. So $K(a,r)\subset A.$

Similarly to $\Bbb R^2$ and to $\Bbb R,$ if $(x,y,z)\in S\subset \Bbb R^3$ and if either (i).. $(x,y,z') \not \in S$ whenever $z'>z$ or (ii).. $(x,y,z')\not \in S$ whenever $z'<z,\;$... then $(x,y,z) \not \in $ int$(S)$.

Remarks. The phrase $0\not \in \{x,y,z\}\land (x\ne 1\land y\ne 2 \land z\ne 3)$ can be written $0\not \in \{x,y,z,1-x,2-y,3-z\}.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).