$$U = \{(x_1,x_2) \in \mathbb{R}^2 : -1 < x_2 \leq 1\}$$
(a) Find all interior points of U. For each interior point, find a value of r for which the open ball lies inside U.
(b) Find all boundary points of U.
(c) Is U an open set? Is U a closed set? Why or why not?
So I know the definitions of boundary points and interior points but I'm not sure how to use the definitions of each to solve for this. First time attempting a problem like this, any guidance would be appreciated.
1 Answer
A drawing of the set U with interior and boundary points
The picture is a drawing of the set U (drawn in purple). The dashed line at $x_2=-1$ is not included in the set U, while the solid line $x_2=1$ is contained in U. Moreover, any point with $x_2$ strictly between -1 and 1 is contained in U. There are three points highlighted inside of U, namely w, y, and z.
Intuitively, y is in the interior of U. In order to show this mathematically, you can use the definition of an interior point that some open ball around y is contained in U. A ball of radius $\epsilon>0$ around y is of the form $B(y; \epsilon)=\{x\in \mathbb{R}^2: \|x-y\|<\epsilon\}$ (in the picture, I drew such a ball). So if $y\in U$ is chosen so that $y_2\in (-1, 1)$, then you can choose $\epsilon$ small enough so that $B(y; \epsilon)\subseteq U$. You'll have to do a bit of work to find such an $\epsilon$.
On the other hand, w and z are on the boundary of U. To show this mathematically, you can use the definition of a boundary point showing that every ball around w or z touches a point in U and a point outside of U. Consider some ball $B(w; \epsilon)$ around $w$ with radius $\epsilon>0$ (in the picture, I drew an example). Note that, since $w_2 = -1$, any choice of $\epsilon$ will result in some point $q\in B(w; \epsilon)$ with $q_2 <-1$ and some point $r\in B(w; \epsilon)$ with $r_2>-1$. Hence $q\not\in U$ but $r\in U$. This shows that $w$ is on the boundary.
