$$U = \{(x_1,x_2) \in \mathbb{R}^2 : -1 < x_2 \leq 1\}$$

(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.

  • $\begingroup$ Umm... you might want to retype what U is. Right now I have no idea what you mean. $\endgroup$ – fleablood Jan 25 '17 at 22:30
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    $\begingroup$ It is still not clear what $U$ is. $\endgroup$ – Jan Jan 25 '17 at 22:32
  • $\begingroup$ Can't seem to get the element of sign and the R^2 to work but I think it's readable now. $\endgroup$ – quwerty Jan 25 '17 at 22:38

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.

  • $\begingroup$ I have a similar diagram and I am using that to prove the boundary points and interior points, so I have that the interior of $$U = \{(x_1,x_2) \ | (-1 < x_1, x_2 < 1)}$$ $\endgroup$ – quwerty Jan 25 '17 at 23:03
  • $\begingroup$ Be careful. The interior of U is not $\{(x_1, x_2) : -1<x_1, x_2<1\}$. For example, the point $(1,0)$ looks to be in the interior of U although it does not fit this definition. Instead, the interior of U is just $\{(x_1, x_2) : -1<x_2<1\}$. $\endgroup$ – JSP Jan 25 '17 at 23:05
  • $\begingroup$ Ah you're right, The origin is excluded from the interior points correct? So I guess I can't really use the picture but I'll have to find an $epsilon$ and go from there. $\endgroup$ – quwerty Jan 25 '17 at 23:08
  • $\begingroup$ Having some trouble figuring out the value of $epsilon$, any hints? $\endgroup$ – quwerty Jan 25 '17 at 23:21
  • $\begingroup$ No, the origin is an interior point. I'll show you why and perhaps that will help you prove something about the interior. Let $y = (0,0)$ and take $\epsilon = 1/2$. I claim that $B(y; \epsilon)\subseteq U$. So take some point in $x\in B(y; \epsilon)$. In order to show that $x\in U$, we need to show that $-1<x_2\leq 1$. By the definition of $y, \epsilon$ and $B(y; \epsilon)$, we have $$\|x-y\| =|(x_1)^2+(x_2)^2| = (x_1)^2+(x_2)^2 < \epsilon = \frac{1}{2}.$$ So it must be the case that $-1<x_2\leq 1$ because otherwise the inequality above would be false. $\endgroup$ – JSP Jan 25 '17 at 23:22

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