0
$\begingroup$

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

(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.

  • 0
    Umm... you might want to retype what U is. Right now I have no idea what you mean. – 2017-01-25
  • 1
    It is still not clear what $U$ is. – 2017-01-25
  • 0
    Can't seem to get the element of sign and the R^2 to work but I think it's readable now. – 2017-01-25

1 Answers 1

0

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.

  • 0
    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 $$U = \{(x_1,x_2) \ | (-1 < x_1, x_2 < 1)}$$ – 2017-01-25
  • 0
    Be careful. The interior of U is not $\{(x_1, x_2) : -1$(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 – 2017-01-25
  • 0
    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. – 2017-01-25
  • 0
    Having some trouble figuring out the value of $epsilon$, any hints? – 2017-01-25
  • 0
    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 $-1$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 – 2017-01-25
  • 0
    So if I can show that (x1,1) and (x1, -1) are greater than epsilon then I have proven that all the interior points are between 1 and -1? – 2017-01-25
  • 0
    Again, you must be careful. We are claiming that $$interior(U) = \{(x_1, x_2): -1$x_1$. So in order to prove our claim, the proof should look like this.... First we will show that $interior(U) \supseteq \{(x_1, x_2): -10$ such that $-1$(x_1, x_2)\in interior(U)$, it is enough to show that $B((x_1, x_2); \epsilon)\subseteq U$. – 2017-01-25
  • 0
    Sorry, I meant to say that it is enough to show that $B((x_1,x_2), \sqrt{\epsilon})\subseteq U$. So take $(q_1, q_2)\in B((x_1, x_2);\sqrt{\epsilon})$. This implies that $$\sqrt{|q_2-x_2|} \leq \sqrt{|q_2-x_2|+|q_1-x_1|} = \|x-q\| <\sqrt{\epsilon}$$ Therefore, $|q_2-x_2|<\epsilon$. Hence, $-1$U$, we see that $(q_1, q_2)$ is therefore inside of $U$. Hence $ B((x_1, x_2);\sqrt{\epsilon})\subseteq U$. By the definition of interior of U, we have that $(q_1, q_2)$ is in the interior of U – 2017-01-25
  • 0
    Oh, I was completely off. I trying to show that x1 is also part of the boundary. If I understand your explanation correctly, I pick 2 random points, take their distance from x1,x2 and then show that the norm of those points is less than an arbitrary epsilon and that would be enough? – 2017-01-26
  • 0
    You are on the right track, but again we need to be careful. To start, in this question note that one point is of the form $(x_1, x_2)$. That is, a point has two coordinates. So neither $x_1$ nor $x_2$ is a point. In order to prove that $\{(x_1, x_2):-10$ so that $B((q_1, q_2); \epsilon)\subseteq U$. Note that $\epsilon$ will most likely depend on the point $(q_1, q_2)$. In the previous message, I did not specify $\epsilon$. – 2017-01-26
  • 0
    Can't I also just choose an arbitrary epsilon as well? And then prove that the norm from (x,y)-(q1,q2) is less than it? The more I understand this the more confused I get. I am having a lot of difficulty as my textbook does not have any examples or a similar question to this so I am king of just going in blind. – 2017-01-26
  • 0
    Let me give an example. I hope this helps. Ex. Let $U = \{(x_1, x_2): x_2\leq 1\}$. Show $interior(U) = \{(x_1, x_2): x_2<1\}$. Solution: Let $(q_1, q_2)\in \{(x_1, x_2): x_2<1\}$. Let $\epsilon = \frac{1-q_2}{2}$. Note that $\epsilon>0$ since $q_2<1$ and $q_2+\epsilon <1$. Consider $B((q_1, q_2), \sqrt{\epsilon})$ -- we want to show this ball is in U. So we need to show for any $(w_1, w_2)\in B((q_1, q_2), \epsilon)$, that $w_2\leq 1$. Note $|q_2-w_2|\leq \sqrt{\epsilon}$ and so $w_2$(w_1, w_2)\in U$ and $B((q_1, q_2), \sqrt{\epsilon})\subseteq interior(U)$. – 2017-01-26
  • 0
    Here is another example: http://math.stackexchange.com/questions/684420/show-that-the-interior-of-a-set-s0-is-open – 2017-01-26
  • 0
    So can I make my $epsilon$ either 1+q or 1-q? Or is there any restriction on that? – 2017-01-26
  • 0
    Also thanks for the example, it's clearing some confusions up. – 2017-01-26
  • 0
    There is no restriction on what you make $\epsilon$ – 2017-01-26