4
$\begingroup$

I know my question will sound stupid, that it should be simple, and I know there are already a lot of questions related to this topic, but I've spent hours on it and I still don't get how to show that a given set is open or not. I'm completely stuck and I don't exactly know how I'm supposed to start and proceed. I already looked for other posts about the topic (like this one, for example, for which I did not manage to understand the answers that were given) but it didn't help.

Here is a set for which I'm supposed to determine if it is open or not: $$U := \{(x_1 , x_2) \in \mathbb{R}^2 : x_2 > \sqrt{|x_1|} \}$$

What I know (and what I would like to use):

  • Let $(X,d)$ be a metric space. A set $U \subset X$ is open if, for all $x \in U$, there exists $r>0$ such that $B(x,r) \subset U$.
  • For a metric space $(X,d)$ with $x \in X$, an open ball $B(x,r)$ is defined as $\{y \in X : d(x,y) < r \}$
  • The metric $d$ is not specified, so I guess it's the Euclidean metric: for $x = (x_1, x_2)$ and $y=(y_1, y_2)$, we have $d(x,y) = \sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2}$.

So, if I understand all this properly, what I'm supposed to show is that, for all $x \in U$, there exists (or not) $r$ such that $B(x,r) \subset U$, i.e. such that the elements $(y_1, y_2)$ of $B(x,r)$ are such that $y_2 > \sqrt{|y_1|}$. And, since these elements are in $B(x,r)$, they are such that $\sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2} < r$.

So, if I get this right, the elements $y$ of $B(x,r)$ should be such that $y_2 > \sqrt{|y_1|}$ and $\sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2} < r$. But then what? I don't understand what I'm supposed to do with all this. How can I "mix" those things together so that I have, at the same time, a convenient $r$ and $y_2 > \sqrt{|y_1|}$?

I'm missing something, but what? Could anyone give me some hints or indications that would help me to get started? I hate to ask for help in such situations, because it looks like I did not even try, but the thing is that I really don't know where to start. Any help would be, therefore, greatly appreciated.

$\endgroup$
  • $\begingroup$ you could simply argue that $f(x_1, x_2) = x_2 - \sqrt{|x_1|}$ is continuous and thus $U= f^{-1}((0,\infty))$ is open... $\endgroup$ – user251257 Apr 18 '17 at 17:38
3
$\begingroup$

Your region is above the line in the graph below. Intuitively, any point above the line is some distance away from it. You can draw a circle of half that radius around the point and stay within your region, so the region is open. It is a little work to find the shortest distance from a given point to the line. If your given point is $(x_3,x_4)$, let us assume $x_3 \gt 0$. The distance to a general point $(x_1,\sqrt{x_1})$on that side of the curve will be $\sqrt{(x_3-x_1)^2+(x_4-\sqrt {x_1})^2}$. You can minimize this over $x_1 \gt 0$ to find the distance.

enter image description here

$\endgroup$
  • $\begingroup$ Minimize the squared distance, it's cleaner. $\endgroup$ – Trevor Gunn Apr 18 '17 at 17:02
  • $\begingroup$ @T.Gunn: good point. $\endgroup$ – Ross Millikan Apr 18 '17 at 17:03
  • $\begingroup$ Thanks for your answer. I have to run to catch a train at the moment so I'll take more time later to study your answer, but right now: what do you mean by "minimize this over $x_1 > 0$" ? $\endgroup$ – justdoit Apr 18 '17 at 17:10
  • $\begingroup$ @alissad The minimization is to compute the distance between the point $(x_3,x_4)$ and the line, and then take $r$ less than that distance, if you want to explicitly give a ball around each point of $U$, contained in $U$. $\endgroup$ – amrsa Apr 18 '17 at 17:39
  • $\begingroup$ @RossMillikan I'm in the train now and I read your answer and comments... I really don't understand what's going on. Why would we consider the point $(x_1, \sqrt{|x_1|})$ since that point is not in our set $U$? And what does it mean to "minimize" the square root? And why would it be cleaner to minimize the squared distance? I'm completely lost. $\endgroup$ – justdoit Apr 18 '17 at 17:42
3
$\begingroup$

I've been playing with this and it seems like the shortest distance between a point in $U$ and the curve $y = \sqrt{|x|}$ is a root of a cubic polynomial. If you would like to avoid solving a cubic polynomial, what you can do instead is try to show that the complement of $U$ is a closed set.

That is, if you have a sequence $(x_n,y_n)$ with $y_n \le \sqrt{|x_n|}$ such that $x_n \to x_0, y_n \to y_0$ then show that $y_0 \le \sqrt{|x_0|}$. This works because the square root function is continuous.

$\endgroup$
  • $\begingroup$ Thanks, that's a good idea. I don't really understand yet why $x_0 \rightarrow a$ or why it should work because of the continuity of the square root function, but it's late here so maybe I'll see things clearer tomorrow. Thanks again and good night! $\endgroup$ – justdoit Apr 18 '17 at 20:40
  • 1
    $\begingroup$ @alissad Whoops. That should have be $x_n \to x_0$. $\endgroup$ – Trevor Gunn Apr 18 '17 at 20:43
  • $\begingroup$ I finally found the time to check what you suggested, and I have a question: why does it work "because the square root function is continuous"? We saw during a lecture that for two convergent sequences $(x_n)$ and $(y_n)$, if $x_n \le y_n$ for all $n$, then $x_0 \le y_0$. And we proved this property without using continuity. I wonder if there is something I'm missing here... $\endgroup$ – justdoit Apr 21 '17 at 10:29
  • $\begingroup$ You need continuity to conclude that if $x_n \to x_0$ then $\sqrt{|x_n|} \to \sqrt{|x_0|}$. $\endgroup$ – Trevor Gunn Apr 21 '17 at 12:31
1
$\begingroup$

There are various tools available for showing that a set is open.

If for every $p\in U$ there exists an open set $N(p)$ such that $p\in N(p)\subset U,$ then $$U=\cup \{ \{p\}: p\in U\}\subset \cup \{N(p):p\in U\}\subset U.$$ So $U=\cup \{N(p):p\in U\}$ is a union of a family of open sets, so $U$ is open.

We may also show that $U$ is the intersection of a finite family of sets, and show that each member of the family is open. This is often useful when $U$ is "complicated".

In your Q we have $U=A\cap B$ where $A=\{(x,y): y^4>x^2\}$ and $B=\{(x,y): y>0\}.$ So if $A$ and $B$ are open then $U$ is open.

We will show that if $p\in A$ then there exists $r>0$ such that the open ball $N(p)=B(p,r)\subset A$, and we will do this similarly for the set $B.$

(I).... For $(x,y)\in A$ let $y^4-x^2=r.$ We have $ r>0.$ There exists $s>0$ such that $y'^4>y^4-r/2$ whenever $|y'-y|<s.$ There exists $t>0$ such that $x'^2<x^2+r/2$ whenever $|x'-x|<t.$ So let $r=\min (s,t).$

We have $r>0$. Now whenever $\sqrt {(x'-x)^2+(y'-y)^2}\;<r$, we have $|x'-x|<r\leq s$ and $|y'-y|<r\leq t$. So $y'^4-x'^2>(y^4-r/2)-(x^2+r/2)=0.$ So $A$ is open.

(II).... For $(x,y)\in B$ let $r=y/2.$ We have $r>0.$ If $\sqrt {(x'-x)^2+(y'-y^2)}\;<r$ then $|y'-y|<r$, and we have $(|y'-y|<r=x/2 \land y>0)\implies y'\geq y-r=y/2>0 $. So $(x',y')\in B.$ So $B$ is open.

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer. This looks like the best way to do it so far, but I don't understand why "There exists $s>0$ such that ${x'}^4 > x^4 - \frac{r}{2}$ whenever $|x'-x|<s$" (and, of course, I don't understand similar sentences with $y$ and $y'$) ; where does that come from? $\endgroup$ – justdoit Apr 19 '17 at 6:32
  • $\begingroup$ It comes from the functions $f(x)=x^4$ and $g(x)=x^2$ being continuous. $\endgroup$ – DanielWainfleet Apr 19 '17 at 7:47
  • $\begingroup$ Thanks, this helps a lot! The only thing that bothers me now is that, actually, the set I have I have to study is not exactly equal to $A \cap B$, since I have to consider the case $x < 0$ as well as the case $x>0$. Would it be possible to use only what you wrote and to conclude something for $x<0$? Not sure about this... $\endgroup$ – justdoit Apr 19 '17 at 9:49
  • $\begingroup$ Sorry.I had my x and y reversed $\endgroup$ – DanielWainfleet Apr 19 '17 at 10:17
  • $\begingroup$ I just realized something else: do we really need to consider $A$ and $B$? Couldn't we just prove that the set $U := \{(x,y) \in \mathbb{R}^2 : y > \sqrt{|x|}\}$ is open by using your proof (I), since $y > \sqrt{|x|}$ $\Leftrightarrow$ $y^4 > x^2$ and since $y > \sqrt{|x|} \Longrightarrow y > 0$ (because we work in $\mathbb{R}$ and therefore $\sqrt{|x|} \ge 0$)? $\endgroup$ – justdoit Apr 19 '17 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.