# How to show that some set is open?

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.

• 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... – user251257 Apr 18 '17 at 17:38

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.

• Minimize the squared distance, it's cleaner. – Trevor Gunn Apr 18 '17 at 17:02
• @T.Gunn: good point. – Ross Millikan Apr 18 '17 at 17:03
• 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 $x_1 > 0$" ? – justdoit Apr 18 '17 at 17:10
• @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 $U$. – amrsa Apr 18 '17 at 17:39
• @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. – justdoit Apr 18 '17 at 17:42

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.

• 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! – justdoit Apr 18 '17 at 20:40
• @alissad Whoops. That should have be $x_n \to x_0$. – Trevor Gunn Apr 18 '17 at 20:43
• 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... – justdoit Apr 21 '17 at 10:29
• You need continuity to conclude that if $x_n \to x_0$ then $\sqrt{|x_n|} \to \sqrt{|x_0|}$. – Trevor Gunn Apr 21 '17 at 12:31

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.

• 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? – justdoit Apr 19 '17 at 6:32
• It comes from the functions $f(x)=x^4$ and $g(x)=x^2$ being continuous. – DanielWainfleet Apr 19 '17 at 7:47
• 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... – justdoit Apr 19 '17 at 9:49
• Sorry.I had my x and y reversed – DanielWainfleet Apr 19 '17 at 10:17
• 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 $\sqrt{|x|} \ge 0$)? – justdoit Apr 19 '17 at 11:17