Proving that a particular set in $\mathbb R^2$ is open.

Question

I have to prove that the subset $A=\{ (x,y)|x^2>y\}$ of $\mathbb R^2$ is open. I graphed it, and I can see that it is open, and I can also see that the boundary is $y=x^2$.

We have not yet learned about continuity of these, so I cannot use the argument that I keep seeing everywhere where it just tells you to argue that $f(x,y)=x^2-y$ is continuous.

I know I have to either show that every point in $A$ is also in $intA$, or use the fact that the boundary $B_A=\{(x,y)|x^2=y\}$.

It seems easier to me to use the boundary as I can make the following argument: If $x^2>y$, then $x^2 \neq y$ and so $B_A\cap A = \emptyset$. Therefore, $A$ is open since (by a proposition in my text) $A$ is open in $\mathbb R^n \iff A \cap B_A=\emptyset$. With this proof though, I am unsure if I have to prove that the boundary of $A$ is in fact $x^2=y$. And if I do have to prove it, I don't know how.

I would really like to see the proof of it being open just by showing that every point in $A$ is also in $intA$, as there are no proofs in my textbook and my professor did not do any examples in lecture.

Answer 1

Let $x^{2} >y$. Choose $\epsilon >0$ such that $\epsilon (1+\epsilon+2|x|)<x^{2}-y$. If the distance from $x',y')$ to $(x,y)$ is less than $\epsilon $ then $|x-x'|<\epsilon $ and $|y-y'|<\epsilon $. Now, $|x'^{2}-x^{2}| <\epsilon (|x'+x|) <\epsilon (2|x|+\epsilon) $. Hence $x'^{2} >x^{2}-\epsilon (2|x|+\epsilon)>y+\epsilon >y'$. This proves that $(x,y)$ is an interior point of $A$.

Answer 2

Draw a picture.

Take a point $p$ in $A$. You have to be able to draw a circle (or a disk) around $p$ that also lies within $A$. If $p$ is near the boundary of $A$, then you’ll have to make the radius of the circle very small in order to keep it inside $A$. All you have to do is figure out what “very small” means.

It might be easier to draw a square around $p$. Once you have done this, you can just make your circle a bit smaller than the square.