# Open set between two curves

Let $U := \{ (x,y) \in \mathbb{R}^2 : x^2 > y \} \;\cap\; \{ (x, y) \in \mathbb{R}^2 : x \geq 0 \}$.

I want to prove (!) that $U$ is an open set. Therefore I made a sketch:

Here $U$ is represented by the blue area. The basic idea I had for the proof was to show that $U$ can be described as an infinite union of open spheres around all real points on the $x$-axis with $x > 0$.

I've tried and have come to the conclusion that there must be an easier way to prove this as it was part of an exam in my university.

EDIT

As stated in the comments, the sketch of $U$ above is wrong as e.g. $(0, -1) \in U$. It should rather look like this:

And then it is obvious that $U$ cannot be open as there is no open surrounding of $(0, -1)$ which is a subset of $U$.

• I have some doubt here…is (0,-1) in U and if it is…is it an interior point? – Li Chun Min Apr 9 '17 at 11:36
• Sorry, your approach, while I can't say if it easier of course, but I agree should prove the result, contrary to what I said earlier. – marshal craft Apr 9 '17 at 11:53
• @Li Chun Min Well, I think you are right, something is wrong with the sketch. – fpmoo Apr 9 '17 at 11:55
• Maybe show there is no $x$ or $y$ max or min? – marshal craft Apr 9 '17 at 12:11
• @marshalcraft Consider the set $[0, 1] \cup [-2, -1] \cup [2, 3] \cup [-4, -3] \cup \dots$. It has no min and no max but it is yet not open. – fpmoo Apr 9 '17 at 12:17

$$U = \{ (x,y) \in \mathbb{R}^2 : x^2 > y \} \;\cap\; \{ (x, y) \in \mathbb{R}^2 : x > 0 \}$$
And $A = \{ (x,y) \in \mathbb{R}^2 : x^2 > y \}$ is open, beacuse $f(x,y) = x^2-y$ is continuous, and $A = f^{-1} (]0,+\infty[)$, hence the inverse image of an open set y a continuous function.
Same with $B = \{ (x, y) \in \mathbb{R}^2 : x > 0 \}$, by considering $g(x,y) = x$ that is also continuous.
• To me it is not clear at all in words what statement you are trying to make about $g$ and the set $B$. Yes $B$ is open. $g$ is boundary, continuous. Now what? – marshal craft Apr 9 '17 at 11:59