$V = \{(x,y)\in \Bbb R^2 : x>0, y>0, xy<1 \}$ is open.

Question

Prove that $V = \{(x,y)\in \Bbb R^2 : x>0, y>0, xy<1 \}$ is open.

I already read two posts

Prove that $S = \{(x,y)\in \Bbb R^2 : x>0, y>0\}$ is open.

and

Show that $S = \{ (x,y) \in \Bbb R^2 : |x| + |y| < 1 \} $ is an open set.

Then I just think of above case, additionally impose $xy<1$. I know $V$ is open, but having trouble prove this using "Open Ball". I think I need constraint equation about $xy<1$, but how I can insert this information on radius?

My simple guess was $r=\min(x,y) <1$, and i tried similar computation as in above link, but having trouble construction of $x'y'<1$.

Answer 1

Let $(a,b) \in V$. Let $0<\delta <\min \{1,a,b,\frac {1-ab} {a+b+1}\}$. If $d((x,y),(a,b)) <\delta$ then $|x-a|<\delta$ and $|y-b| <\delta$. Hence $x =a+(x-a)>a-\delta>0$. Similarly, $y >0$. Now $|xy-ab| \leq |xy-xb| +|xb-ab| \leq x \delta + b \delta <(a+\delta+b)\delta<(a+1+b)\delta$. Hence $xy \leq ab+(a+1+b)\delta<1$.

Answer 2

The functions $p_1,p_2,p: \mathbb{R}^2 \to \mathbb{R}$ defined by $p_1(x,y)=x$, $p_2(x,y)=y$ and $p(x,y)=xy$ are continuous.

Your set $$V = p_1^{-1}[(0,+\infty)] \cap p_2^{-1}[(0,+\infty)] \cap p^{-1}[(-\infty,1)]$$

is a finite intersection of inverse images of open sets under continuous functions, hence open.