4
$\begingroup$

The problem is to prove that the set $A=\{(x,y)\in\mathbb{R}^2: x^2 + y^2 > 4, y < 6\}$ is open in $\mathbb{R}^2$

My attempt:

Let $S=(0,0)$

Let's take an arbitrary $x=(x_1,x_2) \in A$ and define $R = \min\{\lvert{d(x,S)-2}\rvert,\lvert x_2-6 \rvert \}$, so we have an open ball $K(x,R)$ To prove that the set is open we need to prove that for an arbitrary $y=(y_1,y_2) \in K(x,R)$ that $y$ has to be in $A$.

By a few manipulations, mainly using the triangle inequality and the points $S,x,y$ I've managed to prove that $y_1^2+y_2^2 > 4$.

But proving $y_2 < 6$ has proven to be very difficult and I'm left without ideas as to what to do specifically.I have a feeling I'm missing something much simpler here.

Thanks in advance!

$\endgroup$

4 Answers 4

1
$\begingroup$

As $y\in K(x,R)$ $$ y_2<x_2+R\leq x_2+|x_2-6| $$ Now as $x\in A$, $x_2<6$, hence $|x_2-6|=6-x_2$. And we conclude. $$ y_2<x_2+6-x_2=6 $$

I think you did the tricky part (That was actually finding R)

$\endgroup$
1
  • $\begingroup$ The simplest answers are usually the best, can't believe I didn't see that. Thank you very much! $\endgroup$
    – Collapse
    Oct 22, 2017 at 17:16
1
$\begingroup$

Here is an alternative approach.

Notice that $A = B \cap C$, where $B= \{ (x,y) \in \mathbb R^2 :x^2 + y^2 >4 \}$, and $C=\{ (x,y) \in \mathbb R^2 : y < 6 \}$. Since the intersection of two open sets is open, you only need to show that $B$ and $C$ are open.

It may also be easier to show that $B^c$ and $C^c$ are closed.

$\endgroup$
0
$\begingroup$

If $y_2\geq 6$, then $d(x,y)\geq y_2-x_2\geq 6-x_2\geq R$,

$\endgroup$
0
$\begingroup$

\begin{align} 6-x_2 &= 6 - y_2 + (y_2 - x_2)\\ &\le 6 - y_2 + |y_2 - x_2|\\ &\le 6 - y_2 + d\big((y_1,y_2),(x_1,x_2)\big)\\ &< 6 - y_2 + |x_2 - 6|\\ \end{align}

This implies $$6 - y_2 > 6 - x_2 - |x_2 - 6| \ge 6 - x_2 - (6 - x_2) = 0$$ so $y_2 < 6$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .