1
$\begingroup$

Let $N$ be the following subset of $\mathbb{R}^2$: $$N \colon= \{ \ (x,y) \in \mathbb{R}^2 \ \colon \ \vert x \vert < \frac{1}{y^2+1} \ \}.$$ Then intuitively it is apparent that $N$ is open.

How to show this very fact rigorously?

Let $(x_0, y_0) \in N$. Then we have $$\vert x_0 \vert < \frac{1}{y_0^2+1}.$$ So $$- \frac{1}{y_0^2+1} < x_0 < \frac{1}{y_0^2+1}.$$

We now need to find some $\delta > 0$ such that the open ball of radius $\delta$ centered at $(x_0, y_0)$ lies in $N$. How do we choose our $\delta$?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Le be $f:\mathbb{R}^2\to\mathbb{R}$ such $f(x,y)=|x|(y^2+1)$ a continuous function because it is a polynomial, then $N=f^{-1}(\langle -\infty;1\rangle )$ is open because $\langle -\infty;1\rangle $ is open.

$\endgroup$
3
  • $\begingroup$ Felipe Villavicencic, thanks. You're right. But can you please also try to complete my argument? $\endgroup$
    – Saaqib Mahmood
    May 10, 2015 at 17:47
  • $\begingroup$ I'm bad using $\varepsilon-\delta$ method, and I always try to find a continuous function who helps me, but i'll try the $\varepsilon-\delta$ method a little now. $\endgroup$
    – L F
    May 10, 2015 at 17:51
  • $\begingroup$ @SaaqibMahmuud, okey, i tried and get nothing :( , really sorry. $\endgroup$
    – L F
    May 10, 2015 at 19:48

You must log in to answer this question.

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