# Example 7, Sec. 26 in Munkres' TOPOLOGY, 2nd ed: How to show this set to be open?

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 $\delta$?

## 1 Answer

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.

• Felipe Villavicencic, thanks. You're right. But can you please also try to complete my argument? May 10 '15 at 17:47
• 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. May 10 '15 at 17:51
• @SaaqibMahmuud, okey, i tried and get nothing :( , really sorry. May 10 '15 at 19:48