Proving $D := \{ \ (x,y) \in \mathbb{R^2} \ \mid \ 1+x^2+y > 0 \ \}$ is open

Question

Let $D := \{ \ (x,y) \in \mathbb{R^2} \ \mid \ 1+x^2+y > 0 \ \}$ Determine whether $D$ is an open set or closed set in $\mathbb{R^2}$

So far I have that $D$ is open in $\mathbb{R^2}$

Let $p = (p_0, p_1) \in D$ , it follows $1 + p_{0}^{2} + p_{1} > 0$

Let $u = (u_0, u_1) \in B_{\epsilon}(p)$, it follows $$d(p, u) = \|p - u\| = \sqrt{(p_0-u_0)^2 + (p_1-u_1)^2} < \epsilon$$

My problem is coming up with a definition of $\epsilon$ so that I could prove: $1 + u_{0}^{2} + u_{1} > 0$ and therefore $B_{\epsilon}(p) \subseteq D$

Any help would be really appreciated.

Answer 1

The set $D$ is open because, if you define $f\colon\mathbb{R}^2\longrightarrow\mathbb R$ by $f(x,y)=1+x^2+y$, then $f$ is continuous and $D=f^{-1}\bigl((0,+\infty)\bigr)$. Besides, $(0,+\infty)$ is an open subset of $\mathbb R$.

And it is not closed because $\left(\left(0,-1-\frac1n\right)\right)_{n\in\mathbb N}$ is a sequence of elements of $D$ which converges to $(0,-1)$, which is not an element of $D$.