$\{(x,y) \in \mathbb{R^2} : x^2+y^2 > c\}$ is neither open nor closed?

Question

Why $$\{(x,y) \in \mathbb{R^2} : x^2+y^2 > c\} \space (c > 0) $$is neither open nor closed?

Answer 1

That's wrong. It's an open set, as it is the preimage of $(c, \infty) \subseteq \mathbf R$ under the continuous map $f\colon (x,y) \mapsto x^2 + y^2$. It's not closed, as $x_n := (\sqrt c + \frac 1n, 0)$ belongs to the set for every $n \in \mathbf N$, but the limit of the $x_n$, namely $(\sqrt c,0)$, does not.