Prove that the set $ D=\{(x,y): 0<x^2+y^2<4\} $ is open. Apply the definition.

Question

I tried this:

Denoting $P=(x,y)$, we see that $0<x^2+y^2=||P||^2<2^2 \Rightarrow ||P||<2$

Let $(s,t)=Q \in B_{2-||P||}(P) \Rightarrow ||Q - P|| < 2 - ||P||$

And $||Q|| - ||P|| \leq ||Q - P|| \Rightarrow ||Q|| = ||(s,t)|| = \sqrt{s^2 + t^2} < 2$

From here $s^2 + t^2 < 4$

The problem is that i don't know how to restrict that $0<s^2 + t^2$

Answer 1

If $Q\in B_{\lVert P\rVert}(P)$, then $\lVert Q\rVert>0$. So, consider the open ball$$B_{\min\{2-\lVert P\rVert,\lVert P\rVert\}}(P).$$

Answer 2

You need to use the fact that $f(x) = x_1^2 + x_2^2$ is a continuous function i.e. $$\forall x \in \mathbb{R}^2 \forall \epsilon > 0 \exists \delta > 0 : |x-y|< \delta \Rightarrow |f(x)-f(y)|> \epsilon$$