How to prove a set is open
Matthew Martinez 
The problem is to prove $\{(x,y) \mid 2 \lt x^2 + y^2 \lt 4\}$ is open. So I have an arbitrary circle in this set, with a radius greater than $2$ and less than 4 (as given in the problem) and an arbitrary point $(a,b)$ in this arbitrary circle. I want to show that this arbitrary point is in the set, so I have that $2 \lt |a - x| \lt 4$ and $2 \lt |b - y| \lt 4$ since $2 \lt |a - x| \lt x^2 + y^2 \lt 4$ and $2 \lt |b - y| \lt x^2 + y^2 \lt 4$ (I think?). But after some time algebraically manipulating these inequalities, I cannot come to the conclusion that $2 \lt a \lt 4$ and $2 \lt b \lt 4$ which is what I think we want.
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$\begingroup$Hint: Let $S$ be the set of all $(x,y)$ such that $2\lt x^2+y^2\lt 4$. Let $(a,b)\in S$. We want to show that there is a positive $r$ such that the open disk with centre $(a,b)$ and radius $r$ is entirely contained in $S$.
Draw a picture. It is clear that if $r\le \min(\sqrt{a^2+b^2}-\sqrt{2}, \sqrt{4}-\sqrt{a^2+b^2})$ then the open disk with centre $(a,b)$ and radius $r$ is entirely contained in $S$.
$\endgroup$ $\begingroup$Hint: Write it in terms of open disks and (complements of) closed disks: $$\{(x,y) \mid 2 \lt x^2 + y^2 \lt 4\} = \{(x,y) \mid x^2 + y^2 \lt 4\} \cap \{\mathbb{R}^2 \setminus \{(x,y)\mid x^2 + y^2 \leq 2\} $$
$\endgroup$ $\begingroup$In general it works best by finding some general expression for the radius $r$ of the open Ball $B_r(a)$ dependent on $a \in M$. Then we have proven that you can construct a ball around every element that is still contained in $M$.
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