The difficulty is to prove $\(x,y) \mid 2 \lt x^2 + y^2 \lt 4\$ is open. For this reason I have an arbitrary circle in this set, v a radius greater than $2$ and less than 4 (as given in the problem) and also an arbitrary suggest $(a,b)$ in this arbitrarily circle. I desire to show that this arbitrary point is in the set, for this reason I have that $2 \lt |a - x| \lt 4$ and $2 \lt |b - y| \lt 4$ due to the fact that $2 \lt |a - x| \lt x^2 + y^2 \lt 4$ and $2 \lt |b - y| \lt x^2 + y^2 \lt 4$ (I think?). Yet after part time algebraically manipulating these inequalities, ns cannot pertained to the conclusion the $2 \lt a \lt 4$ and $2 \lt b \lt 4$ i beg your pardon is what i think we want.

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Hint: permit $S$ be the collection of all $(x,y)$ such the $2\lt x^2+y^2\lt 4$. Allow $(a,b)\in S$. We want to present that over there is a confident $r$ such the the open up disk v centre $(a,b)$ and also radius $r$ is entirely contained in $S$.

Draw a picture. The is clear that if $r\le \min(\sqrta^2+b^2-\sqrt2, \sqrt4-\sqrta^2+b^2)$ climate the open disk with centre $(a,b)$ and also radius $r$ is entirely contained in $S$.

Hint: create it in state of open disks and also (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 \{\derekwadsworth.combbR^2 \setminus \(x,y)\mid x^2 + y^2 \leq 2\$$

In basic it works best by recognize some general expression for the radius $r$ that the open up Ball $B_r(a)$ dependent on $a \in M$. Then we have proven that you have the right to construct a ball roughly every aspect that is still included in $M$.

But avoid

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