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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.

Noa Even
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mr eyeglasses
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    You seem to be entirely misunderstanding what is being asked. You take a circle in the set, and a point on the circle, and you want to show that it's in the set? Of course it is, by construction - but that's not what you're being asked. (Incidentally, that also is not equivalent to $2<a<4$ and $2<b<4$). – Jack M Feb 19 '14 at 18:27
  • Then I totally misunderstand...what is being asked of me? – mr eyeglasses Feb 19 '14 at 18:37
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    To prove that the set is open. What is the definition of an open set? – Jack M Feb 19 '14 at 18:46
  • If there exists an open ball centered at x for all x such that the ball is in the set – mr eyeglasses Feb 19 '14 at 18:51
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    So what you need to do is let $x$ be a point in the set, and show that there's a ball centered on it that's entirely within the set. Do you see how this is different to what you were trying to do? – Jack M Feb 19 '14 at 18:52
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    I know this is very late, but @JackM - I've also learned it that way too in which you take a point in the circle and want to show it's in the set (unless I'm misunderstanding). With that, you want to determine how small the radius should be according to the shift in that point. – August Oct 01 '14 at 04:46

3 Answers3

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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 $S$.

André Nicolas
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Hint: Write it in terms of open disks and (complements of) closed disks: $$\{(x,y) \mid 2 \lt x^2 + y^2 \lt 4\} = \{(x,y) \mid x^2 + y^2 \lt 4\} \cap \{\mathbb{R}^2 \setminus \{(x,y)\mid x^2 + y^2 \leq 2\} $$

ir7
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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 $M$.

High GPA
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TheFibonacciEffect
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