Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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 at 18:27
    
Then I totally misunderstand...what is being asked of me? –  user130018 Feb 19 at 18:37
1  
To prove that the set is open. What is the definition of an open set? –  Jack M Feb 19 at 18:46
1  
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 at 18:52
1  
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 1 at 4:46

2 Answers 2

up vote 3 down vote accepted

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 $S$.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.