finding the interior of a set Let $X = \{ (x,y) \in R^2 : 1 \leq \sqrt{x^2 + y^2} < 2 \} $. I need to find, the interior, closure, and boundary of $X$. I know that $Int X = \{(x,y) : 1 < \sqrt{x^2 + y^2} < 2 \} $. But im having hard time trying to show this. In fact, my approach is the following:
Let $\underline{x} \in Int X $. We need to find $r >0 $ such that $B(\underline{x}, r ) \subset Int X $. This will show that the mentioned set above is indeed the interior of $X$. But, how can I find such an $r$ ?
Also, the closure is $\overline{X} = \{ (x,y) \in R^2 : 1 \leq \sqrt{x^2 + y^2} \leq 2 \} $. But, how can I show this?
help would be really appreciated. thanks.
 A: Hint: $r = \frac{\min(\sqrt{x^2+y^2}-1,2-\sqrt{x^2+y^2})}{2}$ works for interior. For the closure think about the limit points of $X$.
A: By definition, $(x,y)\in X^0$ if and only if for all open ball centred at $P:(x,y)$, we can find some points in $X$ and in $\bar{X}$ also. Now let $(x,y)\in X^0$, so $$B_{0.5}(P)\cap X\neq\emptyset,~B_{0.5}(P)\cap \bar{X}\neq\emptyset,~x^2+y^2=1$$ and $$B_{1}(P)\cap X\neq\emptyset,~B_{1}(P)\cap \bar{X}\neq\emptyset,~x^2+y^2=4$$ 
A: Try $r=\min\{||z||-1,\ 2-||z||\}$.
To show that each $z$ with $||z||=1$ is in the boundary, consider the point $(1-\delta)z$ for some $\delta<ϵ$. More generally if $ϵ>0$ and $||z||=s$ then $(1-\delta)z$ and $(1+\delta)z$ will be contained within $B_ϵ(z)$ if $\delta<ϵ/s$.
Note that this implies that the map $z\mapsto||z||$ is open, as the image of the open neighborhood $B_ϵ(z)$ contains the neighborhood $(||z||-ϵ,||z||+ϵ)$ of $||z||$
A: For many of these questions a good approach is guess and verify.
If $x \in X$, then $r=1<\|x\|<2$, and so you can find an open ball of radius $\min(r-1,2-r)$ centered at $x$ that lies completely inside $X$. Draw a picture to convince yourself and use the triangle inequality to prove it.
For the closure, note that for any point $x$ with $\|x\| =2$, you can find points in $X$ converging to $x$, so $x$ must be in the closure. 
Similarly,for any point $x$ with $\|x\| =1$, you can find points in $X$ converging to $x$, so $x$ must be in the closure.
So we guess that the closure is $C=\{(x,y)| 1 \le \sqrt{x^2+y^2} \le 2 \}$. We have just shown that $C \subset \bar{X}$.
Since $\phi(x,y) = \sqrt{x^2+y^2}$ is continuous, and $C = \phi^{-1} [1,2]$, we see that $C$ is closed, and clearly $X \subset C$, so we must have $\bar{X} \subset C$.
Combining the two shows that $\bar{X} = C$.
