Computing the union and intersection of family of sets Suppose we are given for all $n \in \mathbb{N} $
$$ X_n = \{ (x,y) \in \mathbb{R} \times \mathbb{R} : n^2 \leq x^2 + y^2 \leq (n+1)^2 \} $$
I am trying to compute $\bigcup_{n \in \mathbb{N} } X_n $ and $\bigcap X_n $
My try: I was trying to draw the various annulus for varying values of $n$. Certainly, I find that $\bigcup X_n $ should be entire plane since this annulus keep expanding as $n$ grows.
As for the intersection, it would just be the smallest annulus. That is 
$ \bigcap X_n = \{ (x,y) : 1 \leq x^2 + y^2 \leq 4 \} $. 
My question is: How can I prove this rigorously? thanks
 A: Attention if $0\notin \Bbb N$ then the union is not the whole plane but the whole plane without the open unit ball induced by the 2-norm $\|\cdot\|_2$ (i.e. $\{(x,y)\mid x^2+y^2< 1\}$).
General strategy for a rigorous proof:  


*

*For the union: Take a vector $(x,y)$ in the region you believe is the union and show that there exists $n$ such that $(x,y)\in X_n$. You also have to show that every $X_n$ is contained in the region in question.

*For the intersection: choose $(x,y)$ in the region you believe is the
intersection and show that $(x,y)\in X_n$ for every $n\in \Bbb N$. You also have to show that the region in question in contained in every $X_n$.
For your particular example:


*

*For the union: note that $\|(x,y)\|_2\in\Bbb R$ and so there exists always $n\in\Bbb N$ such that $n\leq \|(x,y)\|_2\leq n+1$.

*For the intersection: What can you say about $X_1\cap X_{10}\supset \bigcap_{n\in\Bbb N}X_n$?
A: Denote $\Lambda$ the entire plane, i.e $\Lambda = \{(x,y): x \in \mathbb{R}, y \in \mathbb{R}\}$, then as suggested you show $\cup_{n \in \mathbb{N}} X_n=\Lambda \setminus D_1$ ( * ), whereas $D_1 = \{(x,y): x^2+y^2 < 1\}$. We have:
$X_n \subseteq \Lambda \setminus D_1, \forall n \geq 1 \Rightarrow \cup_{n \in \mathbb{N}} X_n \subseteq \Lambda \setminus D_1$. Next if $P = (x,y) \in \Lambda \setminus D_1$, then let $n$ be the smallest natural number such that: $x^2+y^2 \leq n^2$, then we have: $(n-1)^2 < x^2+y^2 \leq n^2 \Rightarrow P \in X_{n-1}\Rightarrow P \in \cup_{n \in \mathbb{N}} X_n\Rightarrow \Lambda \setminus D_1 \subseteq \cup_{n \in \mathbb{N}} X_n$.
Thus ( * ) holds.
I am sure you can do the other part similarly.
