$\lim_{n \rightarrow \infty}\bigcup\limits_{k \ge n} A_k$ of $A_n:=$ open circle with radius $=1 $ and center $(\frac{(-1)^n}{n}, 0)$

Question

$A_n:=$ open circle with radius $=1 $ and center $(\frac{(-1)^n}{n}, 0)$.

What's

a) $\lim_{n \rightarrow \infty}\bigcup\limits_{k \ge n} A_k$ and

b) $\lim_{n \rightarrow \infty}\bigcap\limits_{k \ge n} A_k$?

For b) I would say the open circle with radius $1$ and center $(0,0)$

Answer 1

Hint : for a) the closed circle with radius 1, center $(0,0)$ without the points : $(0,-1)$ and $(0,1)$

for b) as you said.

Sketch of proof of a) : let $(x,y)$ a point such that $x^2+y^2\leq 1$ and $(x,y)\notin \{(0,-1),(0,1)\} $. The more difficult point is if $x^2+y^2=1$ ; in this last case, we have $|x|>0$ (we suppose wlog $x>0$ in the sequel). Then if we denote $\mathcal{C}_{n}$ the open circle of center $\big((-1)^n\frac{1}{n},0\big)$, I claim that there exists $N>0$ such that for all $n>N$, \begin{align*} (x,y)\in \mathcal{C}_{2n} &\quad \Leftrightarrow \quad \bigg(x-\frac{1}{2n}\bigg)^2 + y^2< 1 \\ & \quad \Leftrightarrow \quad \bigg(x-\frac{1}{2n}\bigg)^2 + 1-x^2< 1 \\ & \quad \Leftrightarrow \quad \bigg(x-\frac{1}{2n}\bigg)^2 -x^2< 0 \\ & \quad \Leftrightarrow \quad x>\frac{1}{2n} \end{align*} You have just to take $N=[\frac{1}{2x}]+1$.

The case $x^2+y^2< 1$ is obtain with the same techniques.