Question: Let $\Omega = R^2. A_n$ is the interior of a circle with center at $\left\{\frac{(-1)^n}{n},0 \right\} $ at radius 1.
Find $\lim \sup A_n$ and $\lim \inf A_n$?
My answer is the following;
Let $\Omega = R^2$
As we know, $\lim\sup A_n =\{W: W\in A_n \ \mathrm{for} \ \mathrm{infinite} \ \mathrm{ n}\}$
$\lim\inf A_n =\{W: W\in A_n \ \mathrm{for} \ \mathrm{finite} \ \mathrm{ n}\}$
In this question, there exists a circle with center at $\left\{\frac{(-1)^n}{n},0 \right\} $ at radius 1.
$X^2 + Y^2 =1 \rightarrow x^2 + y^2 =1 \rightarrow (\cos\theta)^2+(\sin\theta)^2 =1 $
let $n=1$, there exists a circle with $(-1,0)$ at radius 1.
$(x-h)^2 +(y-k)^2 =r^2$
$h=1, k=0, r=1$
$(x+1)^2 +(y-0)^2=1^2$
let n=2, there exists a circle with $(1/2, 0)$ at radius 1
$(x-1/2)^2 +(y-0)^2=1^2$
$n=3 \rightarrow (-1/2, 0)$ radius=1
$n=4 \rightarrow (1/4, 0)$
and so on…
$\lim\inf A_n =\{(x,y): x^2+y^2\lt 1\}$
$\lim\sup A_n =\{(x,y): x^2+y^2\le 1\} – \{(0,1), (0,-1)\}$
what i dont understand is a point in the last gray box. How do we obtain these limsup and liminf? please clearly explain the way to get these limsup and liminf
thank you for helping.
Best Answer
Let's show that $\liminf A_n = C_0$ where $C_0 = \{(x,y) : x^2+y^2<1\}$ and $\bar{C}_0 = C_0 \cup \partial C_0$.
Observe that $\liminf A_n$ can be interpreted as the set of the $w \in \mathbb{R}^2$ such that exist a $n$ such that $w \in A_m$ to all $m \ge n$. So, let $w=(w_1,w_2)$ a point of $C_0$. Can you see that the circles "converges" to $C_0$? I mean, can you convince yourself that for this $w$ exist a $n$ such that $w$ is in all circles $A_m$ for $m$ large enough?
But what happens with the points on the border of $C_0$, well, they can belong to all circles at the same time because the centers are changing his position. If you take $w$ in the border of $C_0$ but on the left of the origin, for large $m$ it won't belong to $A_m$ when $m$ is even...
To the $\limsup A_n$ remember that it is the set of the $w$ which belongs to infinitely many $A_m$'s ...
Hope this can help!