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Propositions

Let $ \{ A_{n}: n \in \mathbb{N} \} $ be a sequence of sets such that $A_{n} \rightarrow A $. And let $ \{ B_{n}: n \in \mathbb{N} \} $ be a sequence of sets such that $B_{n} \rightarrow B $.

then

1.$\lim_{n\to \infty} A_{n}\cap B_{n} = A\cap B $.

2.$\lim_{n\to \infty} A_{n}-B_{n} = A-B $.

In order to prove that $C_{n} \to C $, where $ \{ C_{n}:n \in \mathbb{N} \}$ is a sequence of sets, I have to show that

$ \limsup C_{n} = \liminf C_{n} = C $.

i.e,

$\liminf C_{n} = \cup_{N=1}^{\infty} \cap_{n = N}^{\infty} C_{n} = C \ \ \ \ \ $ and

$\limsup C_{n} = \cap_{N=1}^{\infty} \cup_{n = N}^{\infty} C_{n} = C \ \ \ \ \ $

but when I define $C_{n} = A_{n} \cap B_{n} $ or $C_{n} = A_{n} - B_{n} $, I don't know how to compute the $\limsup C_{n}$ and the $\liminf C_{n}$.

Questions:

  1. Do you know a counterexample for this propositions?
  2. Do you know how to prove them?.

thanks in advanced.

TheComander
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  • $A_n \to A$ iff $\chi_{A_n}(x) \to \chi_A(x)$ for every $x$. – Kavi Rama Murthy Jun 23 '21 at 04:59
  • Hope this will help:https://math.stackexchange.com/questions/108273/convergent-sequence-of-sets – A learner Jun 23 '21 at 05:19
  • For proposition (2), which is true, I recommend a direct $\epsilon, N(\epsilon)$ approach. The idea is that for $\epsilon/2, ~\exists N_1$ which places $a_n ~: ~n \geq N_1$ in a neighborhood of radius $(\epsilon/2)$ around $A$. Similarly, $\exists N_2$, re $b_n$ around $B$. Then, set $N = \max(N_1, N_2)$ and use triangle inequality. – user2661923 Jun 23 '21 at 05:23
  • Also,https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – A learner Jun 23 '21 at 05:32

1 Answers1

2

I'll give a general strategy, but reveal the working out in a spoiler block.

First show that $x\in A\cap B$ implies that $x\in \lim\inf A_n\cap B_n$.

Suppose $x\in A\cap B$, then $x\in A=\lim\inf A_n=\bigcup_{m\in\Bbb N}\bigcap_{n\geq m} A_n$. This means that there is some $m\in\Bbb N$ such that $x\in \bigcap_{n\geq m}A_n$. Therefore, for such $m\in \Bbb N$ we have $x\in A_n$ for all $n\geq m$. Similarly, we see that there is some $m'\in \Bbb N$ such that $x\in B_n$ for all $n\geq m'$. Now take $k=\max\{m,m'\}$, then $x\in A_n\cap B_n$ for all $n\geq k$, and thus $x\in \lim\inf A_n\cap B_n$.

Then show that $x\in \lim\sup A_n\cap B_n$ implies that $x\in A\cap B$.

Suppose $x\in \lim\sup A_n\cap B_n=\bigcap_{m\in\Bbb N}\bigcup_{n\geq m} A_n\cap B_n$, then $x\in \bigcup_{n\geq m}A_n\cap B_n$ for every $m\in\Bbb N$. Hence, for any $m\in \Bbb N$ we see that there is some $n\geq m$ such that $x\in A_n\cap B_n$. But then for each $m\in\Bbb N$ there is some $n\geq m$ such that $x\in A_n$, thus $x\in \lim\sup A_n=A$. Similarly, $x\in \lim\sup B_n=B$. Therefore $x\in A\cap B$.

Do you see why the above two parts combine to show that $A\cap B=\lim\sup A_n\cap B_n=\lim\inf A_n\cap B_n$?

$$A\cap B~\subseteq~ \lim\inf A_n\cap B_n~\subseteq~\lim\sup A_n\cap B_n~\subseteq~ A\cap B$$

The same strategy works for proving $A-B=\lim\inf A_n-B_n=\lim\sup A_n-B_n$, but with a twist. I'll leave the working out to you.

Start with "if $x\in A-B$, then $x\in A=\lim\inf A_n$ and $x\notin B=\lim\mathbf{sup}\, B_n$" and show that this leads to $x\in \lim\inf A_n-B_n$.

For the second part, assume that $x\in\lim\sup A_n - B_n$ and prove that this leads to $x\in \lim\sup A_n=A$ and $x\notin \lim\mathbf{inf}\, B_n=B$.

Vsotvep
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