set valued maps (liminf,limisup,...) I want to prove that:
1) $\overline \lim (A_n \cap B_n ) \subset \overline \lim A_n \cap \overline \lim B_n$
2)$\underline \lim (A_n \cap B_n ) \subset \underline \lim A_n \cap \underline \lim B_n$
3)$\overline \lim (A_n \cup B_n ) \subset \overline \lim A_n \cup \overline \lim B_n$
$A_n , B_n$ are sets of a metric space .
for one i known that : 
$x\in \overline\lim(A_n \cap B_n) \Rightarrow$
$ x\in \displaystyle\bigcap_{\varepsilon>0}\bigcap_{N>0}\bigcup_{n\geq N} (A_n\cap B_n)_\varepsilon \Rightarrow \forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(x,A_n \cap B_n)< \varepsilon$
But how to finish ?
 can i say that $A_n\cap B_n \subset A_n$ so $d(x,A_n\cap B_n )\leq d(x,A_n)$ ?
please , thank you 
 A: Let’s unpack your definition of the lim sup a bit. Judging by the work in your question, you define
$$\limsup_nA_n=\bigcap_{\epsilon>0}\bigcap_{m\in\Bbb N}\bigcup_{n\geq m}(A_n)_\epsilon\;,$$
where $(A)_\epsilon=\{x:d(x,A)<\epsilon\}$. Now $x\in\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}(A_n)_\epsilon$ if and only if for each $m\in\Bbb N$ there is an $n\ge m$ such that $x\in(A_n)_\epsilon$. In other words, $x\in\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}(A_n)_\epsilon$ if and only if $x$ belongs to infinitely many of the sets $(A_n)_\epsilon$, and $x\in\limsup_nA_n$ if and only if this is true for each $\epsilon>0$.
We can use this observation to prove a characterization of $\limsup_nA_n$ that will make the problem a bit easier.

Lemma. $x\in\limsup_nA_n$ if and only if there are a sequence $\langle n_k:k\in\Bbb N\rangle$ of natural numbers and a sequence $\langle x_k:k\in\Bbb N\rangle$ such that $x_k\in A_{n_k}$ for each $k\in\Bbb N$ and $\langle x_k:k\in\Bbb N\rangle\to x$.
Proof. Suppose first that such sequences $\langle n_k:k\in\Bbb N\rangle$ and $\langle x_k:k\in\Bbb N\rangle$ exist. Let $\epsilon>0$, and choose $m\in\Bbb N$ so that $d(x_k,x)<\epsilon$ for all $k\ge m$. Then $x\in(A_{n_k})_\epsilon$ for each $k\ge m$, so $x$ is in infinitely many of the sets $(A_n)_\epsilon$. Since this is true for every $\epsilon>0$, $x\in\limsup_nA_n$.
Now suppose that $x\in\limsup_nA_n$. Then $x$ belongs to infinitely many of the sets $(A_n)_1$, so there are an $n_0\in\Bbb N$ and $x_0\in A_n$ such that $d(x,x_0)<1$. Suppose that $m>0$ and we’ve chosen $n_k$ and $x_k\in A_{n_k}$ for all $k<m$. Since $x$ belongs to infinitely many of the sets $(A_n)_{2^{-m}}$, we can choose a natural number $n_m>n_{m-1}$ and a point $x_m\in A_{n_m}$ such that $d(x,x_m)<2^{-m}$. In this way we construct an increasing sequence $\langle n_k:k\in\Bbb N\rangle$ of natural numbers and a sequence $\langle x_k:k\in\Bbb N\rangle$ of points such that $x_k\in A_{n_k}$ and $d(x,x_k)<2^{-k}$ for each $n\in\Bbb N$; clearly $\langle x_k:k\in\Bbb N\rangle\to x$. $\dashv$

You want to prove that $\limsup_n\,(A_n\cap B_n)\subseteq\limsup_nA_n\cap\limsup_nB_n$. Start, as you did, with an arbitrary $x\in\limsup_n\,(A_n\cap B_n)$. By the lemma there are a sequence $\langle n_k:k\in\Bbb N\rangle$ of natural numbers and a sequence $\langle x_k:k\in\Bbb N\rangle$ such that $x_k\in A_{n_k}\cap B_{n_k}$ for each $k\in\Bbb N$ and $\langle x_k:k\in\Bbb N\rangle\to x$. Clearly $x_k\in A_{n_k}$ and $x_k\in B_{n_k}$ for each $k\in\Bbb N$, so the lemma tells us that $x\in\limsup_nA_n$ and $x\in\limsup_nB_n$, i.e., that $x\in\limsup_nA_n\cap\limsup_nB_n$, and the result follows immediately.
You can also use this lemma to prove the third result. I’ll just mention the key point. Suppose that $\langle x_k:k\in\Bbb N\rangle\to x$, where $x_k\in A_{n_k}\cup B_{n_k}$; show that there is a subsequence $\langle x_{k_i}:i\in\Bbb N\rangle\to x$ of $\langle x_k:k\in\Bbb N\rangle\to x$ such that either $x_{k_i}\in A_{n_{k_i}}$ for each $i\in\Bbb N$, or $x_{k_i}\in B_{n_{k_i}}$ for each $i\in\Bbb N$.
For the second problem you may find it useful to prove a similar characterization of $\liminf_nA_n$:

Lemma. $x\in\liminf_nA_n$ if and only if there is a sequence $\langle x_n:n\in\Bbb N\rangle$ such that $x_n\in A_n$ for each $n\in\Bbb N$ and $\langle x_n:n\in\Bbb N\rangle\to x$.

The key idea here is that $x\in\liminf_nA_n$ if and only if for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $x\in(A_n)_\epsilon$ for all $n\ge m$, i.e., $x$ belongs to all but finitely many of the sets $(A_n)_\epsilon$. I’ll leave the rest to you, at least for now, since I did the first problem in such complete detail
