Set-valued maps please how to proove that 
$\displaystyle\underline\lim_{n\rightarrow \infty} A_n =\bigcap_{\varepsilon>0} \bigcup_{N>0}\bigcap_{n\geq N}(A_n)_\varepsilon $
 A: Take $f_n(x)=d(x,A_n)$. By def,
$$\varliminf_{n\rightarrow \infty} A_n =\{x\in E : \lim_{n\to\infty} f_n(x)=0\}$$
and take
$$U(n,\varepsilon)=\{x\in E : |f_n(x)|<\varepsilon\}.$$
So
$$\lim_{n\to\infty} f_n(x)=0 \iff \forall\varepsilon\exists N \forall n>N: |f_n(x) |<\varepsilon \iff \forall\varepsilon\exists N \forall n>N: x\in U(n,\varepsilon).$$
By definition of arbitary intersection, we get
$$\forall n>N : x\in U_n \iff x\in \bigcap _{n>N}U_n$$
and we get
$$ x\in \lim_{n\to\infty} f_n(x)=0 \iff \forall\varepsilon\exists N :  x\in \bigcap _{n>N}U(n,\varepsilon).$$
and by definition of arbitary union
$$\bigcup_{A\in S} A=\{x|\exists A\in S:x\in A\},$$
we obtain
$$\forall\varepsilon\exists N :  x\in \bigcap _{n>N}U(n,\varepsilon) \iff \forall\varepsilon: x\in \bigcup_{N>0} \bigcap _{n>N}U(n,\varepsilon)$$
and use def. of arbitary intersection again, we get
$$\forall\varepsilon: x\in \bigcup_{N>0} \bigcap _{n>N}U(n,\varepsilon) \iff  x\in \bigcap_{\varepsilon>0}\bigcup_{N>0} \bigcap _{n>N}U(n,\varepsilon)$$
A: Hint: use the definition of "$\!\!\!\!\!\!\!\displaystyle\lim_{\quad \quad n\rightarrow \infty}\!\!\!\!\!\!\!\!\mbox{-inf} A_n$" :
$$
\!\!\!\!\!\!\!\displaystyle\lim_{\quad \quad n\rightarrow \infty}\!\!\!\!\!\!\!\!\mbox{-inf} A_n=\bigcup_{N\in\mathbb{N}}\bigcap_{n>N} A_n
$$
and the fact
$$
A_n\subset (A_n)_\epsilon \implies \bigcup_{N\in\mathbb{N}}\bigcap_{n>N} A_n\subset \bigcup_{N\in\mathbb{N}}\bigcap_{n>N} (A_n)_\epsilon.
$$
for all $\epsilon>0$. 
