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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 $

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What is the $(A_n)_\varepsilon$? –  tetori Jan 15 '13 at 15:22
is the $\varepsilon$-neighborhood of the set $A_n$ –  kiroro Jan 15 '13 at 15:31
$x \in (A_n)_\varepsilon \Leftrightarrow d(x,A_n) < \varepsilon$ –  kiroro Jan 15 '13 at 15:33
$\displaystyle\underline\lim_{n\rightarrow \infty} A_n =\lbrace x\in E, \displaystyle\lim_{n\rightarrow \infty}d(x,A_n)=0 \rbrace$, $(E,d)$ is a metric space –  kiroro Jan 15 '13 at 15:51

2 Answers 2

up vote 1 down vote accepted

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 $$\forall\varepsilon: x\in \bigcup_{N>0} \bigcap _{n>N}U(n,\varepsilon) \iff x\in \bigcap_{\varepsilon>0}\bigcup_{N>0} \bigcap _{n>N}U(n,\varepsilon)$$

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thank you ^_^ . –  kiroro Jan 15 '13 at 17:39

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 $\epsilon>0$.

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Please , how can i proov this ? $\!\!\!\!\!\!\!\displaystyle\lim_{\quad \quad n\rightarrow \infty}\!\!\!\!\!\!\!\!\mbox{-inf} A_n=\bigcup_{N\in\mathbb{N}}\bigcap_{n>N} A_n$ –  kiroro Jan 15 '13 at 16:12
$\!\!\!\!\!\!\!\displaystyle\lim_{\quad \quad n\rightarrow \infty}\!\!\!\!\!\!\!\!\mbox{-inf} A_n$ is a notation for $\displaystyle\underline\lim_{n\rightarrow \infty} A_n$. –  MathOverview Jan 15 '13 at 16:14
yes i known ... –  kiroro Jan 15 '13 at 16:16
@kiroro You can prove $A_n\subset (A_n)_\epsilon$ usign the definition of $(A_n)_\epsilon$. And using the definitions of $\bigcap$ and $\bigcup$ for conclude $A_n\subset (A_n)_\epsilon \implies \bigcup_{N\in\mathbb{N}}\bigcap_{n>N} A_n\subset \bigcup_{N\in\mathbb{N}}\bigcap_{n>N} (A_n)_\epsilon$. –  MathOverview Jan 15 '13 at 16:21

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