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Let $A_n$, $B_n$ and $C$ be closed subsets in euclidean space such that \begin{equation} A_n \cap B_n \neq \varnothing, A_n \cap B_n \cap C\neq \varnothing \end{equation} for each $n=1,2,...$, and $\lim_{n\to \infty}A_n=A$, $\lim_{n\to \infty}B_n=B$. If $A\cap C\neq \varnothing$, $B\cap C\neq \varnothing$ and $A\cap B \neq \varnothing$, then can we have $A\cap B\cap C \neq \varnothing$?

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  • $\begingroup$ What is your definition of $\lim A_n=A$? Do you want an example where $A \cap B \cap C $ is not empty or do you want to prove that this set cannot be empty? $\endgroup$ – Kavi Rama Murthy Nov 22 '19 at 5:40
  • $\begingroup$ Your assumptions $A_n\cap B_n\ne\varnothing$ and $A_n\cap B_n\cap C\ne\varnothing$ are redundant, as the latter implies the former. $\endgroup$ – bof Nov 22 '19 at 6:45
  • $\begingroup$ @Kabo Murphy Consider a correspondence $F(x)$, where $x\in E^n$ and $F(x) \in E^m$, the limit notation here is that when $x_n \to x$, $F(x_n) \to F(x)$, imagining $A_n$, $B_n$ as some $F_1(x_n)$ and $F_2(x_n)$. I want to prove this set is not empty... $\endgroup$ – Huaixin Nov 22 '19 at 12:57
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If
$A_n=\{-1\}\cup\{-2\}\cup[n,\infty)$,
$B_n=\{-1\}\cup\{-3\}\cup[n,\infty)$,
$C=\{-2\}\cup\{-3\}\cup[0,\infty)$,
then:
$A_n,B_n,C$ are closed subsets of $\mathbb R$;
$A_n\cap B_n=\{-1\}\cup[n,\infty)\ne\varnothing$;
$A_n\cap B_n\cap C=[n,\infty)\ne\varnothing$;
$A=\lim_{n\to\infty}A_n=\{-1,-2\}$;
$B=\lim_{n\to\infty}B_n=\{-1,-3\}$;
$A\cap C=\{-2\}\ne\varnothing$;
$B\cap C=\{-3\}\ne\varnothing$;
$A\cap B=\{-1\}\ne\varnothing$;
$A\cap B\cap C=\varnothing$.

Does that answer your question?

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  • $\begingroup$ It is very clear, thanks! $\endgroup$ – Huaixin Nov 22 '19 at 12:46

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