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Is it true that $A_n \to A \iff A_{n+p} \to A$ for $p \in \mathbb{N}$, where $A_n, n \geq 0$ are sets?

I can show that "$\implies$" holds, but I'm not making progress with the other implication. Any ideas?

EDIT: $A_n \to A$ means that $\limsup A_n = A = \liminf A_n$

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    What mean $A_n\to A$ ? that $\bigcup_{n\in\mathbb N}A_n=A$ ? – Surb Apr 11 '18 at 10:52
  • $A_n \to A$ isn't standard notation for sets. From the tags on this question I guess you mean $\limsup A_n = A$, or maybe $\liminf A_n = A$. Is this the case? – A.P. Apr 11 '18 at 10:55
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    No, $A_n \to A$ means that the limsup and liminf of $(A_n)$ are both equal to $A$. Sorry if this was unclear. –  Apr 11 '18 at 10:57

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If $x$ belongs to infinitely many elements of $A_{n+p}$, then it belongs to infinitely many elements of $A_n$ and converselly if an element belongs to infinitely many elements of $A_n$ then it belongs to infinitely many elements of $A_{n+p}$. Therefore $\limsup_n A_n=\limsup_n A_{n+p}$.

If an element belongs to all $A_{n+p}$ for $n$ sufficiently large, $\forall n>N$ for some $N$, then so does for all $A_n$ for $n$ sufficiently large just take $n>N+p$. Conversely, if $x$ belongs to all $A_{n}$ for $n$ sufficiently large, then so does for all $A_{n+p}$ for $n$ large. Therefore $\liminf_n A_n=\liminf A_{n+p}$.

The implication when you know that one of the limits exist is a particular case.