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If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define $$\limsup E_k = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right).$$

So here's what I don't understand: I see $\limsup E_k$ as \begin{align*} \limsup E_k & = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots \cup E_\infty)\\ &= (E_1 \cup E_2 \cup \cdots \cup E_\infty) \cap (E_2 \cup E_3 \cup \cdots \cup E_\infty) \cap E_\infty\\ & = E_\infty. \end{align*}

But them I am confused by the statement:

$\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$.

So how should I understand $E_\infty$? Is it infinitely many $E_k$s? So for $j = \infty$, we still get infinitely many sets in $\bigcup_{k=j}^\infty E_k$?

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    $\begingroup$ $E_{\infty}$ isn't defined. Based on your notation, the index $k$ runs over every finite natural number. $\endgroup$
    – user61527
    Commented Aug 21, 2013 at 2:55
  • $\begingroup$ Thank you @T.Bongers - so could you make a correction to make it right? $\endgroup$ Commented Aug 21, 2013 at 3:09
  • $\begingroup$ Some examples to gain intuition. Consider A) $E_k = (0,1)$ for even $k$, $(1,2)$ for odd $k$. B) $E_k=(k,k+1)$. C) $E_k=(-1/k,1)$ $\endgroup$
    – leonbloy
    Commented Aug 21, 2013 at 3:18

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There is no set called $E_\infty$. The notation $\bigcup_{k=j}^\infty E_k$ does not mean $E_j\cup E_{j+1}\cup E_{j+2}\cup\cdots\cup E_\infty$. Rather, it means $E_j\cup E_{j+1}\cup E_{j+2}\cup\cdots$. A point is a member of this set if and only if it is a member of at least one of $E_j, E_{j+1}, E_{j+2}, \ldots$. There is no case in which $j=\infty$.

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  • $\begingroup$ That's very clear and helpful, thank you Michael. $\endgroup$ Commented Aug 21, 2013 at 3:16

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