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I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.

Proof $\liminf E_k \subset \limsup E_k $

If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \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)\\ &= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}

Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k & = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.

Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$.

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It’s not quite right: you’ve actually assumed that $x\in\limsup_kE_k$, so the argument is circular. Recall that

$$\liminf_{k\in\Bbb N}E_k=\bigcup_{n\ge 0}\bigcap_{k\ge n}E_k\;;$$

this means that if $x\in\liminf_{k\in\Bbb N}E_k$, then there is some $n_0\in\Bbb N$ such that $x\in\bigcap\limits_{k\ge n_0}E_k$.

Since $$\limsup_{k\in\Bbb N}E_k=\bigcap_{n\ge 0}\bigcup_{k\ge n}E_k\;,$$ you have to use this somehow to show that $x\in\bigcup\limits_{k\ge n}E_k$ for every $n\in\Bbb N$. That’s not actually very hard: we know that $x\in E_k$ for each $k\ge n_0$, so if we just take $k=\max\{n,n_0\}$ then ... ?

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  • $\begingroup$ Oh Brian, I see I made a big mistake. I expanded the formula and hopefully got it right...? $\endgroup$ – 1LiterTears Aug 21 '13 at 16:04
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    $\begingroup$ @Jellyfish: Much better! Informally it just boils down to saying that a point that is in every $E_k$ from some point on has to be in infinitely many of them. $\endgroup$ – Brian M. Scott Aug 21 '13 at 16:57
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    $\begingroup$ Hello Mr. Scott, Id like to ask why do we need to take $k = \max( n,n_0 ) $ ?? Isnt it true that $E_k \subset \bigcup_{k \geq n} E_k $ for all $k \geq n_0 $ ? $\endgroup$ – user203867 Jan 1 '15 at 9:21
  • $\begingroup$ @Willie: Not necessarily. If $n_0\le k<n$ it’s quite possible that $E_k\nsubseteq\bigcup_{\ell\ge n}E_\ell$. $\endgroup$ – Brian M. Scott Jan 1 '15 at 20:28

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