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This question already has an answer here:

Let $\{ E_n \}_{n \in \mathbb{N} }$ be a sequence of sets in some ambient set $\Omega $. I want to show that

$$ \liminf E_n \subset \limsup E_n $$

My attempt: IF $x \in \liminf E_n = \bigcup_{k=1}^{\infty} \bigcap_{n \geq k} E_n $, then there is some $k_0 \in \mathbb{N}$ so that $x \in \bigcap_{n \geq k_0} E_n $. How can I show that $x \in \bigcap_{k =1}^{\infty} \bigcup_{n \geq k} E_n = \limsup E_n $ ??

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marked as duplicate by user147263, Jonas Meyer, egreg, Asaf Karagila, Ahaan S. Rungta Dec 31 '14 at 22:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Assume $x \in \cap_{n \geq k_0} E_n$. Then

For $1 \leq k \leq k_0$, we have $x \in E_{k_0}$ and hence $x \in \cup_{n \geq k} E_n$.

For $k > k_0$, we have $x \in E_k $ and hence $x \in \cup_{n \geq k} E_n$.

Therefore $x \in \cap_{k \geq 1} \cup_{n \geq k} E_n$.

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  • $\begingroup$ Why do you need to consider two cases? $\endgroup$ – user203867 Jan 2 '15 at 8:22
  • $\begingroup$ Because in the two different cases of $k$ we have to consider different $n$ to give evidence of $x \in \cup_{n \geq k} E_n$. $\endgroup$ – Empiricist Jan 2 '15 at 8:45
  • $\begingroup$ I dont understand this part. Can you explain a bit more please? $\endgroup$ – user203867 Jan 2 '15 at 8:51
  • $\begingroup$ You are asked to show $x \in \cup_{n \geq k} E_n$ for all $k \geq 1$ given that $x \in E_n$ for all $n \geq k_0$. Fix $k$, you need to find $n \geq k$ such that $x \in E_n$. It is trivial that we can pick $n = k$ for $k > k_0$. For $1 \leq k \leq k_0$, this is also easy, we can pick any $n \geq k_0$. The difference is just that we cannot pick $n = k$ in this case. $\endgroup$ – Empiricist Jan 2 '15 at 8:57
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    $\begingroup$ Can you give me feedback on my full solution as I wrote it in my own words. thanks very much: Let $x \in \liminf E_n $. $x \in \bigcap_{n \geq k_0} E_n $ for some $k_0 \in \mathbb{N}$. In particular, $x \in E_n$ for all $n \geq k_0 $. Let $k \geq 1$ be arbitrary. We need to find $n \geq k$ so that $x \in E_n $. Take $n = \max \{k,k_0\} $. It follows that $n \geq k$ and $x \in \bigcup_{n \geq k} E_n $. Since $k \geq 1$ was arbitrary, it follows that $x \in \bigcap_{ k \geq 1} \bigcup_{n \geq k} E_n = \limsup E_n $. $\endgroup$ – user203867 Jan 2 '15 at 9:16