# Inclusions regarding the limsup and liminf of sets: $\liminf E_n \subset \limsup E_n$ [duplicate]

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 $x \in \bigcap_{k =1}^{\infty} \bigcup_{n \geq k} E_n = \limsup E_n$ ??

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

• Remember $\bigcap A_i\subseteq bigcup A_i$ – Kamster Dec 31 '14 at 10:01
• – Martin Sleziak Dec 31 '14 at 10:27

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

• Why do you need to consider two cases? – user203867 Jan 2 '15 at 8:22
• Because in the two different cases of $k$ we have to consider different $n$ to give evidence of $x \in \cup_{n \geq k} E_n$. – Empiricist Jan 2 '15 at 8:45
• I dont understand this part. Can you explain a bit more please? – user203867 Jan 2 '15 at 8:51
• 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. – Empiricist Jan 2 '15 at 8:57
• 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 $x \in \bigcap_{ k \geq 1} \bigcup_{n \geq k} E_n = \limsup E_n$. – user203867 Jan 2 '15 at 9:16