Show that $\lim\inf X_{n}\subseteq \lim\sup X_{n}$ always.

Question

Definition. Let $(X_n)_n$ be a sequence of sets. Define:

$\lim\inf X_{n}=${$x:$ there is a $N$ in $\mathbb{N}$ such that $x\in X_n$ for all $n>N$}

$\lim\sup X_{n}=${$x:$ $x\in X_n$ for infinetly many $n$}

• Show that $\lim\inf X_{n}\subseteq \lim\sup X_{n}$ always.

My idea for proof. Let $x\in\lim\sup X_{n}$, i.e., there is a $N$ in $\mathbb{N}$ such that $x\in X_n$ for all $n>N$. We need to show that $x\in\lim\sup X_{n}$. But, how?

Can you help?

Answer 1

If there is some $N$ such that $x \in X_{n}$ for all $n \geq N$, then there are infinitely many $n$ such that $x \in X_{n}$; note that we already have $n \geq N$ implying $x \in X_{n}$.

The subset $\{ n \in \mathbb{N} \mid n \geq N\}$ of $\mathbb{N}$ is still an infinite set.