Why is does proposition about upper limit set hold?

Question

Let $X_n$ be a set and let $\displaystyle\overline\lim_{n\to \infty} X_n=\bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} X_n$.

Prove that

$ x \in \displaystyle\overline\lim_{n\to \infty} X_n \iff$ There exists $\{ n_k \}_{k=1}^{\infty} \subset \mathbb{N}$ such that $n_k < n_{k+1}$ and $x \in X_{n_k}$ for all $k\in \mathbb{N}. $

My attempt is as follows.

$ x \in \displaystyle\overline\lim_{n\to \infty} X_n \iff x \in \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} X_n \iff $ For all $k \in \mathbb{N},$ there exists $n_k \in \mathbb{N}$ s.t. $n_k \geqq k$ and $x \in X_{n_k}$

Thus, what I have to prove is

For all $k \in \mathbb{N},$ there exists $n_k \in \mathbb{N}$ s.t. $n_k \geqq k$ and $x \in X_{n_k} \iff $ There exists $\{ n_k \}_{k=1}^{\infty} \subset \mathbb{N}$ such that $n_k < n_{k+1}$ and $x \in X_{n_k}$ for all $k\in \mathbb{N}.$

I cannot know why this holds.

I'd like you to give me some ideas.

Answer 1

Let $x \in \limsup_{n \to \infty} X_n$. You can build up the subsequence $\{n_k\}$ one by one.

• For the first term: There exists some $n_1$ such that $x \in X_{n_1}$.
• Suppose you have $n_1 < \cdots < n_k$ already. Since $x \in \limsup_{n \to \infty} X_n \subseteq \bigcup_{n=n_k+1}^\infty X_n$ you can find some $n_{k+1} > n_k$ such that $x \in X_{n_{k+1}}$.

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Suppose $x \in X_{n_k}$ for all $n_k$, and $n_1 < n_2 < \cdots$. It suffices to show $x \in \bigcup_{n = \ell}^\infty X_n$ for any $\ell \ge 1$.

Fix $\ell \ge 1$. Since $\{n_k\}$ increases to infinity, there is some $n_k$ satisfying $n_k \ge \ell$. Then $x \in X_{n_k} \subseteq \bigcup_{n = \ell}^\infty X_n$.