$\limsup_{n \to \infty}\{\frac{X_{n}}{\log{(n)}}\leq\epsilon\}\subseteq \{\liminf_{n \to \infty}\frac{X_{n}}{\log{(n)}}\leq\epsilon\}$

Question

I recently saw an assertion made that

$$\bigcap_{m \in \mathbb N} \bigcup_{n \geq m}\left\{\frac{X_{n}}{\log{(n)}}\leq\epsilon\right\} \subseteq\left\{\liminf_{n \to \infty}\frac{X_{n}}{\log{(n)}}\leq\epsilon\right\}.$$

And I have a few questions:

$1.$ Why is this supposed to be clear, I cannot see it myself, let alone prove it.

$2.$ By definition $\limsup_{n \to \infty}\{\frac{X_{n}}{\log{(n)}}\leq\epsilon\}:=\bigcap_{m \in \mathbb N} \bigcup_{n \geq m}\{\frac{X_{n}}{\log{(n)}}\leq\epsilon\}$. But am I correct in saying the $\limsup_{n \to \infty}$ on the set $\{\frac{X_{n}}{\log{(n)}}\leq\epsilon\}$ is a different definition limit superior/inferior than to the $\liminf_{n \to \infty}$ used in $\{\liminf_{n \to \infty}\frac{X_{n}}{\log{(n)}}\leq\epsilon\}$

With respect to $\{\liminf_{n \to \infty}\frac{X_{n}}{\log{(n)}}\leq\epsilon\}$, we are looking at a subsequence (i.e. the smallest convergent sequence) of $(\frac{X_{n}}{\log{(n)}})_{n}$, namely $(\frac{X_{n_{k}}}{\log{(n_{k})}})_{n_{k}}$ such that $\frac{X_{n_{k}}}{\log{(n_{k})}}\leq \epsilon$

Answer 1

You are right in noting that limsup of sequence of sets is a different notion than lim sup of sequence of numbers. $\lim \sup A_n$ consists of points that belong to infinitely many of the sets $A_n$. Hence LHS contains points for which $\frac {X_n} {\log(n)} \leq \epsilon$ for infinitely many vales of $n$. This means there exist integers $n_1<n_2,\cdots$ such that $\frac {X_{n_k}} {\log(n_k)} \leq \epsilon$ for all $k$. This gives the subsequence you are looking for, so LHS is contained in RHS.