Can anyone prove this question? I tried but I didn't get any I idea, so I hope someone can solve it.
Let $X_n\leq Y_n$ for each $n\in \Bbb N$. Show that $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$.
Please prove this question – thanks.
The definition I have:
Let $X_n$ be a sequence in real number and let $$E=\{x\in \Bbb R^\sharp:(X_{n_k}) \rightarrow x \text{ for some subsequence }(X_{n_k})\text{ of }(X_n)\}$$ for all $n \in \Bbb N$ and $k$ from $1$ to $\infty$. Then by definition $\lim\sup X_n = \sup E$ and $\lim\inf X_n = \inf E$.
Best Answer
HINT: Suppose that $X_n\le Y_n$ for each $n\in\Bbb N$, and suppose, to get a contradiction, that $\liminf_nY_n<\liminf_nX_n$. Then there is a subsequence $\langle Y_{n_k}:k\in\Bbb N\rangle$ that converges to some $y<\liminf_nX_n$. Show that either $\langle X_{n_k}:k\in\Bbb N\rangle\to-\infty$, or $\langle X_{n_k}:k\in\Bbb N\rangle$ has a subsequence that converges to some $x\le y$; both are impossible, since $y<\liminf_nX_n$.
The other inequality can be proved in a very similar way.