$X_n\leq Y_n$ implies $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$

Question

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 $\lim\inf X_n = \inf E$.

Answer 1

The definition of limsup/liminf for a sequence $ \{a_n: n \geq 1 \} $ is given by

$$ \liminf_{ n \to \infty} a_n = \sup_{n \geq 1} \inf_{ k \geq n } a_k \text{ and } \limsup_{n \to \infty} = \inf_{n \geq 1} \sup_{ k \geq n } a_k . $$

Let us show the inequality for $\liminf $, the other being similar. Define, for a sequence $ \{a_n \} $ and $ n \geq 1$, set $$ I_n (a) = \inf \{ a_k : k \geq n \}. $$ Clearly, $ I_n (a) \leq I_{n+1} (a) $ since the infimum is taken over $ \{ k : k \geq n \} $ in the first case which contains the set $ \{ k : k \geq n+1 \}$ over which the infimum is taken in the second case. Thus, $ I_n (a) $ is non-decreasing and hence the limit of $ I_n (a) $ exists and the supremum is the limit (which need not be bounded). Thus, we have, $$ \liminf_{ n \to \infty} a_n = \sup_{n \geq 1} \inf_{ k \geq n } a_k = \lim_{ n \to \infty} I_n (a) . $$ Now, coming to the problem, let $ X_n \leq Y_n $ for all $ n \geq 1$, we claim $$ I_n (X) \leq I_n (Y) \text{ for all } n \geq 1 . $$ Once we show this, the inequality follows from the above relation about $ \liminf $ and limit of the sequence of $ I_n (X) $ and $ I_n (Y) $ respectively.

To prove the claim: fix $ n \geq 1 $ and take any $ \epsilon > 0 $. Then $ I_n (Y) + \epsilon > I_n (Y) $ and hence there is a $ k (\geq n ) $ such that $ Y_k < I_n (Y) + \epsilon $. Thus, we have $ X_k \leq Y_k < I_n (Y) + \epsilon $. Thus, we have $ I_n (X) \leq I_n (Y) + \epsilon $. Since $ \epsilon > 0 $ is arbitrary, we have $ I_n (X) \leq I_n (Y) $ and the claim is proved.

Answer 2

I could have sworn I asked a question like this but I can't find it in my question list. I recall a proof of this question and it went similar to this:

If $u>\lim sup(y_n)$, then there can be only a finite number of $n\in\mathbb{N}$ such that $u<y_n$. Since $x_n\le y_n$ then $\lim sup(x_n)\le u$ and so $\lim sup (x_n)\le \lim sup(y_n)$.

Answer 3

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.

Answer 4

If $\lim\inf X_n=\infty$ then $X_n\rightarrow\infty$ implies $Y_n\rightarrow\infty$. Assume $\infty>\lim\inf X_n>\lim\inf ~Y_n$. For $0<\epsilon<\lim\inf X_n-\lim\inf ~Y_n$, we've $n_0\in \mathbb{N}$ such that $X_{m}>(\lim\inf X_n)-\epsilon/2$ for $m>n_0$. From the choice of $\epsilon$, there is a $n_1>n_0$ such that $(\lim\inf X_n)-\epsilon/2>Y_{n_1}$. But then we get $X_{n_1}>Y_{n_1}$, a contradiction to the assumption! So, we conclude $\lim\inf X_n\leq\lim\inf ~Y_n$. We apply this to the sequences $\lbrace-X_n\rbrace$ and $\lbrace-Y_n\rbrace$ by observing $-X_n\geq-Y_n$ to prove for the $\lim\sup$.