Let $x_n\leq y_n$ for each $n\in \Bbb N$. Then,
$$\liminf x_n \leq \liminf y_n$$
Here is a solution for the above problem ( source:https://math.stackexchange.com/a/934738/199156)
Define $u_n:=\inf\{x_k|k\geq n\}$ and $v_n:=\inf\{y_k|k\geq n\}$.
Then the sequence $(u_n)$ is non-decreasing and satisfies
$\lim_{n\rightarrow\infty}u_n=\liminf x_n$.Likewise $\lim_{n\rightarrow\infty}v_n=\liminf y_n$.
From $x_m\leq y_m$ for each $m$ it follows directly that $u_m\leq v_m$
for each $m$ and consequenly $\liminf x_n=\lim_{n\rightarrow\infty}u_n\leq\lim_{n\rightarrow\infty}v_n=\liminf y_n$
However, I was wondering if it can be done using the definition of $\lim\limits_{n \to \infty} \inf x_{n}=\underset{n \geq 1}{\sup} \underset{k \geq n}{\inf} x_{k}$ instead.
Best Answer
The idea in the linked proof can also be expressed as follows.
As you note, we have $\liminf x_n = \sup_{n \geq 1} \inf_{k \geq n} x_k$. We note that because $x_k \leq y_k$ for each $n$, it must hold that for any $n \in \Bbb N$, we have $$ \inf_{k \geq n} x_k \leq \inf_{k \geq n} y_k. $$ Now, let $X_n = \inf_{k \geq n} x_k$, and similarly $Y_n = \inf_{k \geq n} y_k$. We have established that $X_n \leq Y_n$ holds for each $n$. It follows that $$ \sup_{n \geq 1} X_n \leq \sup_{n \geq 1} Y_n. $$ That is, we have $\liminf x_n \leq \liminf y_n$, which is what we wanted.