Let $x_n\leq y_n$ for each $n\in \Bbb N$. Show that $\liminf x_n \leq \liminf y_n$

Question

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.

Answer 1

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.

Answer 2

Suppose that $x=\liminf_{n\to\infty}x_n>\liminf_{n\to\infty}y_n=y$. Now since $y$ is the limit of a subsequence of $\{y_n\}_n$, it means that after a threshold all the points of that subsequence of $\{y_n\}_n$ are less than $x$. Consider the subsequence in $\{x_n\}_n$ corresponding to these indices, and you'll get a subsequence of $\{x_n\}_n$ all of whose terms are less than $x$ after a threshold. This is a contradiction.