If $X=\{x_n\}, Y = \{y_n\}$ be bounded sequences of real numbers. Then, if $x_n \leq y_n~\forall~n$, then $\lim \inf (x_n) \leq \lim \inf(y_n)$

Question

If $X=\{x_n\}, Y = \{y_n\}$ be bounded sequences of real numbers. Then, if $x_n \leq y_n~\forall~n$, then show that $\lim \inf (x_n) \leq \lim \inf(y_n)$ and $\lim \sup (x_n) \leq \lim \sup (y_n)$

Attempt: $\lim \inf (x_n)$ represents the least limit point in $X$ and $\lim \inf(y_n)$ represents the least limit point of $Y$.

Given that, $~\forall~m, x_m \leq y_m, $ I am still not able to see any compulsion why the least limit point of $x_n$ should be less than the least limit point of $y_n$.

That is, for each index $m, x_m \leq y_m$. But, that doesn't place any restriction on the least limit point of $x_n$ becoming greater than the least limit point of $y_n$?

Where could I be going wrong?

Thank you for your help.

Answer 1

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 $\liminf x_n=\lim_{n\rightarrow\infty}u_n\leq\lim_{n\rightarrow\infty}v_n=\liminf y_n$

Answer 2

For $\liminf$, let $$ u_n=\inf\{x_k:k\ge n\},\quad v_n=\inf\{y_k:k\geq n\}. $$ Then, $u_n$ is a lower bound for $\{y_k:k\geq n\}$ because for each $k\geq n$, we have $u_n\leq x_k\leq y_k$. Because $v_n$ is the greatest lower bound for $\{y_k:k\geq n\}$, we have $$ u_n\leq v_n. $$ Now take the limit $n\to\infty$.

You can apply the same approach for $\limsup$.

Answer 3

You have $x_n' = \inf_{k \ge n} x_k \le y_n$, and $x_n' \uparrow \liminf_n x_n$. Now let $\epsilon>0$, then for $n$ greater than some $N$, we have $x_n' \ge \liminf_n x_n -\epsilon$, and so $y_n \ge \liminf_n x_n -\epsilon$. Noting that the right hand side is a constant, we have $\liminf_n y_n \ge \liminf_n x_n -\epsilon$. Since $\epsilon>0$ was arbitrary, we have the desired result.

Since $\limsup_n a_n = - \liminf_n (-a_n)$, we have the other direction.