can we say that there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that: $ x_{n_k}\leq x\qquad k\geq 1 $

Question

Let $\{x_n\}$ be a positive real sequence such that: $$ \liminf_n x_n< x $$ can we say that there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that: $$ x_{n_k}\leq x\qquad k\geq 1 $$

Answer 1

Having made the edit, the answer is now yes. Let $x_n$ be a sequence such that

$$\liminf_{n\to\infty}x_n=x_l<x$$

In order to cover all bases, first consider the cases where $x_l$ is finite. Denote

$$\frac{x-x_l}{2}=\epsilon$$

By definition, we know

$$\lim_{n\to\infty}\left(\inf_{m\geq n}x_m\right)=x_l$$

This implies there are infinite $x_n$ such that

$$|x_n-x_l|<\epsilon$$

$$-\epsilon<x_n-x_l<\epsilon$$

$$-\epsilon+x_l<x_n<\epsilon+x_l$$

Denote this subsequence $x_{n_k}$. Then

$$x-x_{n_k}> x-(\epsilon+x_l)=x-\left(\frac{x-x_l}{2}+x_l\right)=\frac{x-x_l}{2}>0$$

and we are done. Now, consider the case where $x_l=-\infty$ (Depending on your definition of limit infimum, this is valid). In this case, there is a subsequence $x_{n_k}$ of $x_n$ such that

$$\lim_{k\to\infty}x_{n_k}=-\infty$$

In either case, we conclude there is a subsequence of $x_n$ such that

$$x_{n_k}<x$$