The question I am working on says that $(x_n)_{n \in \mathbb{N}}$ is a bounded sequence in $\mathbb{R}$. The questions says to assume that $x_n >0 $ for all $ n \in \mathbb{N}$. Is then $$\limsup_{n\to \infty}x_n >0$$
Prove or give a counter example.
My thinking goes as follows: since we know that $x_n >0$ then the supremum would also be greater than $0$. As $\limsup_{n\to \infty} x_n = \lim_{n\to \infty} sup\left\{x_m|m>n\right\}$ and the supremum of $\left\{x_m|m>n\right\}$ must also be greater than $0$, then the original statment is true.
Is this correct?
The supremum of a set of positive integers is positive, clearly because the supremum is an upper bound, so is greater than all the elements of the set, which are positive numbers, so it positive.
However, here, we are taking a limit of supremums. Even if all the supremums are positive, the limit of the supremums need not be.
For example, consider $a_n = \frac 1n$. It is a strictly decreasing sequence i.e. $a_m < a_n$ for all $m > n$, so it follows that $\sup \{a_m : m > n\} = a_n$ for all $n \in \mathbb N$. But, $$\limsup a_n = \lim_{n \to \infty} \sup\{a_m : m > n\} = \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac 1n = 0$$
Therefore the statement, made as before, is false. But, what is true, is that $\limsup a_n $ $\color{grey}{\geq}$ $0$. To see this, it is actually enough to see that the limit of a sequence of positive quantities can only be non-negative, proved using a common $\epsilon-\delta$ argument.