Show $\limsup x_n \le \limsup y_n$ provided that $x_n \le y_n$ for all $n \in \mathbb N$
So I have found some proofs for this same question here but I think we have a different definition of $\limsup$ in the class and I couldn't prove the statement with this definition. The definition we have is $$\limsup a_n := \begin{cases} \lim\limits_{n\to\infty} (\sup\limits_{k\ge n}a_k) & \text{if } (a_n)_n \text{ is upper bounded,}\\ +\infty & \text{otherwise.} \end{cases}$$
So I was able to show that if $x_n \le y_n$ for all $n \in \mathbb N$, we will also have $\sup\limits_{k\ge n}x_k \le \sup\limits_{k\ge n}y_k$ for all $n \in \mathbb N$. After saying that, the other proofs that I have seen take the limits of the both sides and directly conclude the result. But with the above definition of $\limsup$, this only shows the case when both sequences are upper bounded. So I have to prove three more cases.
If $y_k$ is not upper bounded and $x_k$ is upper bounded, the inequality will hold since we will have $\limsup y_k=\infty$ and $\limsup x_k=L < \infty$.
Also, it is not possible that $y_k$ is upper bounded but $x_k$ is not upper bounded. Because the upper bound of $y_k$ is obviously an upper bound for $x_k$ too.
So, the last case is when both the sequences are not upper bounded. For this case, I would get something like $\infty \le \infty$ and I don't know what to do. Can you please help on what should I do with this case?
Also, I have mostly seen that $\limsup$ is defined as $\limsup a_n := \lim\limits_{n\to\infty} (\sup\limits_{k\ge n}a_k)$. What are the differences of these two definitions?