Definition: $\limsup x_n$= supremum of all subsequential limits of sequence $(x_n)=\sup E_x$, where $E_x$ is set of all subsequential limits of sequence $(x_n)$. Let $x^*=\limsup x_n$.
Given any two sequences $(a_n)$ and $(b_n)$, it is to be proven that: $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$, provided sum on right side is not of the form $\infty -\infty$. $\tag 1$
I tried to prove it as follows:
Let $c_n=a_n+b_n$ for all $n\in \mathbb N$
Case $1: a^*, b^*, c^* \in \mathbb R$
It follows that the sequences $(a_n),(b_n),(c_n)$ are bounded. And since set of all subsequential limits of a sequence is closed, it follows that $a^*\in E_a, b^*\in E_b, c^*\in E_c\implies$ There exist subsequences $(a_{n_k}), (b_{n_l})$ and $(c_{n_m})$ such that $a_{n_k}\to a^*, b_{n_l}\to b^*, c_{n_m}\to c^* $.
For any $\epsilon \gt 0, \exists s_\epsilon\in \mathbb N$ such that $s\ge s_\epsilon \implies |a_{n_s}-a^*|\lt \epsilon/4 \space, |b_{n_s}-b^*|\lt \epsilon/4 \implies |a_{n_s}+b_{n_s}-a^*-b^*|\lt \epsilon/2$.
Now suppose on the contrary that $c^*\gt a^*+b^*$. We have: for any $\epsilon \gt 0$
$\exists M_\epsilon \in \mathbb N: m\ge M_\epsilon\implies |c_{n_m}-c^*|\lt \epsilon/2.$
Hence for $r\in \mathbb N: r\ge \sup(M_\epsilon, s_\epsilon),$ we have:
$0\lt c^*-a^*-b^*=|c^*-(a_{n_r}+b_{n_r})+ (a_{n_r}+b_{n_r}-a^*-b^*)|\le |c^*-(a_{n_r}+b_{n_r})|+ |(a_{n_r}+b_{n_r}-a^*-b^*)|\lt\epsilon \implies c^*=a^*+b^* $, which is contradiction. Therefore, $c^*\le a^*+b^*$.
Case $2$: $a^*=\infty, b^*\in \mathbb R\cup \{\infty\},$ then the result is trivial.
Case $3$: $b^*=\infty, a^*\in \mathbb R\cup \{\infty\}$, then as in Case $3$, the result is true.
Case $4$: $a^*\in \mathbb R, b^*=-\infty .$
$E_b=\{-\infty\}$ and $ (a_n)$ is bounded by some $M\gt 0$. So, $(c_n)$ is not bounded below as $(b_n)$ is not bounded below.
So for some $M\in \mathbb N$, $n\ge M\implies c_n\lt a^*+b_n\implies c^*\le a^*+b^* $.
Case $5$: $b^*\in \mathbb R, a^*=-\infty .$
Same as case $4$.
So proved.
My question is what happens when right hand side of $(1)$ is $\infty -\infty $?
Consider for example: $(a_n)=(n), (b_n)=(-n)$, clearly $c_n=0\to \limsup c_n=0$. But can we claim that $0\gt \infty -\infty $ (I don't think so as $\infty -\infty $ is undefined.)?. If not, then $(1)$ always holds?
Please help. Thanks.
Best Answer
Your proof is not correct. You have (possibly) distinct subsequences $(a_{n_k})$, $(b_{m_k})$ of the original sequences converging to $a^*$ and $b^*$, and these need not have any indices in common.
Also $\infty-\infty$ is undefined, and nothing can be said about $\limsup (a_n+b_n)$ if the right-hand side of $(1)$ is of that form. It can be $-\infty$, $+\infty$, or any finite real number.
Using your definition of $\limsup$ as the supremum of all subsequential limits, I would argue as follows:
If one of $\limsup a_n$ or $\limsup b_n$ is equal to $+\infty$ then the other cannot be $-\infty$. In that case the right-hand side of $(1)$ is $+\infty$ and the inequality holds.
Therefore is suffices to consider the case that both $\limsup a_n$ and $\limsup b_n$ are a real number or $-\infty$. In particular, both sequences are bounded above.
We can also assume that $c^* = \limsup(a_n+b_n)$ is not $-\infty$.
For every $\epsilon > 0$ there is a convergent subsequence $c_{n_k}$ of $(c_n)$ with $$ \lim_{k \to \infty} (a_{n_k} + b_{n_k}) = \lim_{k \to \infty} c_{n_k} > c^* - \epsilon \, . $$ Now $(a_{n_k})$ and $(b_{n_k})$ are both bounded sequences. Then $(a_{n_k})$ has a convergent subsequence $(a_{n_{k_l}})$ and $(b_{n_{k_l}})$ has a convergent subsequence $(b_{n_{k_{l_m}}})$. Then $$ c^* - \epsilon < \lim_{m \to \infty} (a_{n_{k_{l_m}}} + b_{n_{k_{l_m}}}) = \lim_{m \to \infty} a_{n_{k_{l_m}}} + \lim_{m \to \infty} b_{n_{k_{l_m}}} \le a^* + b^* \, . $$ This holds for all $\epsilon > 0$, which implies that $c^* \le a^* + b^*$.
Remark: The difference to your proof is that we started with a convergent subsequence $(c_{n_k})$ of $(c_n)$ and then chose “sub-sub-sequences” of $(a_n)$ and $(b_n)$ which are “simultaneously” convergent.