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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.

Koro
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2 Answers2

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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.

Martin R
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  • I think it is correct but what is wrong in my proof is not clear to me yet. In my case 1, I note that there is a subsequence $a_{n_k}$, which converges to $a^$. Indices are $n_1<n_2<...,<n_k,... $. Similarly, $b_{m_l}$ is a convergent subsequence which converges to $b^$. Indices are $m_1<m_2<...<...$. So this is what is problematic? – Koro Feb 28 '21 at 12:10
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    @Koro: Yes. For example, what if the indices for $a$ are $1, 3, 5, 7, \ldots$ and the indices for $b$ are $2, 4, 6, 8, \ldots$? Your proof claims that there are “common subsequences” $(a_{n_s})$ and $(b_{n_s})$ such that $a_{n_s}+b_{n_s} \to a^+b^$, but that need not be the case. – Martin R Feb 28 '21 at 12:13
  • In your answer, what if cgt. subsequence of $a_{n_k}$ has indices $n_{k_{l}}$ such that $4<8<12<...<$.Similarly, for cgt subsequence of $b_{n_k}$ we may have indices $n_{k_{s}}$ like this $2,6,18,...$. Now both indices are not common. – Koro Feb 28 '21 at 22:35
  • But I observe that it will not be the case as $(a_{n_k} + b_{n_k}) $ is cgt. And subsequence $a_{n_{k_l}}$ is cgt. therefore $ b_{n_{k_l}}= (a_{n_{k_l}} + b_{n_{k_l}})- (a_{n_{k_l}})$ is also cgt. – Koro Feb 28 '21 at 22:57
  • Can you please have a look at my answer below? Thanks in advance – Koro Feb 28 '21 at 23:15
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I am posting an answer based on inputs by Mr. Martin R:

If exactly one of $a^*$ or $b^*$ is infinite, we are done. If both $a^*$, $b^*$ are infinity of same sign, then also we are done.

We consider the case, when $a^*, b^*$ and $c^*$ are finite. Since $c^*\in E_c, \exists$ subsequence $c_{n_k}\to c^*$. That is $c_{n_k}=a_{n_k}+b_{n_k} \to c^* $.

$(a_{n_k})$ is bounded and therefore has convergent subsequence $(a_{n_{k_l}}): (a_{n_{k_l}})\to \alpha$. Therefore, $b_{n_{k_l}}=(a_{n_{k_l}}+b_{n_{k_l}})-(a_{n_{k_l}})\to c^*-\alpha$.

Now, $a_{n_{k_l}}+b_{n_{k_l}}\to (\alpha)+(c^*-\alpha)\le a^*+b^*\implies c^*\le a^*+b^*$.
Proved.

Koro
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    That is correct and works even for the case that $a^$ or $b^$ are $-\infty$. It is quite similar to my approach, only that you (correctly) observed that $(b_{n_{k_l}})$ is already convergent, so that it is not necessary to chose another subsequence. – Martin R Mar 01 '21 at 04:01
  • @MartinR: Thanks a lot – Koro Mar 01 '21 at 04:10