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Let $(x_n)$ and $(y_n)$ be bounded sequences. Then there exist real numbers $M$ and $N$ such that $x_n \leq M$ and $y_n \leq N$ for all $n$. Let $L = \limsup(x_n)$ and $K = \limsup(y_n)$. Then for any $\epsilon > 0$, there exists an integer $N_1$ such that for all $n \geq N_1$, we have $$ L - \epsilon < x_n \leq L + \epsilon. $$ Similarly, there exists an integer $N_2$ such that for all $n \geq N_2$, we have $$ K - \epsilon < y_n \leq K + \epsilon. $$ Let $N_3 = max(N_1,N_2)$. Then for all $n\ge N_3$, we have $$ (L+K)-2\epsilon < x_n+y_n\le (L+K)+2\epsilon. $$

Taking the limit superior of both sides as n goes to infinity gives us: $$ (L+K)-2\epsilon\le limsup(x_n+y_n)\le (L+K)+2\epsilon. $$

Since $\epsilon>0$ was arbitrary, it follows that: $$limsup(x_n+y_n)\le limsup(x_n)+limsup(y_n).$$

Can anyone see if my proof is alright?

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    $\begingroup$ Abbreviating words obscures. The internet is big enough for the word "bounded." $\endgroup$
    – Thomas Andrews
    Commented Mar 17, 2023 at 4:13
  • $\begingroup$ This doesn't seem right. How can you conclude that each $x_n$ beyond $N_1$ is between $L - \epsilon$ and $L + \epsilon$ - this is infact wrong. What you can say is that $\sup_{m \ge n} x_m$ for $n \ge N_1$ is between $L - \epsilon$ and $L + \epsilon$ . $\endgroup$
    – Anon
    Commented Mar 17, 2023 at 4:22
  • $\begingroup$ For $L-\epsilon<x_n$, I apply the property of supremum. For $L+\epsilon$, I am not sure whether it is strictly larger or nonstrict. $\endgroup$
    – user1101956
    Commented Mar 17, 2023 at 4:26
  • $\begingroup$ @PowerPointTrenton No, if $\sup_{m \ge n} x_m$ is between $L - \epsilon$ and $L + \epsilon$, then the sequence $x_n$ lies in $(-\infty , L + \epsilon)$. $\endgroup$
    – Anon
    Commented Mar 17, 2023 at 4:31

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You can show that forall $n \in \mathbb{N}$, $$ \sup_{m \ge n} (x_m + y_m) \le \sup_{m \ge n} x_m + \sup_{m \ge n} y_m$$ by simply showing the $\sup_{m \ge n} x_m + \sup_{m \ge n} y_m$ is an upper bound of the set $\{ x_m + y_m | m \ge n \}$. And then just take limit on both sides, you know it exists in $\mathbb{R}$ because the sequences are given to be bounded, and the result follows by linearity of limit over sums.

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  • $\begingroup$ Thank you. That helps a lot! $\endgroup$
    – user1101956
    Commented Mar 17, 2023 at 8:15

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