# Suppose $(x_n)$ and $(y_n)$ bounded, prove $\limsup(x_n+y_n)\le \limsup(x_n)+\limsup(y_n).$

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?

• Abbreviating words obscures. The internet is big enough for the word "bounded." Commented Mar 17, 2023 at 4:13
• 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 $L + \epsilon$ .
– Anon
Commented Mar 17, 2023 at 4:22
• 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.
– user1101956
Commented Mar 17, 2023 at 4:26
• @PowerPointTrenton No, if $\sup_{m \ge n} x_m$ is between $L - \epsilon$ and $L + \epsilon$, then the sequence $x_n$ lies in $(-\infty , L + \epsilon)$.
– Anon
Commented Mar 17, 2023 at 4:31

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.