# Real Analysis - More limsups

## Homework Statement

Show that limsup(s_n + t_n) <= limsup(s_n) + limsup(t_n) for bounded sequences (s_n) and (t_n).

## The Attempt at a Solution

My book gives a hint that says to first show that sup{s_n + t_n : n > N} <= sup{s_n : n > N} + sup{t_n : n > N}. I'm not really even sure how to do that. Any ideas? Thanks!

tiny-tim
Homework Helper
My book gives a hint that says to first show that sup{s_n + t_n : n > N} <= sup{s_n : n > N} + sup{t_n : n > N}. I'm not really even sure how to do that. Any ideas? Thanks!

Hi steelphantom! sup{} can really only apply to a finite set, so I think the book must mean show that sup{s_n + t_n : n < N} ≤ sup{s_n : n < N} + sup{t_n : n < N}.

Can you do that? HallsofIvy
Homework Helper
?? What makes you say the "sup" only applies to a finite set? If, for example, sn= 1/n, Then sup{sn: n> N} is 1/N.

quasar987
Homework Helper
Gold Member
tiny-tim, what do you mean by "sup{} can really only apply to a finite set"? For instance, sup{[0,1]}=1, yet [0,1] is an infinite set.

steelphantom, ok, so clearly the book wants you to use the characterizations of limsup

$$\limsup_{n\rightarrow +\infty}x_n=\lim_{N\rightarrow+\infty}\left(\sup_{n> N}x_n\right)$$

Following the hint of the book, try to show that more generally, for A, B two sets of real numbers,

$$\sup(A+B)\leq \sup(A)+\sup(B)$$

Let $\{a_n+b_n\}_{n\in\mathbb{N}}$ be a sequence in A+B converging to sup(A+B) [show such a sequence must exist if you haven't done it already]. Then clearly, $a_n\leq\sup(A)$ and $b_n\leq\sup(B)$ for all n, hence.... (you finish)

tiny-tim, what do you mean by "sup{} can really only apply to a finite set"? For instance, sup{[0,1]}=1, yet [0,1] is an infinite set.

steelphantom, ok, so clearly the book wants you to use the characterizations of limsup

$$\limsup_{n\rightarrow +\infty}x_n=\lim_{N\rightarrow+\infty}\left(\sup_{n> N}x_n\right)$$

Following the hint of the book, try to show that more generally, for A, B two sets of real numbers,

$$\sup(A+B)\leq \sup(A)+\sup(B)$$

Let $\{a_n+b_n\}_{n\in\mathbb{N}}$ be a sequence in A+B converging to sup(A+B) [show such a sequence must exist if you haven't done it already]. Then clearly, $a_n\leq\sup(A)$ and $b_n\leq\sup(B)$ for all n, hence.... (you finish)

Thanks for the help! Let's see if I got it...

limsup(a_n + b_n) = sup(A + B) = lim(a_n + b_n) = lim(a_n) + lim(b_n). Since (a_n), (b_n) are <= sup(A), sup(B), respectively, then so are lim(a_n) and lim(b_n). So we have limsup(a_n + b_n) <= sup(A) + sup(B). But we have sup(A) = sup(a_n) = limsup(a_n) and sup(B) = sup(b_n) = limsup(b_n).

Finally, we get limsup(a_n + b_n) <= limsup(a_n) + limsup(b_n).

Is this correct, or am I assuming too much? Thanks again.

quasar987
First concentrate on proving the little lemma I outlined for you (namely, for A, B two arbitrary sets of real numbers, $\sup(A+B)\leq \sup(A)+\sup(B)$), and then worry about solving your problem involving limsups of the sequence s_n and t_n.