Proving the Lemma for Bounded Sets in Real Analysis

In summary, the hint in the book is 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!
  • #1
steelphantom
159
0

Homework Statement


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

Homework Equations




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!
 
Physics news on Phys.org
  • #2
steelphantom said:
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! :smile:

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? :smile:
 
  • #3
?? 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.
 
  • #4
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

[tex]\limsup_{n\rightarrow +\infty}x_n=\lim_{N\rightarrow+\infty}\left(\sup_{n> N}x_n\right)[/tex]

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

[tex]\sup(A+B)\leq \sup(A)+\sup(B)[/tex]

Let [itex]\{a_n+b_n\}_{n\in\mathbb{N}}[/itex] 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, [itex]a_n\leq\sup(A)[/itex] and [itex]b_n\leq\sup(B)[/itex] for all n, hence... (you finish)
 
  • #5
quasar987 said:
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

[tex]\limsup_{n\rightarrow +\infty}x_n=\lim_{N\rightarrow+\infty}\left(\sup_{n> N}x_n\right)[/tex]

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

[tex]\sup(A+B)\leq \sup(A)+\sup(B)[/tex]

Let [itex]\{a_n+b_n\}_{n\in\mathbb{N}}[/itex] 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, [itex]a_n\leq\sup(A)[/itex] and [itex]b_n\leq\sup(B)[/itex] 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.
 
  • #6
You are assuming way too much and complicating things way too much. And I also think you'Re confusing the problems at hand.

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

What is a limsup in real analysis?

A limsup, or limit superior, is a concept in real analysis that represents the largest limit point of a sequence. It is denoted as lim sup(a_n) and can be calculated as the supremum of the set of all accumulation points of the sequence.

How is a limsup different from a limit?

A limit in real analysis represents the exact value that a sequence approaches as n tends to infinity. On the other hand, a limsup represents the largest limit point of a sequence, which may not necessarily be the exact limit.

What is the significance of limsup in real analysis?

The limsup is useful in determining the behavior of a sequence as n tends to infinity. It helps in identifying important properties of the sequence, such as convergence, oscillation, and boundedness.

Can a sequence have multiple limsups?

Yes, a sequence can have multiple limsups. If a sequence has more than one limit point, then each limit point can be considered as a limsup of the sequence.

How is the limsup related to the liminf in real analysis?

The limsup and liminf are two important concepts in real analysis that represent the largest and smallest limit points of a sequence, respectively. They are related in that the limsup is the upper bound of the liminf, and the difference between them represents the oscillation of the sequence.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top