Bounded sets, Limits superior and convergence

  • #1
(Hey guys and gals!)

Homework Statement


Given a bounded set x_n and for any y_n the following condition holds:
[tex] \limsup_{n \rightarrow ∞}(x_n+y_n) = \limsup(x_n)+\limsup(y_n) [/tex]
Show that x_n converges.

Homework Equations



Definition of limsup(x_n) = L:
[tex] \forall \epsilon > 0 \mid \exists N \mid \forall n > N \mid x_n < L +\epsilon [/tex]

The Attempt at a Solution


I started by defining the limsups as follows:
Let limsup(y_n) = a; limsup(x_n) = b;
According to the given conditions limsup(x_n+y_n) = a + b(see above).
Therefore:
[tex]
\forall \epsilon > 0 \mid \exists N_1 \mid \forall n > N_1 \mid x_n+y_n < (a+b) + \epsilon \\
\mbox{Taking the same Epsilon:}\\
\exists N_2 \mid \forall n > N_2 \mid y_n < a + \epsilon \\
n>N \mid N = \max(N_1, N_2)
[/tex]
Now I arrive at the following congruent inequalities(to be subtracted by their transitive property):
[tex]
x_n+y_n < (a+b) + \epsilon \\
y_n < a + \epsilon \Rightarrow \\
x_n < b
[/tex]
Now, I claim that since I am given that x_n is bounded, by arriving at that final inequality, I've effectively discovered its upper bound .
Therefore:
[tex]
x_n<b
[/tex]
But, since b is the limit superior, it is also the smallest possible, real upper bound.
And yet, here I'm stuck.
On the one hand, the above relation means that -b is the largest possible lower bound , or the limit inferior.
But that means that limsup(x_n)≠liminf(x_n) which would imply that x_n does not converge.
Where should I turn next?
Is there perhaps a better way to look at this?
Very thankful for your attention,
Daniel
 

Answers and Replies

  • #2
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I would consider a specific y_n. With the right choice, it is nearly trivial.

Your definition for limsup looks strange - every L' > L would satisfy it as well. And (-1)^n would not have the correct 1 as limsup.
 
  • #3
jbunniii
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I started by defining the limsups as follows:
Let limsup(y_n) = a; limsup(x_n) = b;
According to the given conditions limsup(x_n+y_n) = a + b(see above).
Therefore:
[tex]
\forall \epsilon > 0 \mid \exists N_1 \mid \forall n > N_1 \mid x_n+y_n < (a+b) + \epsilon \\
\mbox{Taking the same Epsilon:}\\
\exists N_2 \mid \forall n > N_2 \mid y_n < a + \epsilon \\
n>N \mid N = \max(N_1, N_2)
[/tex]
Now I arrive at the following congruent inequalities(to be subtracted by their transitive property):
[tex]
x_n+y_n < (a+b) + \epsilon \\
y_n < a + \epsilon \Rightarrow \\
x_n < b
[/tex]
Now, I claim that since I am given that x_n is bounded, by arriving at that final inequality, I've effectively discovered its upper bound .
Therefore:
[tex]
x_n<b
[/tex]
No, this isn't right at all. It's possible to have a sequence ##(x_n)## with a lim sup ##b## such that ##x_n > b## for all ##n##. For example, if ##x_n = 1/n##, then ##\lim \sup x_n = 0##, but ##x_n > 0## for all ##n##.
 
  • #4
Firstly thanks for your replies.
If you can follow my reasoning(which I know can be sometimes tortuous), I deduced that x_n < b only after taking both x_n+y_n<(a+b)+ε & the fact that y_n < a + eps.
jbunn, I can clearly see your example, you're absolutely right; the problem is that I obtained this inequality solely through the definition of the limsup and the given statement re x_n being bound.
I am clearly at loss, now, as to how to proceed. Any pointer in this regard would be extremely helpful.
**
Mfb, that sounds promising. Are you sure I'm allowed to take a discrete y_n(i.e. not arbitrary)?
Obviously it occurred to me(please don't take me as quite so daft ;)) that if y_n = -x_n, then the given relation reduces to:
limsup(x_n+y_n) = limsup(x_n+(-x_n)) = 0 = limsup(x_n)-liminf(x_n) -->limsup(x_n)=liminf(x_n) <-> x_n converges.
But what if y_n isn't such? The proof should touch, I believe, on the general case.
In any event, I am rightly stumped now. Do you have any suggestions?
I would greatly appreciate any help,
And I am grateful for your time, and useful advice,
Daniel
 
  • #5
jbunniii
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(Hey guys and gals!)

Homework Statement


Given a bounded set x_n and for any y_n the following condition holds:
[tex] \limsup_{n \rightarrow ∞}(x_n+y_n) = \limsup(x_n)+\limsup(y_n) [/tex]
Show that x_n converges.
Are you sure I'm allowed to take a discrete y_n(i.e. not arbitrary)?
Obviously it occurred to me(please don't take me as quite so daft ;)) that if y_n = -x_n, then the given relation reduces to:
limsup(x_n+y_n) = limsup(x_n+(-x_n)) = 0 = limsup(x_n)-liminf(x_n) -->limsup(x_n)=liminf(x_n) <-> x_n converges.
But what if y_n isn't such? The proof should touch, I believe, on the general case.
In any event, I am rightly stumped now. Do you have any suggestions?
I would greatly appreciate any help,
And I am grateful for your time, and useful advice,
Daniel
Well, it does say that the condition holds for any (mathematician-speak for every) ##y_n##, so I'm sure mfb is correct: you can choose any specific ##y_n## you like to get the job done.

Do you see where your proof implicitly uses the fact that ##(x_n)## is bounded?
 
  • #6
Jbun, thanks again.
I used the condition to state that after I arrived at x_n < b, for n > N, it is the limsup that forms the eventual, smallest possible, upper bound, in other words, the
limsup(x_n) = inf(sup(x_N, x_(N+1), ..., x_n)).
Therefore, I had thought(as you say, mistakenly) that since x_n < b, for those n > N, x_n can construed bound by b(in that region, at least).
If that's the case, and since I can eliminate a finite number of outlying elements of x_n(preceeding N), all of x_n, after that N, lies squarely(or equals either) between -b, and b.
Then I considered that by writing it thus(subtracting b, taking the absolute value of both sides):
|x_n-b| => 0, I can firmly state that for any(!) ε>0, this: |x_n-b|<ε holds, since the former conditions is true iff 0 >= x_n-b >= 0 ->|x_n-b|=0, and |x_n-b| < ε is the implicit definition of a limit of a sequence.
The "N" in this case, would be the matching "N" to the above relation...
Is this getting me anywhere?
As you say, the better way is to simply take y_n=-x_n. I tried keep it as generalised as possible, without discretisation.
Of course, if you consider it valid, then it's surely quite trivial.
Still, what do you think?
Is there a better method for the general case?(if at all?)
Thanks again!
Much beholden,
Daniel
 

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