Bounded sets, Limits superior and convergence

[danielakkerma]

(Hey guys and gals!)

Homework Statement

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

Homework Equations

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

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:
<br /> \forall \epsilon &gt; 0 \mid \exists N_1 \mid \forall n &gt; N_1 \mid x_n+y_n &lt; (a+b) + \epsilon \\<br /> \mbox{Taking the same Epsilon:}\\<br /> \exists N_2 \mid \forall n &gt; N_2 \mid y_n &lt; a + \epsilon \\<br /> n&gt;N \mid N = \max(N_1, N_2)<br />
Now I arrive at the following congruent inequalities(to be subtracted by their transitive property):
<br /> x_n+y_n &lt; (a+b) + \epsilon \\<br /> y_n &lt; a + \epsilon \Rightarrow \\<br /> x_n &lt; b<br />
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:
<br /> x_n&lt;b<br />
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

 

[mfb]

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.

 

[jbunniii]

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:
<br /> \forall \epsilon &gt; 0 \mid \exists N_1 \mid \forall n &gt; N_1 \mid x_n+y_n &lt; (a+b) + \epsilon \\<br /> \mbox{Taking the same Epsilon:}\\<br /> \exists N_2 \mid \forall n &gt; N_2 \mid y_n &lt; a + \epsilon \\<br /> n&gt;N \mid N = \max(N_1, N_2)<br />
Now I arrive at the following congruent inequalities(to be subtracted by their transitive property):
<br /> x_n+y_n &lt; (a+b) + \epsilon \\<br /> y_n &lt; a + \epsilon \Rightarrow \\<br /> x_n &lt; b<br />
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:
<br /> x_n&lt;b<br />

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 $n$.

 

[danielakkerma]

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

 

[jbunniii]

(Hey guys and gals!)

Homework Statement

Given a bounded set x_n and for any y_n the following condition holds:
\limsup_{n \rightarrow ∞}(x_n+y_n) = \limsup(x_n)+\limsup(y_n)
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?

 

[danielakkerma]

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