- #1
danielakkerma
- 231
- 0
(Hey guys and gals!)
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.
Definition of limsup(x_n) = L:
[tex] \forall \epsilon > 0 \mid \exists N \mid \forall n > N \mid x_n < L +\epsilon [/tex]
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
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