B Continuation of sequence of supremums question

  • B
  • Thread starter Thread starter Eclair_de_XII
  • Start date Start date
  • Tags Tags
    Sequence
Eclair_de_XII
Messages
1,082
Reaction score
91
TL;DR Summary
Let ##\{a_n\}\subset \mathbb{R}##. Let ##A_k\equiv \sup\{a_n:n\geq k\}## and suppose that ##A_k\rightarrow \lambda##. Prove:
(a) There is a sequence ##B_n## with the properties:
(i) ##B_n<A_n## for all ##n##
(ii) ##B_n## is an increasing sequence
(iii) ##B_n\rightarrow \lambda##
(b) There are infinitely many ##k## such that ##a_n\in [B_n,A_n]##
(c) There exists a subsequence ##\{a_{n_k}\}## that converges to ##\lambda##
(d) The limit of a subsequence ##\{a_{n_k}\}## is at most ##\lambda##
First, it must be justified that ##A_n## is decreasing and is bounded from below by the point it converges to. See the other topic

(1) ##A_n## is decreasing
By definition, ##A_n## is the supremum of the set containing all ##a_k## where ##k\geq n+1## and the set containing ##a_n##. Hence, it must be an upper bound for ##A_{n+1}##. By definition, ##A_n\geq A_{n+1}##.

(2) ##A_n\geq \lambda##
If not, then choose ##k## such that ##A_k<\lambda-\epsilon## for some ##\epsilon>0##. Since ##A_n## is decreasing, it follows that for ##n\geq k##, no matter how large the interval of convergence is, ##A_n## must lie outside of it. This contradicts the fact that it converges. Hence, ##A_n\geq \lambda##.

Note: This is just a rephrasing of Mr. Office Shredder's post from that topic I mentioned.

===(a)===
Define ##B=2\lambda-A_n##.
==(i)
This follows from the fact that ##A_n## is bounded from below by ##\lambda##
##A_n\geq \lambda##
##-A_n\leq -\lambda##
##2\lambda-A_n\leq 2\lambda-\lambda##
##2\lambda-A_n\leq \lambda##
##B_n\leq \lambda\leq A_n##

==(ii)
This follows from the fact that ##A_n## is decreasing.
##A_{n+1}\leq A_n##
##-A_{n+1}\geq -A_n##
##2\lambda-A_{n+1}\geq 2\lambda-A_n##
##B_{n+1}\geq B_n##

==(iii)
Finally, if ##A_n\rightarrow \lambda##, then for any ##\epsilon>0##, there is some integer ##N>0## such that if ##n\geq N##:

##|A_n-\lambda|<\epsilon##
##-\epsilon<\lambda-A_n<\epsilon##
##\lambda-\epsilon<2\lambda-A_n<\lambda+\epsilon##
##\lambda-\epsilon<B_n<\lambda+\epsilon##
##-\epsilon<B_n-\lambda<\epsilon##
##|B_n-\lambda|<\epsilon##

===(b)===
Let ##\epsilon>0## and let ##A_n## be fixed. By definition, ##A_n\geq a_k## whenever ##k\geq n##. Hence, there are infinitely many points of this form below ##A_n##. If we can show that ##a_k\geq B_n##, for ##k\geq n##, we will have our result.

Suppose that it is not. Then ##B_n## must be an upper bound for ##\{a_k:k\geq n\}##. Since ##B_n<A_n##, this contradicts the fact that ##A_n## is the least upper bound. Hence, ##a_k\geq B## for ##k\geq n##.

===(c)===
Let ##\epsilon>0##. Then there is ##N\in \mathbb{N}## such that if ##n\geq N##:

##-\epsilon+\lambda<A_n<\epsilon+\lambda##

Choose ##\epsilon_1\in (0,\epsilon)##. Then there is ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. Now choose an ##\epsilon_2>0## smaller than ##\epsilon_1## such that ##a_{n_1}\notin (A_n-\epsilon_2,A_n]##. For example, choose ##\epsilon_2\in (0,A_n-a_{n_1})##. Then there must be some ##n_2\geq n## such that ##a_{n_2}\in (A_n-\epsilon_2,A_n]##.

With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##
##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##
##-\epsilon<a_{n_k}-\lambda<\epsilon##
##|a_{n_k}-\lambda|<\epsilon##

===(d)===
Let some subsequence ##a_{n_k}## converge to ##\gamma\in\mathbb{R}##. Suppose that ##\gamma>\lambda##. Choose ##m\in\mathbb{N}## such that ##\gamma>A_m>\lambda##. Set ##\epsilon=\gamma-A_m##. Let ##N\in\mathbb{N}##. Then whenever ##n\geq N##, specifically, when ##n\geq m##:

\begin{align*}
|\gamma-a_n|&=&\gamma-a_n\\
&\geq&\gamma-A_m\\
&=&\epsilon
\end{align*}
 
Last edited:
Physics news on Phys.org
Your proof for b is wrong. Consider the sequence ##a_k=0## if k is odd, and when ##k=2m## is even, ##a_{2m}=1+1/m##. Then ##A_{2m}=A_{2m-1}=1+1/m##, ##\lambda=1##, ##B_{2m}=B_{2m-1}=1-1/m##, but it is certainly not true that ##a_k\geq B_n## for ##k\geq n##, since when ##k## is odd ##a_k## is zero. Just because ##B_n## is larger than some points in a set doesn't mean it's larger than all the points in a set.

I thought your choice for ##B_n## is fairly clever and I agree with all your proofs for (a).
 
What about if I used the fact that ##A_n## is a supremum for the subsequence ##\{a_k:k\geq n\}##?

For any positive ##\epsilon_1##, there must exist some ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. In particular, choose ##\epsilon=A_n-B_n##. Now choose ##\epsilon_2\in(0,\epsilon_1)## such that ##(A_n-\epsilon_2,A_n]## excludes the ##a_{n_1}##. Now there is some ##n_2\geq n## such that ##a_{n_2}\in(A_n-\epsilon_2,A_n]##. Repeat this process in order construct a subsequence contained in ##[B_n,A_n] ## that converges to ##A_n##.
 
Yes, that looks like what you want.

One thing to keep in mind for this, while your choice for ##B_n## is clever, it's also unnecessary. Any increasing sequence that converges to ##\lambda## will also work here.For part (c) this sentence stood out.

Eclair_de_XII said:
With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##
##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##
##-\epsilon<a_{n_k}-\lambda<\epsilon##
##|a_{n_k}-\lambda|<\epsilon##

The sentence you wrote obviously disproves what you are trying to do. It sounds like your sequence ##a_{n_k}## converges to ##A_n##, but that means it doesn't converge to ##\lambda## unless they happen to coincide.

You need to actually look at a sequence of ##A_k##s and for each one pick a ##a_{n_k}## that is really close to ##A_k## in order to get convergence to ##\lambda##.
 
Office_Shredder said:
You need to actually look at a sequence of ##A_k## s and for each one pick a ##a_{n_k}## that is really close to ##A_k## in order to get convergence to ##\lambda##.

I figure I could choose some ##A_k## at random, set ##\epsilon=A_k-\lambda##, then choose an ##n_1>k## such that ##a_{n_1}\in (A_k-\epsilon,A_k]##. From then on, I'd pick some ##k_1>k## such that ##a_{n_1}>A_{k_1}##. Set ##\epsilon=A_{k_1}-\lambda##. Then choose ##a_{n_2}\in (A_{k_1}-\epsilon,A_{k_1}]##. Now find a ##k_2>k_1## such that ##A_{k_2}<a_{n_2}##. And so on.

Basically, I'm just choosing some ##A_k## at random, choosing some ##a_{n_k}## between ##\lambda## and ##A_k##, then choosing an element of the sequence of supremums smaller than the chosen ##a_{n_k}##. In this way, I obtain a subsequence of ##\{A_n\}## and a subsequence ##\{a_{n_k}\}## that decreases alongside ##A_n##. It sounds like it would converge by the Sandwich Theorem but this book didn't cover that yet, so it wouldn't be valid to use it, I should think.
 
Yeah I think you can just say it like this, without relying on any theorems.

For every ##k##, pick ##A_{n_k}## such that ##|\lambda - A_{n_k}| < 1/(2k)##, and then pick ##a_{n_k}## such that ##|A_{n_k}-a_{n_k}| < 1/(2k)##. Then by the triangle inequality ##|\lambda - a_{n_k}| < 1/k##. So ##a_{n_k}## converges to ##\lambda##.

Your proof for d doesn't look complete, I'm not sure what the conclusion is there. You also just drop the || from your first inequality but you don't actually know what the sign of ##\gamma-a_n## is, so probably you should think about it a bit more.
 
Office_Shredder said:
but you don't actually know what the sign of ##\gamma-a_n## is, so probably you should think about it a bit more.

I disagree. By definition, ##A_m\geq a_n## for all ##n\geq m##. By hypothesis, ##\gamma>A_m\geq a_n##. My conclusion is that if there is a subsequence ##a_{n_k}## that converges to some number, then that number cannot exceed ##\lambda##.
 
Last edited:
Oh you're right, the proof works. You shouldn't even mention ##N##, just delete all references to it.
 
Right-o, right-o. In my defense, I was trying to recite the negation of the epsilon-delta definition of the convergence of sequences of real numbers. As always, thank you for your valuable input, besides.
 
Last edited:
  • #10
If you really want to hammer it home I would write something like

Let ##N=m,##. Then for all ##n\geq N ##, we have ##|\gamma - a_n| \geq \epsilon##
 
  • Like
Likes Eclair_de_XII
Back
Top