Continuation of sequence of supremums question

[Eclair_de_XII]

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*}

 

[Office_Shredder]

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).

 

[Eclair_de_XII]

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 $A_n$.

 

[Office_Shredder]

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.

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 $\lambda$.

 

[Eclair_de_XII]

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.

 

[Office_Shredder]

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.

 

[Eclair_de_XII]

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 $\lambda$.

 

[Office_Shredder]

Oh you're right, the proof works. You shouldn't even mention $N$, just delete all references to it.

 

[Eclair_de_XII]

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.

 

[Office_Shredder]

If you really want to hammer it home I would write something like

Let $N=m,$. Then for all $n\geq N$, we have $|\gamma - a_n| \geq \epsilon$