Question about set containing subsequential limits of bounded sequence

[Eclair_de_XII]

TL;DR

Let $\{x_n\}_{n\in\mathbb{N}}$ be a bounded sequence and let $E$ be the set of subsequential limits of that sequence. Assume that $E$ is non-empty. Prove that $E$ is bounded and contains both its supremum and its infimum.

Let $L\in E$. By definition, there is a subsequence $\{x_{n_k}\}_{k\in\mathbb{N}}$ that converges to $L$. There is a natural number $N$ s.t. if $n_k\geq N$, $L\in(x_{n_k}-1,x_{n_k}+1)\subset(\inf\{x_n\}-1,\sup\{x_n\}+1)$. Hence, $E$ is a bounded set.

If $E$ is a finite set, then it has no accumulation points, which means that it must contain both its infimum and supremum.

Now assume otherwise.

Denote $x=\sup E$, and denote the subsequence of $\{x_n\}$ converging to $x$ as $\{x_{m_k}\}_{k\in\mathbb{N}}$.

Let $\epsilon>0$. There is $N\in\mathbb{N}$ s.t. for integers $n> N$, $x-L_n<\epsilon$.

If the set $\{x_n\}\cap(x-L_N,x)$ is finite, then each point $L_n$ in the set could not be an accumulation point. For each $n>N$, simply choose $\delta=L_n-L_N$; any ball of radius $\delta$ centered at $L_n$ would contain only finitely many points $x_n$. It follows that each point $L_n$ must be converged onto point-wise. However, as there are only finitely many points $x_n\in(x-L_N,x)$, and infinitely many subsequential limits $L_n\in(x-L_N,x)$, this is impossible.

Therefore, there infinitely many points of $\{x_n\}$ in the interval $(x-L_N,x)$. Consequently, there must exist a subsequence of $\{x_n\}$ that converges to $x$. By definition, $x\in E$.

 

[fresh_42]

Summary:: Let $\{x_n\}_{n\in\mathbb{N}}$ be a bounded sequence and let $E$ be the set of subsequential limits of that sequence. Assume that $E$ is non-empty. Prove that $E$ is bounded and contains both its supremum and its infimum.

Let $L\in E$. By definition, there is a subsequence $\{x_{n_k}\}_{k\in\mathbb{N}}$ that converges to $L$. There is a natural number $N$ s.t. if $n_k\geq N$, $L\in(x_{n_k}-1,x_{n_k}+1)\subset(\inf\{x_n\}-1,\sup\{x_n\}+1)$. Hence, $E$ is a bounded set.

If $E$ is a finite set, then it has no accumulation points, which means that it must contain both its infimum and supremum.

Now assume otherwise.

Denote $x=\sup E$, and denote the subsequence of $\{x_n\}$ converging to $x$ as $\{x_{m_k}\}_{k\in\mathbb{N}}$.

Why does $\{x_{m_k}\}_{k\in\mathbb{N}}$ exist? You use $x\in E$ to show that $x=\sup E \in E$!
You have to a) find a sequence $(y_k)_{k\in \mathbb{N}}$ in $E$ that converges to $x,$ because all you know is that $x$ is the lowest upper bound for $E$. Where do you get a sequence from? Then b) consider all subsequences of $(x_n)_{n\in \mathbb{N}}$ which converge to the elements $y_k$.

Let $\epsilon>0$. There is $N\in\mathbb{N}$ s.t. for integers $n> N$, $x-L_n<\epsilon$.

If the set $\{x_n\}\cap(x-L_N,x)$ is finite, then each point $L_n$ in the set could not be an accumulation point. For each $n>N$, simply choose $\delta=L_n-L_N$; any ball of radius $\delta$ centered at $L_n$ would contain only finitely many points $x_n$. It follows that each point $L_n$ must be converged onto point-wise. However, as there are only finitely many points $x_n\in(x-L_N,x)$, and infinitely many subsequential limits $L_n\in(x-L_N,x)$, this is impossible.

Therefore, there infinitely many points of $\{x_n\}$ in the interval $(x-L_N,x)$. Consequently, there must exist a subsequence of $\{x_n\}$ that converges to $x$. By definition, $x\in E$.

 

[Eclair_de_XII]

Where do you get a sequence from?

The fact that $x$ is an accumulation point of $E$. I reasoned that there must exist infinitely many subsequential limits within a given neighborhood of $x$, which I erroneously denoted as $x_n$ instead of $L_n$; also, this sequence $\{L_n\}$ would converge to $x$, which I used in my explanation of why there must exist infinitely many points of $\{x_n\}$ in that open interval $(x-L_N,x)$.

 

[fresh_42]

The fact that $x$ is an accumulation point of $E$.

Why? If $E=\{1/n\, : \,n\in \mathbb{N}\}$ then $\sup E =1$ but it is no accumulation point.

I reasoned that there must exist infinitely many subsequential limits within a given neighborhood of $x$, which I erroneously denoted as $x_n$ instead of $L_n$; also, this sequence $\{L_n\}$ would converge to $x$, which I used in my explanation of why there must exist infinitely many points of $\{x_n\}$ in that open interval $(x-L_N,x)$.

 

[Eclair_de_XII]

Why?

Let's say we have a bounded set $E$. Its supremum $x$ must either belong to the set or be an accumulation point thereof.

If it is not the former, then let $\epsilon>0$. There is a point $e_1\in E$ s.t. $e_1\in(x-\epsilon,x)$. Otherwise, $x-\epsilon$ is an upper bound of $E$, contradictory to the definition of $x$. For similar reasons, there is a point $e_2\in E$ s.t. $e_2\in (e_1,x)$. We construct a sequence $\{e_n\}_{n\in\mathbb{N}}$ that converges to $x$ this way. Hence, $x$ must be an accumulation point of $E$.

Now suppose it is not the latter. Then there is a number $\epsilon>0$ s.t. $|(x-\epsilon,x]\cap E|<\infty$. If $x\notin E$, then we can find an open ball around $x$ that contains no elements of $E$. But this would mean that the left end-point is an upper-bound for $E$, which is not possible. Hence, $x\in E$.

In the second line, I considered the case where $x$ is not an accumulation point of $E$. Now, I'm considering the other case.

 

[fresh_42]

Let's say we have a bounded set $E$. Its supremum $x$ must either belong to the set or be an accumulation point thereof.

If it is not the former, then let $\epsilon>0$. There is a point $e_1\in E$ s.t. $e_1\in(x-\epsilon,x)$. Otherwise, $x-\epsilon$ is an upper bound of $E$, contradictory to the definition of $x$. For similar reasons, there is a point $e_2\in E$ s.t. $e_2\in (e_1,x)$. We construct a sequence $\{e_n\}_{n\in\mathbb{N}}$ that converges to $x$ this way. Hence, $x$ must be an accumulation point of $E$.

Now suppose it is not the latter. Then there is a number $\epsilon>0$ s.t. $|(x-\epsilon,x]\cap E|<\infty$. If $x\notin E$, then we can find an open ball around $x$ that contains no elements of $E$. But this would mean that the left end-point is an upper-bound for $E$, which is not possible. Hence, $x\in E$.

In the second line, I considered the case where $x$ is not an accumulation point of $E$. Now, I'm considering the other case.

This sounds better. Always take one step after the other. It doesn't matter whether $E$ is finite or not.

You have already shown that $E$ is bounded (by choosing $\varepsilon =1$ in the convergence definition for the subsequence that converges to $L$.)

Let $L\in E$. By definition, there is a subsequence $\{x_{n_k}\}_{k\in\mathbb{N}}$ that converges to $L$. There is a natural number $N$ s.t. if $n_k\geq N$, $L\in(x_{n_k}-1,x_{n_k}+1)\subset(\inf\{x_n\}-1,\sup\{x_n\}+1)$. Hence, $E$ is a bounded set.

I had to scribble the line
$$
\inf -1 \leq x_n -1 \leq \sup -1<\sup +1 \wedge \inf -1<\inf +1\leq x_n+1\leq \sup +1 \Longrightarrow \inf -1 \leq x_n\pm 1\leq \sup +1
$$
so that would have helped me, but o.k.

Then you defined $x:=\sup E$. So if $x\in E$ we are done. Otherwise $x\in \bar E$ and we find a sequence $(y_n)_{n\in \mathbb{N}}\subseteq E$ with $\lim_{n \to \infty}y_n=x.$

I am not certain what $L_n$ and $L$ are here in the second part, so let me take them as $L_m=y_m$ and $L=x:=\sup E.$

We have $y_m\in E$ for all $m$, i.e. by definition: there are subsequences $(x_{n_{k(m)}})_{k(m)\in \mathbb{N}} \subseteq (x_n)_{n\in \mathbb{N}}$ such that $\lim_{k \to \infty} x_{n_{k(m)}} = y_m$

Now how can we use all these subsequences $(x_{n_{k(1)}}),(x_{n_{k(2)}}),\ldots,(x_{n_{k(m)}}),\ldots$ that converge to $y_1,y_2,\ldots,y_m,\ldots$ to see that the limit of the $y_m$ is in $E$?

I think your idea (from post $1$) was the right one, but I have difficulties reading it. You use only one series $L_n$ which are not really defined, whereas we have infinitely many of them.

I think that the assumption $x\not\in E$ can be used to find a contradiction. $x\not\in E$ means that there is an $\varepsilon_0 >0$ such that there are only finitely many other points of $E$ in $I_{\varepsilon_0} :=(x-\varepsilon_0 ,x+\varepsilon_0).$

Now we have to use the $y_m$ to show that infinitely many of them are in this interval $I_{\varepsilon_0}.$ This can indeed be done with your idea: choose $y_m$ close enough ($m$ large enough) such that all of them are in $I_{\varepsilon_0}$ (don't just say "the set", define it).

Remark:
1) I cannot count all the times I first typed "it", "a", "this" or "that" and replaced it again by what I really meant! E.g. my previous paragraph first was

... find a contradiction. It means that there is an $\varepsilon_0 >0$ such that there ...

before I corrected it to

... find a contradiction. $x\not\in E$ means that there is an $\varepsilon_0 >0$ such that there ...

2) If you use a specific constant, as in our case here $\varepsilon_0$, or in the first part of your proof with $\varepsilon =1$, then it is useful to give it an index because everybody is so used to "any $\varepsilon$, small and positive, but any" that it is helpful to note (by the index) that we have chosen a specific one.

Summary: looks good, but you could improve on readability.

 

[Eclair_de_XII]

It doesn't matter whether $E$ is finite or not.

I see. I need only consider the case when the supremum of $E$ is not an accumulation point, which may be the case even if the cardinality of $E$ is infinite.

I had to scribble the line

I merely used the inequalities:

\begin{eqnarray}
\inf\{x_n\}\leq x_n\\
\sup\{x_n\}\geq x_n\\
x_{n_k}-1<L<x_{n_k}+1\\
-1+\inf\{x_n\}\leq x_{n_k}-1<L<x_{n_k}+1\leq \sup\{x_n\}+1
\end{eqnarray}

I am not certain what $L_n$ and $L$ are here in the second part, so let me take them as $L_m=y_m$ and $L=x:=\sup E.$

$L_n$ is an element of the sequence of subsequential limits that must converge to $x$ since the latter is an accumulation point of $E$. $L$ is just used in the first part, to show that $E$ is bounded. It has nothing to do with the proof that $x\in E$.

I think your idea (from post $1$) was the right one, but I have difficulties reading it. You use only one series $L_n$ which are not really defined, whereas we have infinitely many of them.

I think that the assumption $x\not\in E$ can be used to find a contradiction. $x\not\in E$ means that there is an $\varepsilon_0 >0$ such that there are only finitely many other points of $E$ in $I_{\varepsilon_0} :=(x-\varepsilon_0 ,x+\varepsilon_0).$

Now we have to use the $y_m$ to show that infinitely many of them are in this interval $I_{\varepsilon_0}.$ This can indeed be done with your idea: choose $y_m$ close enough ($m$ large enough) such that all of them are in $I_{\varepsilon_0}$ (don't just say "the set", define it).

Remark:
1) I cannot count all the times I first typed "it", "a", "this" or "that" and replaced it again by what I really meant! E.g. my previous paragraph first was before I corrected it to
2) If you use a specific constant, as in our case here $\varepsilon_0$, or in the first part of your proof with $\varepsilon =1$, then it is useful to give it an index because everybody is so used to "any $\varepsilon$, small and positive, but any" that it is helpful to note (by the index) that we have chosen a specific one.

Summary: looks good, but you could improve on readability.

Thanks for the feedback, besides.