# Question about set containing subsequential limits of bounded sequence

• B
• Eclair_de_XII
In summary: What's the reason for that?The reason for this is to show that the elements of the subsequence are within a specific range that is bounded by the infimum and supremum of the original sequence. This is necessary in order to prove that the subsequence converges to the limit point.
Eclair_de_XII
TL;DR 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}}##.

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 said:
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##.

Eclair_de_XII said:
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 said:
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)##.

Eclair_de_XII said:
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.
Eclair_de_XII said:
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 said:
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.

Eclair_de_XII said:
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##.)
Eclair_de_XII said:
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
fresh_42 said:
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.

fresh_42 said:
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}

fresh_42 said:
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##.

fresh_42 said:
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.

## 1. What is a subsequential limit of a bounded sequence?

A subsequential limit of a bounded sequence is a value that the sequence approaches as the index of the sequence increases. It may or may not be a part of the sequence itself.

## 2. How do you determine the set containing subsequential limits of a bounded sequence?

To determine the set containing subsequential limits of a bounded sequence, you can take the limit of the sequence as the index approaches infinity. This will give you the set of all possible values that the sequence may approach as it continues.

## 3. Can a bounded sequence have more than one subsequential limit?

Yes, a bounded sequence can have more than one subsequential limit. This occurs when the sequence oscillates between two or more values as the index increases.

## 4. Is the set containing subsequential limits of a bounded sequence always bounded?

No, the set containing subsequential limits of a bounded sequence is not always bounded. It can be unbounded if the sequence has no limit or if it has infinitely many subsequential limits.

## 5. How is the concept of subsequential limits related to the concept of convergence?

Subsequential limits are closely related to the concept of convergence. If a sequence has a single subsequential limit, it is said to converge to that value. If a sequence has more than one subsequential limit, it does not converge. Additionally, a sequence that does not have a subsequential limit does not converge either.

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