Question about set containing subsequential limits of bounded sequence

  • #1
Eclair_de_XII
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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##.
 

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  • #2
fresh_42
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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##.
 
  • #3
Eclair_de_XII
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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)##.
 
  • #4
fresh_42
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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)##.
 
  • #5
Eclair_de_XII
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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.
 
  • #6
fresh_42
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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.
 
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  • #7
Eclair_de_XII
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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.
 

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