$liminf E_{n_0+n} = liminflimits E_{n}$ and $limsup E_{n_0+n} = limsup E_{n}$ where $E_n$ is a decreasing...
I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.
In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)
My attempt at a proof is as follows:
Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$
Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$
But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.
I however couldn't prove the reverse inclusion in my attempt below:
Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$
Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$
I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.
I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.
Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...
Original Text:
The main part of my question comes from here:
Theorem 1.26:
And finally, Lemma 1.7:
Hopefully someone can shed some light!
real-analysis measure-theory elementary-set-theory limsup-and-liminf
asked Jan 4 at 0:18
Darius
878
add a comment |
I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.
In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)
My attempt at a proof is as follows:
Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$
Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$
But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.
I however couldn't prove the reverse inclusion in my attempt below:
Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$
Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$
I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.
I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.
Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...
Original Text:
The main part of my question comes from here:
Theorem 1.26:
And finally, Lemma 1.7:
Hopefully someone can shed some light!
real-analysis measure-theory elementary-set-theory limsup-and-liminf
asked Jan 4 at 0:18
Darius
878
2
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago
add a comment |
I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.
In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)
My attempt at a proof is as follows:
Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$
Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$
But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.
I however couldn't prove the reverse inclusion in my attempt below:
Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$
Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$
I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.
I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.
Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...
Original Text:
The main part of my question comes from here:
Theorem 1.26:
And finally, Lemma 1.7:
Hopefully someone can shed some light!
real-analysis measure-theory elementary-set-theory limsup-and-liminf
asked Jan 4 at 0:18
Darius
878
I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.
In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)
My attempt at a proof is as follows:
Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$
Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$
But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.
I however couldn't prove the reverse inclusion in my attempt below:
Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$
Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$
I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.
I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.
Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...
Original Text:
The main part of my question comes from here:
Theorem 1.26:
And finally, Lemma 1.7:
Hopefully someone can shed some light!
real-analysis measure-theory elementary-set-theory limsup-and-liminf
real-analysis measure-theory elementary-set-theory limsup-and-liminf
asked Jan 4 at 0:18
Darius
878
asked Jan 4 at 0:18
Darius
878
edited Jan 4 at 0:38
asked Jan 4 at 0:18
Darius
878
asked Jan 4 at 0:18
Darius
878
asked Jan 4 at 0:18
Darius
878
878
2
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago
add a comment |
2
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago
2
2
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.
Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.
answered Jan 4 at 5:54
Henno Brandsma
105k347114
add a comment |
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Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.
Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.
answered Jan 4 at 5:54
Henno Brandsma
105k347114
add a comment |
Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.
Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.
answered Jan 4 at 5:54
Henno Brandsma
105k347114
add a comment |
Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.
Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.
answered Jan 4 at 5:54
Henno Brandsma
105k347114
Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.
Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.
answered Jan 4 at 5:54
Henno Brandsma
105k347114
edited 2 days ago
answered Jan 4 at 5:54
Henno Brandsma
105k347114
answered Jan 4 at 5:54
Henno Brandsma
105k347114
answered Jan 4 at 5:54
Henno Brandsma
105k347114
105k347114
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2
If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01
Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56
$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20
@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago