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Prove that subsequence converges to limsup
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Given a sequence of real numbers, $ x_n _n=1^infty$, let $alpha =$ limsup$x_n$ and $beta = $ liminf$x_n$.
Prove that there exists a subsequence $ x_n_k$ that converges to $alpha$ as $k rightarrow infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
real-analysis limits limsup-and-liminf
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
asked Nov 25 '13 at 22:15
user12279
9519
 |Â
show 2 more comments
up vote
3
down vote
favorite
Given a sequence of real numbers, $ x_n _n=1^infty$, let $alpha =$ limsup$x_n$ and $beta = $ liminf$x_n$.
Prove that there exists a subsequence $ x_n_k$ that converges to $alpha$ as $k rightarrow infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
real-analysis limits limsup-and-liminf
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
asked Nov 25 '13 at 22:15
user12279
9519
Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
1
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
1
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a sequence of real numbers, $ x_n _n=1^infty$, let $alpha =$ limsup$x_n$ and $beta = $ liminf$x_n$.
Prove that there exists a subsequence $ x_n_k$ that converges to $alpha$ as $k rightarrow infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
real-analysis limits limsup-and-liminf
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
asked Nov 25 '13 at 22:15
user12279
9519
Given a sequence of real numbers, $ x_n _n=1^infty$, let $alpha =$ limsup$x_n$ and $beta = $ liminf$x_n$.
Prove that there exists a subsequence $ x_n_k$ that converges to $alpha$ as $k rightarrow infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
real-analysis limits limsup-and-liminf
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
asked Nov 25 '13 at 22:15
user12279
9519
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
edited Aug 20 '14 at 9:08
Martin Sleziak
43.5k6113259
43.5k6113259
asked Nov 25 '13 at 22:15
user12279
9519
asked Nov 25 '13 at 22:15
user12279
9519
asked Nov 25 '13 at 22:15
user12279
9519
9519
Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
1
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
1
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31
 |Â
show 2 more comments
Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
1
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
1
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31
Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
1
1
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
1
1
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Since $alpha=limsup x_n$, by the definition of $limsup$, there is some $x_n_1$ with $|x_n_1-alpha|<1over 2$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_n_2$ with $|x_n_2-alpha|<1over 2^2$. Continuing, for each $kinmathbbN$, there is some $x_n_k$ with $|x_n_k-alpha|<1over 2^k$.
Then $x_n_ksubset x_n$ and $x_n_kto alpha$ as $ktoinfty$.
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
add a comment |Â
up vote
1
down vote
I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it.
Let us define the terms first.
$limsup s_n :=lim_N rightarrow infty sups_n:n>N$
Let $A_N := s_n:n>N$, $v_N := sup A_N$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that
$A_i subseteq A_j$ when $i>j$.
$therefore sup A_i leq sup A_j Rightarrow v_N$ is a non-increasing(monotonous) sequence.
Thus, $lim sup s_n$ is the infimum of the set of $v_N_N=1^infty$.
$inf v_N = limsup s_n$ ----------------------(d1)
$therefore exists v_N$ s.t. $v_N < lim sup s_n + epsilon forall epsilon > 0$ -------------------- (1)
Using the definition of
$v_N$,
$exists s_n$ s.t. $s_n > v_N - alpha forall alpha > 0$ ---------------(2)
Using (d1) and (1),
$lim sup s_n leq v_N < lim sup s_n + epsilon forall epsilon>0 $ -------------(a)
Using (2) and previous definition of $v_N$,
$v_N geq s_n > v_N - alpha forall alpha > 0$ ---------------- (b)
Using (a) and (b),
$ limsup s_n - alpha leq v_N - alpha < s_n leq v_N < lim sup s_n + epsilon$ where $alpha,epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $lim_Nrightarrow inftyv_N$, i.e. $lim sup s_n$.
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since $alpha=limsup x_n$, by the definition of $limsup$, there is some $x_n_1$ with $|x_n_1-alpha|<1over 2$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_n_2$ with $|x_n_2-alpha|<1over 2^2$. Continuing, for each $kinmathbbN$, there is some $x_n_k$ with $|x_n_k-alpha|<1over 2^k$.
Then $x_n_ksubset x_n$ and $x_n_kto alpha$ as $ktoinfty$.
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
add a comment |Â
up vote
6
down vote
accepted
Since $alpha=limsup x_n$, by the definition of $limsup$, there is some $x_n_1$ with $|x_n_1-alpha|<1over 2$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_n_2$ with $|x_n_2-alpha|<1over 2^2$. Continuing, for each $kinmathbbN$, there is some $x_n_k$ with $|x_n_k-alpha|<1over 2^k$.
Then $x_n_ksubset x_n$ and $x_n_kto alpha$ as $ktoinfty$.
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since $alpha=limsup x_n$, by the definition of $limsup$, there is some $x_n_1$ with $|x_n_1-alpha|<1over 2$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_n_2$ with $|x_n_2-alpha|<1over 2^2$. Continuing, for each $kinmathbbN$, there is some $x_n_k$ with $|x_n_k-alpha|<1over 2^k$.
Then $x_n_ksubset x_n$ and $x_n_kto alpha$ as $ktoinfty$.
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
Since $alpha=limsup x_n$, by the definition of $limsup$, there is some $x_n_1$ with $|x_n_1-alpha|<1over 2$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_n_2$ with $|x_n_2-alpha|<1over 2^2$. Continuing, for each $kinmathbbN$, there is some $x_n_k$ with $|x_n_k-alpha|<1over 2^k$.
Then $x_n_ksubset x_n$ and $x_n_kto alpha$ as $ktoinfty$.
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
answered Nov 25 '13 at 22:49
JohnD
11.7k32055
11.7k32055
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
add a comment |Â
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
2
2
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
How can you be sure that each time you reduce the size of the interval around $limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen?
– hallaplay835
Jan 27 '14 at 15:42
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
Because $alpha$ is by definition the $limsup$ of the sequence, and thus is a limit point of the sequence.
– JohnD
Jan 27 '14 at 16:02
2
2
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
I don't understand what you're saying. $alpha := limsup x_n = lim_N to infty supx_n mid n geq N $ by definition. The fact that for each $N in mathbb N$, $supx_n mid n geq N $ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence.
– hallaplay835
Jan 27 '14 at 16:23
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
$limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of $x_n$: for any $varepsilon>0$, there are infinitely $x_n$ within $varepsilon$ of $alpha$. See this.
– JohnD
Jan 27 '14 at 17:05
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = sups_i:i>N$. You could from this that $|v_n_1 - alpha| < epsilon$ and go on to choose some $epsilon$. But you have rather said that this $v_n_1$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $sup$ of a set exists in it, which may or may not be true. Also, isn't your proof essentially selecting a subsequence $s_n_k$ from the set $A_N = s_i: i > N$ so that: $lim sup A_N geq s_n_k geq lim inf A_N$
– SPRajagopal
Aug 19 '14 at 13:50
add a comment |Â
up vote
1
down vote
I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it.
Let us define the terms first.
$limsup s_n :=lim_N rightarrow infty sups_n:n>N$
Let $A_N := s_n:n>N$, $v_N := sup A_N$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that
$A_i subseteq A_j$ when $i>j$.
$therefore sup A_i leq sup A_j Rightarrow v_N$ is a non-increasing(monotonous) sequence.
Thus, $lim sup s_n$ is the infimum of the set of $v_N_N=1^infty$.
$inf v_N = limsup s_n$ ----------------------(d1)
$therefore exists v_N$ s.t. $v_N < lim sup s_n + epsilon forall epsilon > 0$ -------------------- (1)
Using the definition of
$v_N$,
$exists s_n$ s.t. $s_n > v_N - alpha forall alpha > 0$ ---------------(2)
Using (d1) and (1),
$lim sup s_n leq v_N < lim sup s_n + epsilon forall epsilon>0 $ -------------(a)
Using (2) and previous definition of $v_N$,
$v_N geq s_n > v_N - alpha forall alpha > 0$ ---------------- (b)
Using (a) and (b),
$ limsup s_n - alpha leq v_N - alpha < s_n leq v_N < lim sup s_n + epsilon$ where $alpha,epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $lim_Nrightarrow inftyv_N$, i.e. $lim sup s_n$.
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
add a comment |Â
up vote
1
down vote
I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it.
Let us define the terms first.
$limsup s_n :=lim_N rightarrow infty sups_n:n>N$
Let $A_N := s_n:n>N$, $v_N := sup A_N$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that
$A_i subseteq A_j$ when $i>j$.
$therefore sup A_i leq sup A_j Rightarrow v_N$ is a non-increasing(monotonous) sequence.
Thus, $lim sup s_n$ is the infimum of the set of $v_N_N=1^infty$.
$inf v_N = limsup s_n$ ----------------------(d1)
$therefore exists v_N$ s.t. $v_N < lim sup s_n + epsilon forall epsilon > 0$ -------------------- (1)
Using the definition of
$v_N$,
$exists s_n$ s.t. $s_n > v_N - alpha forall alpha > 0$ ---------------(2)
Using (d1) and (1),
$lim sup s_n leq v_N < lim sup s_n + epsilon forall epsilon>0 $ -------------(a)
Using (2) and previous definition of $v_N$,
$v_N geq s_n > v_N - alpha forall alpha > 0$ ---------------- (b)
Using (a) and (b),
$ limsup s_n - alpha leq v_N - alpha < s_n leq v_N < lim sup s_n + epsilon$ where $alpha,epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $lim_Nrightarrow inftyv_N$, i.e. $lim sup s_n$.
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it.
Let us define the terms first.
$limsup s_n :=lim_N rightarrow infty sups_n:n>N$
Let $A_N := s_n:n>N$, $v_N := sup A_N$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that
$A_i subseteq A_j$ when $i>j$.
$therefore sup A_i leq sup A_j Rightarrow v_N$ is a non-increasing(monotonous) sequence.
Thus, $lim sup s_n$ is the infimum of the set of $v_N_N=1^infty$.
$inf v_N = limsup s_n$ ----------------------(d1)
$therefore exists v_N$ s.t. $v_N < lim sup s_n + epsilon forall epsilon > 0$ -------------------- (1)
Using the definition of
$v_N$,
$exists s_n$ s.t. $s_n > v_N - alpha forall alpha > 0$ ---------------(2)
Using (d1) and (1),
$lim sup s_n leq v_N < lim sup s_n + epsilon forall epsilon>0 $ -------------(a)
Using (2) and previous definition of $v_N$,
$v_N geq s_n > v_N - alpha forall alpha > 0$ ---------------- (b)
Using (a) and (b),
$ limsup s_n - alpha leq v_N - alpha < s_n leq v_N < lim sup s_n + epsilon$ where $alpha,epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $lim_Nrightarrow inftyv_N$, i.e. $lim sup s_n$.
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it.
Let us define the terms first.
$limsup s_n :=lim_N rightarrow infty sups_n:n>N$
Let $A_N := s_n:n>N$, $v_N := sup A_N$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that
$A_i subseteq A_j$ when $i>j$.
$therefore sup A_i leq sup A_j Rightarrow v_N$ is a non-increasing(monotonous) sequence.
Thus, $lim sup s_n$ is the infimum of the set of $v_N_N=1^infty$.
$inf v_N = limsup s_n$ ----------------------(d1)
$therefore exists v_N$ s.t. $v_N < lim sup s_n + epsilon forall epsilon > 0$ -------------------- (1)
Using the definition of
$v_N$,
$exists s_n$ s.t. $s_n > v_N - alpha forall alpha > 0$ ---------------(2)
Using (d1) and (1),
$lim sup s_n leq v_N < lim sup s_n + epsilon forall epsilon>0 $ -------------(a)
Using (2) and previous definition of $v_N$,
$v_N geq s_n > v_N - alpha forall alpha > 0$ ---------------- (b)
Using (a) and (b),
$ limsup s_n - alpha leq v_N - alpha < s_n leq v_N < lim sup s_n + epsilon$ where $alpha,epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $lim_Nrightarrow inftyv_N$, i.e. $lim sup s_n$.
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
answered Aug 20 '14 at 7:58
SPRajagopal
2641314
2641314
add a comment |Â
add a comment |Â
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Think about what it means to say $limsup x_n=alpha$, i.e. the definition of the $limsup$ of a sequence.
– JohnD
Nov 25 '13 at 22:18
@JohnD Since $alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $alpha$ and similarly, all values of $x_n$ are strictly larger than $beta$ since $beta$ is an infimum?
– user12279
Nov 25 '13 at 22:25
No, that's not correct. $alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence.
– JohnD
Nov 25 '13 at 22:37
1
Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions.
– JohnD
Nov 26 '13 at 17:10
1
No problem. Glad to help. It should be helpful to think of $limsup$ as the "largest limit/cluster point" and $liminf$ as "smallest limit/cluster point".
– JohnD
Nov 27 '13 at 14:31