Existence of a subsequence converging to limsup












1












$begingroup$


Let $(a_n)$ be a bounded sequence of real numbers, and define $$beta_n = sup { a_k : k geq n }. $$ This sequence converges to a limit,
$$lim_{n to infty} beta_n = limsup a_n.$$
I'm interested in proving the existence of a subsequence of $(a_n)$ that converges to $limsup a_n.$ I've read over the other posts concerning this and gave a satisfactory proof similar to these arguments, but I'm wondering if the following argument could be used as well. Note that in my study of analysis I've already proven the Bolzano-Weirstass theorem in $mathbb{R}$. Here goes.



If $beta_n$ is a bounded real sequence, Bolzano-Weierstrass implies that it has a convergent subsequence $(beta_{n_j})$. Since $lim beta_n$ exists and is equal to $limsup a_n,$ then $(beta_{n_j})$ converges to $limsup a_n.$



I'm not certain if this argument works because I want a subsequence of $(a_n).$ Is there a way to argue that $(beta_{n_j})$ is a subsequence of $(a_n)$?






real-analysis sequences-and-series limsup-and-liminf







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
    $endgroup$
    – Surb
    Dec 16 '18 at 21:47












  • $begingroup$
    math.stackexchange.com/questions/581128/…
    $endgroup$
    – parsiad
    Dec 16 '18 at 22:09
















1












$begingroup$


Let $(a_n)$ be a bounded sequence of real numbers, and define $$beta_n = sup { a_k : k geq n }. $$ This sequence converges to a limit,
$$lim_{n to infty} beta_n = limsup a_n.$$
I'm interested in proving the existence of a subsequence of $(a_n)$ that converges to $limsup a_n.$ I've read over the other posts concerning this and gave a satisfactory proof similar to these arguments, but I'm wondering if the following argument could be used as well. Note that in my study of analysis I've already proven the Bolzano-Weirstass theorem in $mathbb{R}$. Here goes.



If $beta_n$ is a bounded real sequence, Bolzano-Weierstrass implies that it has a convergent subsequence $(beta_{n_j})$. Since $lim beta_n$ exists and is equal to $limsup a_n,$ then $(beta_{n_j})$ converges to $limsup a_n.$



I'm not certain if this argument works because I want a subsequence of $(a_n).$ Is there a way to argue that $(beta_{n_j})$ is a subsequence of $(a_n)$?






real-analysis sequences-and-series limsup-and-liminf







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
    $endgroup$
    – Surb
    Dec 16 '18 at 21:47












  • $begingroup$
    math.stackexchange.com/questions/581128/…
    $endgroup$
    – parsiad
    Dec 16 '18 at 22:09














1












1








1


1



$begingroup$


Let $(a_n)$ be a bounded sequence of real numbers, and define $$beta_n = sup { a_k : k geq n }. $$ This sequence converges to a limit,
$$lim_{n to infty} beta_n = limsup a_n.$$
I'm interested in proving the existence of a subsequence of $(a_n)$ that converges to $limsup a_n.$ I've read over the other posts concerning this and gave a satisfactory proof similar to these arguments, but I'm wondering if the following argument could be used as well. Note that in my study of analysis I've already proven the Bolzano-Weirstass theorem in $mathbb{R}$. Here goes.



If $beta_n$ is a bounded real sequence, Bolzano-Weierstrass implies that it has a convergent subsequence $(beta_{n_j})$. Since $lim beta_n$ exists and is equal to $limsup a_n,$ then $(beta_{n_j})$ converges to $limsup a_n.$



I'm not certain if this argument works because I want a subsequence of $(a_n).$ Is there a way to argue that $(beta_{n_j})$ is a subsequence of $(a_n)$?






real-analysis sequences-and-series limsup-and-liminf







share|cite|improve this question









$endgroup$




Let $(a_n)$ be a bounded sequence of real numbers, and define $$beta_n = sup { a_k : k geq n }. $$ This sequence converges to a limit,
$$lim_{n to infty} beta_n = limsup a_n.$$
I'm interested in proving the existence of a subsequence of $(a_n)$ that converges to $limsup a_n.$ I've read over the other posts concerning this and gave a satisfactory proof similar to these arguments, but I'm wondering if the following argument could be used as well. Note that in my study of analysis I've already proven the Bolzano-Weirstass theorem in $mathbb{R}$. Here goes.



If $beta_n$ is a bounded real sequence, Bolzano-Weierstrass implies that it has a convergent subsequence $(beta_{n_j})$. Since $lim beta_n$ exists and is equal to $limsup a_n,$ then $(beta_{n_j})$ converges to $limsup a_n.$



I'm not certain if this argument works because I want a subsequence of $(a_n).$ Is there a way to argue that $(beta_{n_j})$ is a subsequence of $(a_n)$?






real-analysis sequences-and-series limsup-and-liminf




real-analysis sequences-and-series limsup-and-liminf






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 21:42









dovedove

83




83








  • 3




    $begingroup$
    No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
    $endgroup$
    – Surb
    Dec 16 '18 at 21:47












  • $begingroup$
    math.stackexchange.com/questions/581128/…
    $endgroup$
    – parsiad
    Dec 16 '18 at 22:09














  • 3




    $begingroup$
    No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
    $endgroup$
    – Surb
    Dec 16 '18 at 21:47












  • $begingroup$
    math.stackexchange.com/questions/581128/…
    $endgroup$
    – parsiad
    Dec 16 '18 at 22:09








3




3




$begingroup$
No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
$endgroup$
– Surb
Dec 16 '18 at 21:47






$begingroup$
No because $(beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(beta _n)$ since it already converge ?
$endgroup$
– Surb
Dec 16 '18 at 21:47














$begingroup$
math.stackexchange.com/questions/581128/…
$endgroup$
– parsiad
Dec 16 '18 at 22:09




$begingroup$
math.stackexchange.com/questions/581128/…
$endgroup$
– parsiad
Dec 16 '18 at 22:09










2 Answers
2






active

oldest

votes


















0












$begingroup$

No, it doesn't work because $beta_n$ needs not be an element of $(a_k)$.

We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>beta_n-frac1n$.

Then we get $Lleftarrow beta_n-frac1n<a_{k_n}lebeta_n to L$, where $L=limbeta_n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great, thanks, I thought so! Appreciate the speedy response.
    $endgroup$
    – dove
    Dec 17 '18 at 0:23



















0












$begingroup$

In context:



$b_n= sup (a_k| k ge n)$



$A:= lim_{n rightarrow infty} b_n=lim sup a_n.$



Show that $A$ is a limit point of $(a_n)_{n in mathbb{N}}$.



1)Let $epsilon >0$ be given.



There is a $N$ such that for $n ge N$



$|b_n -A| lt epsilon/2.$



2) By definition of $b_n = sup (a_k|k ge n)$ there is a



$a_{n_k}$ such that for $k ge k_0$



$|a_{n_k}-b_n| lt epsilon/2.$



For $n ge N$, and $k ge k_0$



3) $|a_{n_k}-A| le$



$ |a_{n_k}-b_n| +|b_n-A| lt epsilon$, i.e.



$A$ is a limit point of $(a_n)_{n in mathbb{N}}$, and $a_{n_k}$ converges to.$A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
    $endgroup$
    – dove
    Dec 17 '18 at 18:26












  • $begingroup$
    dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
    $endgroup$
    – Peter Szilas
    Dec 17 '18 at 19:22











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2 Answers
2






active

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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

No, it doesn't work because $beta_n$ needs not be an element of $(a_k)$.

We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>beta_n-frac1n$.

Then we get $Lleftarrow beta_n-frac1n<a_{k_n}lebeta_n to L$, where $L=limbeta_n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great, thanks, I thought so! Appreciate the speedy response.
    $endgroup$
    – dove
    Dec 17 '18 at 0:23
















0












$begingroup$

No, it doesn't work because $beta_n$ needs not be an element of $(a_k)$.

We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>beta_n-frac1n$.

Then we get $Lleftarrow beta_n-frac1n<a_{k_n}lebeta_n to L$, where $L=limbeta_n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great, thanks, I thought so! Appreciate the speedy response.
    $endgroup$
    – dove
    Dec 17 '18 at 0:23














0












0








0





$begingroup$

No, it doesn't work because $beta_n$ needs not be an element of $(a_k)$.

We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>beta_n-frac1n$.

Then we get $Lleftarrow beta_n-frac1n<a_{k_n}lebeta_n to L$, where $L=limbeta_n$.






share|cite|improve this answer









$endgroup$



No, it doesn't work because $beta_n$ needs not be an element of $(a_k)$.

We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>beta_n-frac1n$.

Then we get $Lleftarrow beta_n-frac1n<a_{k_n}lebeta_n to L$, where $L=limbeta_n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 22:01









BerciBerci

61.4k23674




61.4k23674












  • $begingroup$
    Great, thanks, I thought so! Appreciate the speedy response.
    $endgroup$
    – dove
    Dec 17 '18 at 0:23


















  • $begingroup$
    Great, thanks, I thought so! Appreciate the speedy response.
    $endgroup$
    – dove
    Dec 17 '18 at 0:23
















$begingroup$
Great, thanks, I thought so! Appreciate the speedy response.
$endgroup$
– dove
Dec 17 '18 at 0:23




$begingroup$
Great, thanks, I thought so! Appreciate the speedy response.
$endgroup$
– dove
Dec 17 '18 at 0:23











0












$begingroup$

In context:



$b_n= sup (a_k| k ge n)$



$A:= lim_{n rightarrow infty} b_n=lim sup a_n.$



Show that $A$ is a limit point of $(a_n)_{n in mathbb{N}}$.



1)Let $epsilon >0$ be given.



There is a $N$ such that for $n ge N$



$|b_n -A| lt epsilon/2.$



2) By definition of $b_n = sup (a_k|k ge n)$ there is a



$a_{n_k}$ such that for $k ge k_0$



$|a_{n_k}-b_n| lt epsilon/2.$



For $n ge N$, and $k ge k_0$



3) $|a_{n_k}-A| le$



$ |a_{n_k}-b_n| +|b_n-A| lt epsilon$, i.e.



$A$ is a limit point of $(a_n)_{n in mathbb{N}}$, and $a_{n_k}$ converges to.$A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
    $endgroup$
    – dove
    Dec 17 '18 at 18:26












  • $begingroup$
    dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
    $endgroup$
    – Peter Szilas
    Dec 17 '18 at 19:22
















0












$begingroup$

In context:



$b_n= sup (a_k| k ge n)$



$A:= lim_{n rightarrow infty} b_n=lim sup a_n.$



Show that $A$ is a limit point of $(a_n)_{n in mathbb{N}}$.



1)Let $epsilon >0$ be given.



There is a $N$ such that for $n ge N$



$|b_n -A| lt epsilon/2.$



2) By definition of $b_n = sup (a_k|k ge n)$ there is a



$a_{n_k}$ such that for $k ge k_0$



$|a_{n_k}-b_n| lt epsilon/2.$



For $n ge N$, and $k ge k_0$



3) $|a_{n_k}-A| le$



$ |a_{n_k}-b_n| +|b_n-A| lt epsilon$, i.e.



$A$ is a limit point of $(a_n)_{n in mathbb{N}}$, and $a_{n_k}$ converges to.$A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
    $endgroup$
    – dove
    Dec 17 '18 at 18:26












  • $begingroup$
    dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
    $endgroup$
    – Peter Szilas
    Dec 17 '18 at 19:22














0












0








0





$begingroup$

In context:



$b_n= sup (a_k| k ge n)$



$A:= lim_{n rightarrow infty} b_n=lim sup a_n.$



Show that $A$ is a limit point of $(a_n)_{n in mathbb{N}}$.



1)Let $epsilon >0$ be given.



There is a $N$ such that for $n ge N$



$|b_n -A| lt epsilon/2.$



2) By definition of $b_n = sup (a_k|k ge n)$ there is a



$a_{n_k}$ such that for $k ge k_0$



$|a_{n_k}-b_n| lt epsilon/2.$



For $n ge N$, and $k ge k_0$



3) $|a_{n_k}-A| le$



$ |a_{n_k}-b_n| +|b_n-A| lt epsilon$, i.e.



$A$ is a limit point of $(a_n)_{n in mathbb{N}}$, and $a_{n_k}$ converges to.$A$.






share|cite|improve this answer









$endgroup$



In context:



$b_n= sup (a_k| k ge n)$



$A:= lim_{n rightarrow infty} b_n=lim sup a_n.$



Show that $A$ is a limit point of $(a_n)_{n in mathbb{N}}$.



1)Let $epsilon >0$ be given.



There is a $N$ such that for $n ge N$



$|b_n -A| lt epsilon/2.$



2) By definition of $b_n = sup (a_k|k ge n)$ there is a



$a_{n_k}$ such that for $k ge k_0$



$|a_{n_k}-b_n| lt epsilon/2.$



For $n ge N$, and $k ge k_0$



3) $|a_{n_k}-A| le$



$ |a_{n_k}-b_n| +|b_n-A| lt epsilon$, i.e.



$A$ is a limit point of $(a_n)_{n in mathbb{N}}$, and $a_{n_k}$ converges to.$A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 8:26









Peter SzilasPeter Szilas

11.5k2822




11.5k2822












  • $begingroup$
    Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
    $endgroup$
    – dove
    Dec 17 '18 at 18:26












  • $begingroup$
    dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
    $endgroup$
    – Peter Szilas
    Dec 17 '18 at 19:22


















  • $begingroup$
    Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
    $endgroup$
    – dove
    Dec 17 '18 at 18:26












  • $begingroup$
    dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
    $endgroup$
    – Peter Szilas
    Dec 17 '18 at 19:22
















$begingroup$
Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
$endgroup$
– dove
Dec 17 '18 at 18:26






$begingroup$
Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$?
$endgroup$
– dove
Dec 17 '18 at 18:26














$begingroup$
dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
$endgroup$
– Peter Szilas
Dec 17 '18 at 19:22




$begingroup$
dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings.
$endgroup$
– Peter Szilas
Dec 17 '18 at 19:22


















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Cáceres

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Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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