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Set of partial limits

0

$begingroup$

For the sequences $$x_n = sin(n) $$ $$x_n = cos(n),; x_n=cos(n^2) $$ the set of partial limits will be the segment $[-1,1]$

What will be the necessary and sufficient condition that the set of partial limits of a sequence $x_n$ is a segment $[l,L]$? What conditions should these sequences meet?

real-analysis

share|cite|improve this question

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the set of partial limits of a sequence $x_n$ be a segment $[l,L]$ we must have ${x_n : ninmathbb{N}}$ dense in $[l,L]$.
  $endgroup$
  – Rodrigo Dias
  Mar 22 at 21:09
  

  
  
  
  

add a comment | 

0

$begingroup$

For the sequences $$x_n = sin(n) $$ $$x_n = cos(n),; x_n=cos(n^2) $$ the set of partial limits will be the segment $[-1,1]$

What will be the necessary and sufficient condition that the set of partial limits of a sequence $x_n$ is a segment $[l,L]$? What conditions should these sequences meet?

real-analysis

share|cite|improve this question

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the set of partial limits of a sequence $x_n$ be a segment $[l,L]$ we must have ${x_n : ninmathbb{N}}$ dense in $[l,L]$.
  $endgroup$
  – Rodrigo Dias
  Mar 22 at 21:09
  

  
  
  
  

add a comment | 

$begingroup$

For the sequences $$x_n = sin(n) $$ $$x_n = cos(n),; x_n=cos(n^2) $$ the set of partial limits will be the segment $[-1,1]$

What will be the necessary and sufficient condition that the set of partial limits of a sequence $x_n$ is a segment $[l,L]$? What conditions should these sequences meet?

real-analysis

share|cite|improve this question

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

$endgroup$

share|cite|improve this question

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

share|cite|improve this question

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

edited Mar 22 at 21:34

Bernard

124k741117

edited Mar 22 at 21:34

Bernard

124k741117

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

asked Mar 22 at 21:03

MorrisonFJMorrisonFJ

1

1 Answer
1

active

oldest

votes

0

$begingroup$

It would be easier to first think of a necessary and sufficient condition for a specific point $x$ to be a partial limit, and then expand the condition to fit an entire interval. For a point $x$ to be a partial limit of a sequence, we want to have that in any neighborhood of $x$, there is some member of ${x_n}$. Meaning - $x$ is in the closure of ${x_n}$. We can now genaralize this - if we want the set of partial limits of ${x_n}$ to be exactly $[l,L]$, we want the previous condition to be satisfied to any point in the given interval, and for no other point. This means that $[l,L]$ is the closure of ${x_n}$.

share|cite|improve this answer

answered Mar 22 at 21:58

GSoferGSofer

8051315

$endgroup$

add a comment | 

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0

$begingroup$

It would be easier to first think of a necessary and sufficient condition for a specific point $x$ to be a partial limit, and then expand the condition to fit an entire interval. For a point $x$ to be a partial limit of a sequence, we want to have that in any neighborhood of $x$, there is some member of ${x_n}$. Meaning - $x$ is in the closure of ${x_n}$. We can now genaralize this - if we want the set of partial limits of ${x_n}$ to be exactly $[l,L]$, we want the previous condition to be satisfied to any point in the given interval, and for no other point. This means that $[l,L]$ is the closure of ${x_n}$.

share|cite|improve this answer

answered Mar 22 at 21:58

GSoferGSofer

8051315

$endgroup$

add a comment | 

0

$begingroup$

It would be easier to first think of a necessary and sufficient condition for a specific point $x$ to be a partial limit, and then expand the condition to fit an entire interval. For a point $x$ to be a partial limit of a sequence, we want to have that in any neighborhood of $x$, there is some member of ${x_n}$. Meaning - $x$ is in the closure of ${x_n}$. We can now genaralize this - if we want the set of partial limits of ${x_n}$ to be exactly $[l,L]$, we want the previous condition to be satisfied to any point in the given interval, and for no other point. This means that $[l,L]$ is the closure of ${x_n}$.

share|cite|improve this answer

answered Mar 22 at 21:58

GSoferGSofer

8051315

$endgroup$

add a comment | 

$begingroup$

It would be easier to first think of a necessary and sufficient condition for a specific point $x$ to be a partial limit, and then expand the condition to fit an entire interval. For a point $x$ to be a partial limit of a sequence, we want to have that in any neighborhood of $x$, there is some member of ${x_n}$. Meaning - $x$ is in the closure of ${x_n}$. We can now genaralize this - if we want the set of partial limits of ${x_n}$ to be exactly $[l,L]$, we want the previous condition to be satisfied to any point in the given interval, and for no other point. This means that $[l,L]$ is the closure of ${x_n}$.

share|cite|improve this answer

answered Mar 22 at 21:58

GSoferGSofer

8051315

$endgroup$

share|cite|improve this answer

answered Mar 22 at 21:58

GSoferGSofer

8051315

answered Mar 22 at 21:58

GSoferGSofer

8051315

answered Mar 22 at 21:58

GSoferGSofer

8051315