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If all proper subsequences converge to same limit then the sequence converges.

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5

3

$begingroup$

Let ${X_n}_n$ be a sequence bounded sequence. Its convergent proper subsequences converge to the same limit $ell$.
I want to prove that ${X_n}_n$ converges to $ell$.

Notice that proper subsequences are all the sequences except for the sequence itself.
Is it enough to say that ${X_{2n}}$ and ${X_{2n+1}}$ are convergent to $l$ then ${X_n}$ is convergent to $ell$?

calculus real-analysis sequences-and-series analysis convergence

share|cite|improve this question

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Actually one does not assume that $(x_{2n})$ and $(x_{2n+1})$ converge to $ell$ since the hypothesis concerns only the convergent subsequences, not every subsequence.
  $endgroup$
  – Did
  Oct 1 '17 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/397978/…
  $endgroup$
  – Guy Fsone
  Nov 11 '17 at 16:32
  

  
  
  
  

add a comment | 

5

3

$begingroup$

Let ${X_n}_n$ be a sequence bounded sequence. Its convergent proper subsequences converge to the same limit $ell$.
I want to prove that ${X_n}_n$ converges to $ell$.

Notice that proper subsequences are all the sequences except for the sequence itself.
Is it enough to say that ${X_{2n}}$ and ${X_{2n+1}}$ are convergent to $l$ then ${X_n}$ is convergent to $ell$?

calculus real-analysis sequences-and-series analysis convergence

share|cite|improve this question

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Actually one does not assume that $(x_{2n})$ and $(x_{2n+1})$ converge to $ell$ since the hypothesis concerns only the convergent subsequences, not every subsequence.
  $endgroup$
  – Did
  Oct 1 '17 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/397978/…
  $endgroup$
  – Guy Fsone
  Nov 11 '17 at 16:32
  

  
  
  
  

add a comment | 

$begingroup$

Let ${X_n}_n$ be a sequence bounded sequence. Its convergent proper subsequences converge to the same limit $ell$.
I want to prove that ${X_n}_n$ converges to $ell$.

Notice that proper subsequences are all the sequences except for the sequence itself.
Is it enough to say that ${X_{2n}}$ and ${X_{2n+1}}$ are convergent to $l$ then ${X_n}$ is convergent to $ell$?

calculus real-analysis sequences-and-series analysis convergence

share|cite|improve this question

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

$endgroup$

share|cite|improve this question

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

share|cite|improve this question

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

edited Nov 11 '17 at 16:33

Guy Fsone

17.2k43074

asked Sep 30 '17 at 12:52

KarenKaren

927

asked Sep 30 '17 at 12:52

KarenKaren

927

1 Answer
1

active

oldest

votes

4

$begingroup$

Proof by contradiction

Suppose that ${X_n}$ does not converge to $ell$. Then, there is $varepsilon_0>0$ such that $$forall Ninmathbb N,exists n=n(N) : n>N~~~and ~~~ |X_n -ell|>varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -ell|>varepsilon_0 $$
Taking successively $N_{k+1}> max{N_k, n_k,k+1}$ there exists $n_{k+1}>N_{k+1}$ such that,

$$ |X_{ n_{k+1}} -ell|>varepsilon_0 $$

It is easy to see that, ${X_{ n_k}}_k$ is a subsequence of ${X_{ n}}_n$
since
$$ n_k< n_{k+1} quad i.e ~~text{the map }~~kmapsto n_k~~~text{Is one-to-one}$$

However, $$forall k,~~ |X_{ n_{k}} -ell|>varepsilon_0 qquad text{and}~~~{X_{ n_{k}} }~~~text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists ${X_{ n_{k_p} }}_p$ subsequence of ${X_{ n_{k} }}_k$ which converges to some limit $ell_1 $
but ${X_{ n_{k_p} }}_pto ell_1$ is also a congering subsequence of ${X_n}_n$

By assumption, $ell=ell_1$ that is together with the fact ${X_{ n_{k_p} }}_p$ is a subsequence of ${X_{ n_{k} }}_k$ we have

$$0=lim_{ptoinfty } |X_{ n_{k_p} }-ell|>varepsilon_0>0~~~text{which is a CONTRADICTION}$$

Note that
$$forall p,~~|X_{ n_{k_p} }-ell|>varepsilon_0$$
Since
$$forall k,~~|X_{ n_{k}} -ell|>varepsilon_0$$

share|cite|improve this answer

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why can one assume that ℓ=ℓ1? Doesn't that lead to loss of generality?
  $endgroup$
  – Ahmad Lamaa
  Sep 23 '18 at 20:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I did not assume. read the question again. $ell$ is unique with a specific property which is fulfilled by $ell_1$
  $endgroup$
  – Guy Fsone
  Sep 26 '18 at 22:58
  

  
  
  
  

add a comment | 

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4

$begingroup$

Proof by contradiction

Suppose that ${X_n}$ does not converge to $ell$. Then, there is $varepsilon_0>0$ such that $$forall Ninmathbb N,exists n=n(N) : n>N~~~and ~~~ |X_n -ell|>varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -ell|>varepsilon_0 $$
Taking successively $N_{k+1}> max{N_k, n_k,k+1}$ there exists $n_{k+1}>N_{k+1}$ such that,

$$ |X_{ n_{k+1}} -ell|>varepsilon_0 $$

It is easy to see that, ${X_{ n_k}}_k$ is a subsequence of ${X_{ n}}_n$
since
$$ n_k< n_{k+1} quad i.e ~~text{the map }~~kmapsto n_k~~~text{Is one-to-one}$$

However, $$forall k,~~ |X_{ n_{k}} -ell|>varepsilon_0 qquad text{and}~~~{X_{ n_{k}} }~~~text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists ${X_{ n_{k_p} }}_p$ subsequence of ${X_{ n_{k} }}_k$ which converges to some limit $ell_1 $
but ${X_{ n_{k_p} }}_pto ell_1$ is also a congering subsequence of ${X_n}_n$

By assumption, $ell=ell_1$ that is together with the fact ${X_{ n_{k_p} }}_p$ is a subsequence of ${X_{ n_{k} }}_k$ we have

$$0=lim_{ptoinfty } |X_{ n_{k_p} }-ell|>varepsilon_0>0~~~text{which is a CONTRADICTION}$$

Note that
$$forall p,~~|X_{ n_{k_p} }-ell|>varepsilon_0$$
Since
$$forall k,~~|X_{ n_{k}} -ell|>varepsilon_0$$

share|cite|improve this answer

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why can one assume that ℓ=ℓ1? Doesn't that lead to loss of generality?
  $endgroup$
  – Ahmad Lamaa
  Sep 23 '18 at 20:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I did not assume. read the question again. $ell$ is unique with a specific property which is fulfilled by $ell_1$
  $endgroup$
  – Guy Fsone
  Sep 26 '18 at 22:58
  

  
  
  
  

add a comment | 

4

$begingroup$

Proof by contradiction

Suppose that ${X_n}$ does not converge to $ell$. Then, there is $varepsilon_0>0$ such that $$forall Ninmathbb N,exists n=n(N) : n>N~~~and ~~~ |X_n -ell|>varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -ell|>varepsilon_0 $$
Taking successively $N_{k+1}> max{N_k, n_k,k+1}$ there exists $n_{k+1}>N_{k+1}$ such that,

$$ |X_{ n_{k+1}} -ell|>varepsilon_0 $$

It is easy to see that, ${X_{ n_k}}_k$ is a subsequence of ${X_{ n}}_n$
since
$$ n_k< n_{k+1} quad i.e ~~text{the map }~~kmapsto n_k~~~text{Is one-to-one}$$

However, $$forall k,~~ |X_{ n_{k}} -ell|>varepsilon_0 qquad text{and}~~~{X_{ n_{k}} }~~~text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists ${X_{ n_{k_p} }}_p$ subsequence of ${X_{ n_{k} }}_k$ which converges to some limit $ell_1 $
but ${X_{ n_{k_p} }}_pto ell_1$ is also a congering subsequence of ${X_n}_n$

By assumption, $ell=ell_1$ that is together with the fact ${X_{ n_{k_p} }}_p$ is a subsequence of ${X_{ n_{k} }}_k$ we have

$$0=lim_{ptoinfty } |X_{ n_{k_p} }-ell|>varepsilon_0>0~~~text{which is a CONTRADICTION}$$

Note that
$$forall p,~~|X_{ n_{k_p} }-ell|>varepsilon_0$$
Since
$$forall k,~~|X_{ n_{k}} -ell|>varepsilon_0$$

share|cite|improve this answer

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why can one assume that ℓ=ℓ1? Doesn't that lead to loss of generality?
  $endgroup$
  – Ahmad Lamaa
  Sep 23 '18 at 20:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I did not assume. read the question again. $ell$ is unique with a specific property which is fulfilled by $ell_1$
  $endgroup$
  – Guy Fsone
  Sep 26 '18 at 22:58
  

  
  
  
  

add a comment | 

$begingroup$

Proof by contradiction

Suppose that ${X_n}$ does not converge to $ell$. Then, there is $varepsilon_0>0$ such that $$forall Ninmathbb N,exists n=n(N) : n>N~~~and ~~~ |X_n -ell|>varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -ell|>varepsilon_0 $$
Taking successively $N_{k+1}> max{N_k, n_k,k+1}$ there exists $n_{k+1}>N_{k+1}$ such that,

$$ |X_{ n_{k+1}} -ell|>varepsilon_0 $$

It is easy to see that, ${X_{ n_k}}_k$ is a subsequence of ${X_{ n}}_n$
since
$$ n_k< n_{k+1} quad i.e ~~text{the map }~~kmapsto n_k~~~text{Is one-to-one}$$

However, $$forall k,~~ |X_{ n_{k}} -ell|>varepsilon_0 qquad text{and}~~~{X_{ n_{k}} }~~~text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists ${X_{ n_{k_p} }}_p$ subsequence of ${X_{ n_{k} }}_k$ which converges to some limit $ell_1 $
but ${X_{ n_{k_p} }}_pto ell_1$ is also a congering subsequence of ${X_n}_n$

By assumption, $ell=ell_1$ that is together with the fact ${X_{ n_{k_p} }}_p$ is a subsequence of ${X_{ n_{k} }}_k$ we have

$$0=lim_{ptoinfty } |X_{ n_{k_p} }-ell|>varepsilon_0>0~~~text{which is a CONTRADICTION}$$

Note that
$$forall p,~~|X_{ n_{k_p} }-ell|>varepsilon_0$$
Since
$$forall k,~~|X_{ n_{k}} -ell|>varepsilon_0$$

share|cite|improve this answer

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

$endgroup$

share|cite|improve this answer

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

edited yesterday

416E64726577

534

edited yesterday

416E64726577

534

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074

answered Sep 30 '17 at 13:21

Guy FsoneGuy Fsone

17.2k43074