Prove that $C$ is Banach.












0












$begingroup$


$C={x_n lvert x_n converges } $



Let $x^n in C$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)



which means $x_n rightarrow x$



Now to show that $xin C$



$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$



This gives us $xrightarrow u$



So $xin C$



Is this Correct?






functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
    $endgroup$
    – Mindlack
    Jan 5 at 23:09






  • 1




    $begingroup$
    The second part seems flawed.. What is $u$ there?
    $endgroup$
    – Berci
    Jan 5 at 23:10










  • $begingroup$
    I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
    $endgroup$
    – user3482749
    Jan 5 at 23:10












  • $begingroup$
    @Berci Please check now.
    $endgroup$
    – Hitman
    Jan 5 at 23:30










  • $begingroup$
    @Mindlack it is not a repost.
    $endgroup$
    – Hitman
    Jan 5 at 23:30
















0












$begingroup$


$C={x_n lvert x_n converges } $



Let $x^n in C$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)



which means $x_n rightarrow x$



Now to show that $xin C$



$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$



This gives us $xrightarrow u$



So $xin C$



Is this Correct?






functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
    $endgroup$
    – Mindlack
    Jan 5 at 23:09






  • 1




    $begingroup$
    The second part seems flawed.. What is $u$ there?
    $endgroup$
    – Berci
    Jan 5 at 23:10










  • $begingroup$
    I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
    $endgroup$
    – user3482749
    Jan 5 at 23:10












  • $begingroup$
    @Berci Please check now.
    $endgroup$
    – Hitman
    Jan 5 at 23:30










  • $begingroup$
    @Mindlack it is not a repost.
    $endgroup$
    – Hitman
    Jan 5 at 23:30














0












0








0


1



$begingroup$


$C={x_n lvert x_n converges } $



Let $x^n in C$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)



which means $x_n rightarrow x$



Now to show that $xin C$



$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$



This gives us $xrightarrow u$



So $xin C$



Is this Correct?






functional-analysis







share|cite|improve this question











$endgroup$




$C={x_n lvert x_n converges } $



Let $x^n in C$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m >N $ $$ lVert x^n -x^mrVert< frac {epsilon} {3} $$
we know that for every k $$lvert u^n -u^mrvert le sup_{ige 1} lvert x^n_i -x^m_irvert < frac {epsilon} {3} $$
So $u^n$ is Cauchy in $mathbb R$ which is Banach so $$u^n rightarrow u in mathbb R or lvert u^n -urvert < frac {epsilon} {3} $$
by this we can say that $lVert x^n -xrVert = sup lvert x^n_i -x_i rvert < frac {epsilon} {3} $ $ $ ($epsilon>0, nge Nin mathbb N$)



which means $x_n rightarrow x$



Now to show that $xin C$



$$lvert x-urvert le lvert x^n-xrvert+ lvert x^n-u^nrvert +lvert u^n-urvert <frac {epsilon} {3}+frac {epsilon} {3}+frac {epsilon} {3}=epsilon$$



This gives us $xrightarrow u$



So $xin C$



Is this Correct?






functional-analysis




functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 0:22







Hitman

















asked Jan 5 at 23:05









HitmanHitman

1939




1939












  • $begingroup$
    You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
    $endgroup$
    – Mindlack
    Jan 5 at 23:09






  • 1




    $begingroup$
    The second part seems flawed.. What is $u$ there?
    $endgroup$
    – Berci
    Jan 5 at 23:10










  • $begingroup$
    I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
    $endgroup$
    – user3482749
    Jan 5 at 23:10












  • $begingroup$
    @Berci Please check now.
    $endgroup$
    – Hitman
    Jan 5 at 23:30










  • $begingroup$
    @Mindlack it is not a repost.
    $endgroup$
    – Hitman
    Jan 5 at 23:30


















  • $begingroup$
    You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
    $endgroup$
    – Mindlack
    Jan 5 at 23:09






  • 1




    $begingroup$
    The second part seems flawed.. What is $u$ there?
    $endgroup$
    – Berci
    Jan 5 at 23:10










  • $begingroup$
    I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
    $endgroup$
    – user3482749
    Jan 5 at 23:10












  • $begingroup$
    @Berci Please check now.
    $endgroup$
    – Hitman
    Jan 5 at 23:30










  • $begingroup$
    @Mindlack it is not a repost.
    $endgroup$
    – Hitman
    Jan 5 at 23:30
















$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09




$begingroup$
You will not get a faster answer by reposting, even less if you do not explain, say, your notations.
$endgroup$
– Mindlack
Jan 5 at 23:09




1




1




$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10




$begingroup$
The second part seems flawed.. What is $u$ there?
$endgroup$
– Berci
Jan 5 at 23:10












$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10






$begingroup$
I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct.
$endgroup$
– user3482749
Jan 5 at 23:10














$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30




$begingroup$
@Berci Please check now.
$endgroup$
– Hitman
Jan 5 at 23:30












$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30




$begingroup$
@Mindlack it is not a repost.
$endgroup$
– Hitman
Jan 5 at 23:30










1 Answer
1






active

oldest

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0












$begingroup$

The second part is not correct: $x$ should converge to a real number.



A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you please check again. I have edited the question.
    $endgroup$
    – Hitman
    Jan 5 at 23:48












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

The second part is not correct: $x$ should converge to a real number.



A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you please check again. I have edited the question.
    $endgroup$
    – Hitman
    Jan 5 at 23:48
















0












$begingroup$

The second part is not correct: $x$ should converge to a real number.



A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you please check again. I have edited the question.
    $endgroup$
    – Hitman
    Jan 5 at 23:48














0












0








0





$begingroup$

The second part is not correct: $x$ should converge to a real number.



A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.






share|cite|improve this answer









$endgroup$



The second part is not correct: $x$ should converge to a real number.



A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(lim x^n)$ converges to its limit.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 23:39









BerciBerci

61.9k23776




61.9k23776












  • $begingroup$
    Could you please check again. I have edited the question.
    $endgroup$
    – Hitman
    Jan 5 at 23:48


















  • $begingroup$
    Could you please check again. I have edited the question.
    $endgroup$
    – Hitman
    Jan 5 at 23:48
















$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48




$begingroup$
Could you please check again. I have edited the question.
$endgroup$
– Hitman
Jan 5 at 23:48


















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