Sequence of function on $mathbb{R}$ Cauchy iff convergent












-1












$begingroup$



Theorem: Let $(f_n)$ be a sequence of functions on $I subseteq mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.




Here, I only prove "$Longleftarrow$" since the converse is very straightforward.



Proof attempt: Suppose $left(f_n right)$ cauchy, then $forall epsilon >0 : forall x in I : exists N_o in mathbb{N}: forall m,n in mathbb{N}$ $$n,m geq N_0 implies mid f_n(x)-f_m(x) mid < frac{epsilon}{3} $$



Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$mid f_n(x)-f_{N_0}(x) mid < frac{epsilon}{3} implies mid f_n(x) mid < frac{epsilon}{3}+M$$
and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) to f$. Then $exists N_1 in mathbb{N}:$
$$n_k geq N_1 implies mid f_{n_{k}}(x)-f(x) mid<frac{epsilon}{3} $$
Combining the terms yields $$mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) mid leq mid f_{n_{k}}(x)-f(x) mid + mid f_n(x)-f_{N_0}(x) mid <frac{2epsilon }{3}$$
Now, let $n_k geq N_0$. Then $$mid f_n(x)-f(x)mid <frac{2epsilon }{3}+mid f_{N_0}(x)-f_{n_k}(x)mid $$
But since $n_k geq N_0$, $mid f_{N_0}(x)-f_{n_k}(x)mid < frac{epsilon}{3}$ so $mid f_n(x)-f(x)mid <epsilon$.
And we can conclude that $(f_n) to f$
$square$






real-analysis proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:04










  • $begingroup$
    @Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
    $endgroup$
    – BigbearZzz
    Dec 21 '18 at 17:11










  • $begingroup$
    Yes, that's pretty much everything what can be said about this question.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:14












  • $begingroup$
    @Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
    $endgroup$
    – Sei Sakata
    Dec 21 '18 at 17:35










  • $begingroup$
    Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
    $endgroup$
    – Mark
    Dec 21 '18 at 18:07
















-1












$begingroup$



Theorem: Let $(f_n)$ be a sequence of functions on $I subseteq mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.




Here, I only prove "$Longleftarrow$" since the converse is very straightforward.



Proof attempt: Suppose $left(f_n right)$ cauchy, then $forall epsilon >0 : forall x in I : exists N_o in mathbb{N}: forall m,n in mathbb{N}$ $$n,m geq N_0 implies mid f_n(x)-f_m(x) mid < frac{epsilon}{3} $$



Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$mid f_n(x)-f_{N_0}(x) mid < frac{epsilon}{3} implies mid f_n(x) mid < frac{epsilon}{3}+M$$
and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) to f$. Then $exists N_1 in mathbb{N}:$
$$n_k geq N_1 implies mid f_{n_{k}}(x)-f(x) mid<frac{epsilon}{3} $$
Combining the terms yields $$mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) mid leq mid f_{n_{k}}(x)-f(x) mid + mid f_n(x)-f_{N_0}(x) mid <frac{2epsilon }{3}$$
Now, let $n_k geq N_0$. Then $$mid f_n(x)-f(x)mid <frac{2epsilon }{3}+mid f_{N_0}(x)-f_{n_k}(x)mid $$
But since $n_k geq N_0$, $mid f_{N_0}(x)-f_{n_k}(x)mid < frac{epsilon}{3}$ so $mid f_n(x)-f(x)mid <epsilon$.
And we can conclude that $(f_n) to f$
$square$






real-analysis proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:04










  • $begingroup$
    @Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
    $endgroup$
    – BigbearZzz
    Dec 21 '18 at 17:11










  • $begingroup$
    Yes, that's pretty much everything what can be said about this question.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:14












  • $begingroup$
    @Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
    $endgroup$
    – Sei Sakata
    Dec 21 '18 at 17:35










  • $begingroup$
    Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
    $endgroup$
    – Mark
    Dec 21 '18 at 18:07














-1












-1








-1





$begingroup$



Theorem: Let $(f_n)$ be a sequence of functions on $I subseteq mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.




Here, I only prove "$Longleftarrow$" since the converse is very straightforward.



Proof attempt: Suppose $left(f_n right)$ cauchy, then $forall epsilon >0 : forall x in I : exists N_o in mathbb{N}: forall m,n in mathbb{N}$ $$n,m geq N_0 implies mid f_n(x)-f_m(x) mid < frac{epsilon}{3} $$



Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$mid f_n(x)-f_{N_0}(x) mid < frac{epsilon}{3} implies mid f_n(x) mid < frac{epsilon}{3}+M$$
and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) to f$. Then $exists N_1 in mathbb{N}:$
$$n_k geq N_1 implies mid f_{n_{k}}(x)-f(x) mid<frac{epsilon}{3} $$
Combining the terms yields $$mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) mid leq mid f_{n_{k}}(x)-f(x) mid + mid f_n(x)-f_{N_0}(x) mid <frac{2epsilon }{3}$$
Now, let $n_k geq N_0$. Then $$mid f_n(x)-f(x)mid <frac{2epsilon }{3}+mid f_{N_0}(x)-f_{n_k}(x)mid $$
But since $n_k geq N_0$, $mid f_{N_0}(x)-f_{n_k}(x)mid < frac{epsilon}{3}$ so $mid f_n(x)-f(x)mid <epsilon$.
And we can conclude that $(f_n) to f$
$square$






real-analysis proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$





Theorem: Let $(f_n)$ be a sequence of functions on $I subseteq mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.




Here, I only prove "$Longleftarrow$" since the converse is very straightforward.



Proof attempt: Suppose $left(f_n right)$ cauchy, then $forall epsilon >0 : forall x in I : exists N_o in mathbb{N}: forall m,n in mathbb{N}$ $$n,m geq N_0 implies mid f_n(x)-f_m(x) mid < frac{epsilon}{3} $$



Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$mid f_n(x)-f_{N_0}(x) mid < frac{epsilon}{3} implies mid f_n(x) mid < frac{epsilon}{3}+M$$
and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) to f$. Then $exists N_1 in mathbb{N}:$
$$n_k geq N_1 implies mid f_{n_{k}}(x)-f(x) mid<frac{epsilon}{3} $$
Combining the terms yields $$mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) mid leq mid f_{n_{k}}(x)-f(x) mid + mid f_n(x)-f_{N_0}(x) mid <frac{2epsilon }{3}$$
Now, let $n_k geq N_0$. Then $$mid f_n(x)-f(x)mid <frac{2epsilon }{3}+mid f_{N_0}(x)-f_{n_k}(x)mid $$
But since $n_k geq N_0$, $mid f_{N_0}(x)-f_{n_k}(x)mid < frac{epsilon}{3}$ so $mid f_n(x)-f(x)mid <epsilon$.
And we can conclude that $(f_n) to f$
$square$






real-analysis proof-verification cauchy-sequences




real-analysis proof-verification cauchy-sequences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:48







Sei Sakata

















asked Dec 21 '18 at 16:56









Sei SakataSei Sakata

10810




10810












  • $begingroup$
    You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:04










  • $begingroup$
    @Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
    $endgroup$
    – BigbearZzz
    Dec 21 '18 at 17:11










  • $begingroup$
    Yes, that's pretty much everything what can be said about this question.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:14












  • $begingroup$
    @Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
    $endgroup$
    – Sei Sakata
    Dec 21 '18 at 17:35










  • $begingroup$
    Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
    $endgroup$
    – Mark
    Dec 21 '18 at 18:07


















  • $begingroup$
    You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:04










  • $begingroup$
    @Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
    $endgroup$
    – BigbearZzz
    Dec 21 '18 at 17:11










  • $begingroup$
    Yes, that's pretty much everything what can be said about this question.
    $endgroup$
    – Mark
    Dec 21 '18 at 17:14












  • $begingroup$
    @Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
    $endgroup$
    – Sei Sakata
    Dec 21 '18 at 17:35










  • $begingroup$
    Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
    $endgroup$
    – Mark
    Dec 21 '18 at 18:07
















$begingroup$
You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
$endgroup$
– Mark
Dec 21 '18 at 17:04




$begingroup$
You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $mathbb{R}$ iff it is Cauchy.
$endgroup$
– Mark
Dec 21 '18 at 17:04












$begingroup$
@Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:11




$begingroup$
@Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:11












$begingroup$
Yes, that's pretty much everything what can be said about this question.
$endgroup$
– Mark
Dec 21 '18 at 17:14






$begingroup$
Yes, that's pretty much everything what can be said about this question.
$endgroup$
– Mark
Dec 21 '18 at 17:14














$begingroup$
@Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
$endgroup$
– Sei Sakata
Dec 21 '18 at 17:35




$begingroup$
@Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/…
$endgroup$
– Sei Sakata
Dec 21 '18 at 17:35












$begingroup$
Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
$endgroup$
– Mark
Dec 21 '18 at 18:07




$begingroup$
Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit.
$endgroup$
– Mark
Dec 21 '18 at 18:07










1 Answer
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It seems fine to me except that you should take $n_k geq N_1$ in the last part of the argument instead.



The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.






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    $begingroup$

    It seems fine to me except that you should take $n_k geq N_1$ in the last part of the argument instead.



    The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It seems fine to me except that you should take $n_k geq N_1$ in the last part of the argument instead.



      The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.






      share|cite|improve this answer









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        0












        0








        0





        $begingroup$

        It seems fine to me except that you should take $n_k geq N_1$ in the last part of the argument instead.



        The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.






        share|cite|improve this answer









        $endgroup$



        It seems fine to me except that you should take $n_k geq N_1$ in the last part of the argument instead.



        The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 17:09









        BigbearZzzBigbearZzz

        9,07421653




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