Bounded sequence of functions implies convergent subsequence












0












$begingroup$


Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks



The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":



Theorem 3.6:
If ${p_n}$ is a sequence in a compact metric space X, then some subsequence of ${p_n}$ converges to a point of X



my attempt at the proof










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$endgroup$








  • 1




    $begingroup$
    1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
    $endgroup$
    – angryavian
    Jan 6 at 22:31






  • 1




    $begingroup$
    Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
    $endgroup$
    – Nate Eldredge
    Jan 6 at 22:32












  • $begingroup$
    @NateEldredge my bad. makes a lot of sense!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:33












  • $begingroup$
    @angryavian thanks!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35
















0












$begingroup$


Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks



The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":



Theorem 3.6:
If ${p_n}$ is a sequence in a compact metric space X, then some subsequence of ${p_n}$ converges to a point of X



my attempt at the proof










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
    $endgroup$
    – angryavian
    Jan 6 at 22:31






  • 1




    $begingroup$
    Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
    $endgroup$
    – Nate Eldredge
    Jan 6 at 22:32












  • $begingroup$
    @NateEldredge my bad. makes a lot of sense!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:33












  • $begingroup$
    @angryavian thanks!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35














0












0








0





$begingroup$


Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks



The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":



Theorem 3.6:
If ${p_n}$ is a sequence in a compact metric space X, then some subsequence of ${p_n}$ converges to a point of X



my attempt at the proof










share|cite|improve this question









$endgroup$




Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks



The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":



Theorem 3.6:
If ${p_n}$ is a sequence in a compact metric space X, then some subsequence of ${p_n}$ converges to a point of X



my attempt at the proof







real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 22:25









Kaan YolseverKaan Yolsever

1139




1139








  • 1




    $begingroup$
    1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
    $endgroup$
    – angryavian
    Jan 6 at 22:31






  • 1




    $begingroup$
    Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
    $endgroup$
    – Nate Eldredge
    Jan 6 at 22:32












  • $begingroup$
    @NateEldredge my bad. makes a lot of sense!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:33












  • $begingroup$
    @angryavian thanks!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35














  • 1




    $begingroup$
    1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
    $endgroup$
    – angryavian
    Jan 6 at 22:31






  • 1




    $begingroup$
    Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
    $endgroup$
    – Nate Eldredge
    Jan 6 at 22:32












  • $begingroup$
    @NateEldredge my bad. makes a lot of sense!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:33












  • $begingroup$
    @angryavian thanks!!
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35








1




1




$begingroup$
1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
$endgroup$
– angryavian
Jan 6 at 22:31




$begingroup$
1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$.
$endgroup$
– angryavian
Jan 6 at 22:31




1




1




$begingroup$
Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
$endgroup$
– Nate Eldredge
Jan 6 at 22:32






$begingroup$
Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder.
$endgroup$
– Nate Eldredge
Jan 6 at 22:32














$begingroup$
@NateEldredge my bad. makes a lot of sense!!
$endgroup$
– Kaan Yolsever
Jan 6 at 22:33






$begingroup$
@NateEldredge my bad. makes a lot of sense!!
$endgroup$
– Kaan Yolsever
Jan 6 at 22:33














$begingroup$
@angryavian thanks!!
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35




$begingroup$
@angryavian thanks!!
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.



More precisely, your compactness argument is as follows:



Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $varphi_x : mathbb{N} rightarrow mathbb{N}$ such that $(f_{varphi_x(n)}(x))_n$ is convergent.



But if $y$ is another real number, there is no reason why $(f_{varphi_x(n)}(y))_n$ is convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    isn't the argument valid for any x in the domain [a,b]?
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:29












  • $begingroup$
    Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
    $endgroup$
    – Mindlack
    Jan 6 at 22:34












  • $begingroup$
    thanks. understood! i'll accept your answer when i can
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.



More precisely, your compactness argument is as follows:



Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $varphi_x : mathbb{N} rightarrow mathbb{N}$ such that $(f_{varphi_x(n)}(x))_n$ is convergent.



But if $y$ is another real number, there is no reason why $(f_{varphi_x(n)}(y))_n$ is convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    isn't the argument valid for any x in the domain [a,b]?
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:29












  • $begingroup$
    Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
    $endgroup$
    – Mindlack
    Jan 6 at 22:34












  • $begingroup$
    thanks. understood! i'll accept your answer when i can
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35
















1












$begingroup$

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.



More precisely, your compactness argument is as follows:



Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $varphi_x : mathbb{N} rightarrow mathbb{N}$ such that $(f_{varphi_x(n)}(x))_n$ is convergent.



But if $y$ is another real number, there is no reason why $(f_{varphi_x(n)}(y))_n$ is convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    isn't the argument valid for any x in the domain [a,b]?
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:29












  • $begingroup$
    Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
    $endgroup$
    – Mindlack
    Jan 6 at 22:34












  • $begingroup$
    thanks. understood! i'll accept your answer when i can
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35














1












1








1





$begingroup$

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.



More precisely, your compactness argument is as follows:



Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $varphi_x : mathbb{N} rightarrow mathbb{N}$ such that $(f_{varphi_x(n)}(x))_n$ is convergent.



But if $y$ is another real number, there is no reason why $(f_{varphi_x(n)}(y))_n$ is convergent.






share|cite|improve this answer











$endgroup$



You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.



More precisely, your compactness argument is as follows:



Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $varphi_x : mathbb{N} rightarrow mathbb{N}$ such that $(f_{varphi_x(n)}(x))_n$ is convergent.



But if $y$ is another real number, there is no reason why $(f_{varphi_x(n)}(y))_n$ is convergent.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 22:31

























answered Jan 6 at 22:29









MindlackMindlack

4,360210




4,360210












  • $begingroup$
    isn't the argument valid for any x in the domain [a,b]?
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:29












  • $begingroup$
    Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
    $endgroup$
    – Mindlack
    Jan 6 at 22:34












  • $begingroup$
    thanks. understood! i'll accept your answer when i can
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35


















  • $begingroup$
    isn't the argument valid for any x in the domain [a,b]?
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:29












  • $begingroup$
    Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
    $endgroup$
    – Mindlack
    Jan 6 at 22:34












  • $begingroup$
    thanks. understood! i'll accept your answer when i can
    $endgroup$
    – Kaan Yolsever
    Jan 6 at 22:35
















$begingroup$
isn't the argument valid for any x in the domain [a,b]?
$endgroup$
– Kaan Yolsever
Jan 6 at 22:29






$begingroup$
isn't the argument valid for any x in the domain [a,b]?
$endgroup$
– Kaan Yolsever
Jan 6 at 22:29














$begingroup$
Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
$endgroup$
– Mindlack
Jan 6 at 22:34






$begingroup$
Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not.
$endgroup$
– Mindlack
Jan 6 at 22:34














$begingroup$
thanks. understood! i'll accept your answer when i can
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35




$begingroup$
thanks. understood! i'll accept your answer when i can
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35


















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