Bounded sequence of functions implies convergent subsequence
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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
real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
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add a comment |
$begingroup$
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
real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
$endgroup$
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 $x$.
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– angryavian
Jan 6 at 22:31
1
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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.
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– Nate Eldredge
Jan 6 at 22:32
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@NateEldredge my bad. makes a lot of sense!!
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– Kaan Yolsever
Jan 6 at 22:33
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@angryavian thanks!!
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35
add a comment |
$begingroup$
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
real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
$endgroup$
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
real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function
real-analysis sequences-and-series convergence pointwise-convergence sequence-of-function
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
asked Jan 6 at 22:25
Kaan YolseverKaan Yolsever
1139
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
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@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
add a comment |
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
add a comment |
1 Answer
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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.
answered Jan 6 at 22:29
MindlackMindlack
4,360210
$endgroup$
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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.
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– Mindlack
Jan 6 at 22:34
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thanks. understood! i'll accept your answer when i can
$endgroup$
– Kaan Yolsever
Jan 6 at 22:35
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered Jan 6 at 22:29
MindlackMindlack
4,360210
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 6 at 22:29
MindlackMindlack
4,360210
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 6 at 22:29
MindlackMindlack
4,360210
$endgroup$
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.
answered Jan 6 at 22:29
MindlackMindlack
4,360210
edited Jan 6 at 22:31
answered Jan 6 at 22:29
MindlackMindlack
4,360210
answered Jan 6 at 22:29
MindlackMindlack
4,360210
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
add a comment |
$begingroup$
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
add a comment |
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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