Mixing
Walter Rudin “Principles of Mathematical Analysis” Definition 3.16, Theorem 3.17. I cannot understand.
$begingroup$
I am reading Walter Rudin's "Principles of Mathematical Analysis".
There are the following definition and theorem and its proof in this book.
Rudin didn't prove that $E neq emptyset$.
Why?
Rudin wrote "If $s^* = -infty$, then $E$ contains only one element" in the following proof.
But, if $E = emptyset$, then $s^* = -infty$ and $E$ contains no element.
So, I think Rudin needs to prove that $E neq emptyset$.
I cannot understand Rudin's proof.
Definition 3.16:
Let ${ s_n }$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} rightarrow x$ for some subsequence ${s_{n_k}}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+infty$, $-infty$.
Put $$s^* = sup E,$$ $$s_* = inf E.$$
Theorem 3.17:
Let ${s_n }$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* in E$.
(b) If $x> s^*$, there is an integer $N$ such that $n geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Of course, an analogous result is true for $s_*$.
Proof:
(a)
if $s^* = +infty$, then $E$ is not bounded above; hence ${s_n}$ is not bounded above, and there is a subsequence ${s_{n_k}}$ such that $s_{n_k} to +infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^* = -infty$, then $E$ contains only one element, namely $-infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n to -infty$.
This establishes (a) in all cases.
calculus analysis limsup-and-liminf
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
$endgroup$
add a comment |
$begingroup$
I am reading Walter Rudin's "Principles of Mathematical Analysis".
There are the following definition and theorem and its proof in this book.
Rudin didn't prove that $E neq emptyset$.
Why?
Rudin wrote "If $s^* = -infty$, then $E$ contains only one element" in the following proof.
But, if $E = emptyset$, then $s^* = -infty$ and $E$ contains no element.
So, I think Rudin needs to prove that $E neq emptyset$.
I cannot understand Rudin's proof.
Definition 3.16:
Let ${ s_n }$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} rightarrow x$ for some subsequence ${s_{n_k}}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+infty$, $-infty$.
Put $$s^* = sup E,$$ $$s_* = inf E.$$
Theorem 3.17:
Let ${s_n }$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* in E$.
(b) If $x> s^*$, there is an integer $N$ such that $n geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Of course, an analogous result is true for $s_*$.
Proof:
(a)
if $s^* = +infty$, then $E$ is not bounded above; hence ${s_n}$ is not bounded above, and there is a subsequence ${s_{n_k}}$ such that $s_{n_k} to +infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^* = -infty$, then $E$ contains only one element, namely $-infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n to -infty$.
This establishes (a) in all cases.
calculus analysis limsup-and-liminf
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
$endgroup$
3
$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47
add a comment |
$begingroup$
I am reading Walter Rudin's "Principles of Mathematical Analysis".
There are the following definition and theorem and its proof in this book.
Rudin didn't prove that $E neq emptyset$.
Why?
Rudin wrote "If $s^* = -infty$, then $E$ contains only one element" in the following proof.
But, if $E = emptyset$, then $s^* = -infty$ and $E$ contains no element.
So, I think Rudin needs to prove that $E neq emptyset$.
I cannot understand Rudin's proof.
Definition 3.16:
Let ${ s_n }$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} rightarrow x$ for some subsequence ${s_{n_k}}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+infty$, $-infty$.
Put $$s^* = sup E,$$ $$s_* = inf E.$$
Theorem 3.17:
Let ${s_n }$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* in E$.
(b) If $x> s^*$, there is an integer $N$ such that $n geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Of course, an analogous result is true for $s_*$.
Proof:
(a)
if $s^* = +infty$, then $E$ is not bounded above; hence ${s_n}$ is not bounded above, and there is a subsequence ${s_{n_k}}$ such that $s_{n_k} to +infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^* = -infty$, then $E$ contains only one element, namely $-infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n to -infty$.
This establishes (a) in all cases.
calculus analysis limsup-and-liminf
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
$endgroup$
I am reading Walter Rudin's "Principles of Mathematical Analysis".
There are the following definition and theorem and its proof in this book.
Rudin didn't prove that $E neq emptyset$.
Why?
Rudin wrote "If $s^* = -infty$, then $E$ contains only one element" in the following proof.
But, if $E = emptyset$, then $s^* = -infty$ and $E$ contains no element.
So, I think Rudin needs to prove that $E neq emptyset$.
I cannot understand Rudin's proof.
Definition 3.16:
Let ${ s_n }$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} rightarrow x$ for some subsequence ${s_{n_k}}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+infty$, $-infty$.
Put $$s^* = sup E,$$ $$s_* = inf E.$$
Theorem 3.17:
Let ${s_n }$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* in E$.
(b) If $x> s^*$, there is an integer $N$ such that $n geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Of course, an analogous result is true for $s_*$.
Proof:
(a)
if $s^* = +infty$, then $E$ is not bounded above; hence ${s_n}$ is not bounded above, and there is a subsequence ${s_{n_k}}$ such that $s_{n_k} to +infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^* = -infty$, then $E$ contains only one element, namely $-infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n to -infty$.
This establishes (a) in all cases.
calculus analysis limsup-and-liminf
calculus analysis limsup-and-liminf
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
asked Jan 19 at 10:13
tchappy hatchappy ha
767412
767412
3
$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47
add a comment |
3
$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47
3
3
$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Every sequence in $overline{mathbb{R}}$ has a convergent subsequence.
If the sequence is bounded, this is trivial by Bolzano's theorem.
Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ infty$. If it is unbounded below, you can find a subsequence that converges to $-infty$.
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
$endgroup$
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
add a comment |
$begingroup$
Note that $s^* = sup E$. So if $s^* =-infty$, then $E$ cannot contain any other $x in mathbb R cup {+ infty}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-infty$, which is not possible. So $E = {-infty}$, since $s^*$ exists only when $E$ is non empty.
If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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votes
$begingroup$
Every sequence in $overline{mathbb{R}}$ has a convergent subsequence.
If the sequence is bounded, this is trivial by Bolzano's theorem.
Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ infty$. If it is unbounded below, you can find a subsequence that converges to $-infty$.
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
$endgroup$
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
add a comment |
$begingroup$
Every sequence in $overline{mathbb{R}}$ has a convergent subsequence.
If the sequence is bounded, this is trivial by Bolzano's theorem.
Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ infty$. If it is unbounded below, you can find a subsequence that converges to $-infty$.
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
$endgroup$
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
add a comment |
$begingroup$
Every sequence in $overline{mathbb{R}}$ has a convergent subsequence.
If the sequence is bounded, this is trivial by Bolzano's theorem.
Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ infty$. If it is unbounded below, you can find a subsequence that converges to $-infty$.
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
$endgroup$
Every sequence in $overline{mathbb{R}}$ has a convergent subsequence.
If the sequence is bounded, this is trivial by Bolzano's theorem.
Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ infty$. If it is unbounded below, you can find a subsequence that converges to $-infty$.
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
answered Jan 19 at 10:35
Math_QEDMath_QED
7,67031452
7,67031452
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
add a comment |
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Thank you very much, Math_QED.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
$begingroup$
Glad to help you!
$endgroup$
– Math_QED
Jan 20 at 8:14
add a comment |
$begingroup$
Note that $s^* = sup E$. So if $s^* =-infty$, then $E$ cannot contain any other $x in mathbb R cup {+ infty}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-infty$, which is not possible. So $E = {-infty}$, since $s^*$ exists only when $E$ is non empty.
If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
add a comment |
$begingroup$
Note that $s^* = sup E$. So if $s^* =-infty$, then $E$ cannot contain any other $x in mathbb R cup {+ infty}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-infty$, which is not possible. So $E = {-infty}$, since $s^*$ exists only when $E$ is non empty.
If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
add a comment |
$begingroup$
Note that $s^* = sup E$. So if $s^* =-infty$, then $E$ cannot contain any other $x in mathbb R cup {+ infty}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-infty$, which is not possible. So $E = {-infty}$, since $s^*$ exists only when $E$ is non empty.
If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
Note that $s^* = sup E$. So if $s^* =-infty$, then $E$ cannot contain any other $x in mathbb R cup {+ infty}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-infty$, which is not possible. So $E = {-infty}$, since $s^*$ exists only when $E$ is non empty.
If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
edited Jan 19 at 10:26
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 19 at 10:20
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
38.9k33477
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
add a comment |
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
Thank you very much, астон вілла олоф мэллбэрг.
$endgroup$
– tchappy ha
Jan 20 at 2:46
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 10:06
add a comment |
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$begingroup$
$E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-infty$ implies $E={-infty}$.
$endgroup$
– Lord Shark the Unknown
Jan 19 at 10:17
$begingroup$
Thank you very much, Lord Shark the Unknown.
$endgroup$
– tchappy ha
Jan 20 at 2:47