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Rudin 2.10 (b) Example
Rudin Theorem $1.11$Nested Interval Property implies Axiom of CompletenessEquivalent definitions of Lebesgue Measurability (Rudin and Royden)Baby Rudin Problem 2.29False proofs claiming that $mathbbQ$ is uncountable.Supremum of closed setsIs Abbott's Proof of the Uncountabilty of Real Numbers too strong?Existence of Nth root, by Rudin soft questionProof of uncountability of irrationals without using completeness of real numbersbaby rudin chapter 2 problem 21(b)- help with alternative method
$begingroup$
Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_x$ be the set of real numbers $y$ such that $0< y< x$. Then
$bigcap_x in A E_x$ is empty.
The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_x$ if $x < y$. Hence $y notin bigcap_x epsilon A E_x$
I didn't understand the proof & also here is my understanding.
Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.
Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.
Please explain how rudin got this result ?
real-analysis set-theory
edited 3 hours ago
Lucas Corrêa
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_x$ be the set of real numbers $y$ such that $0< y< x$. Then
$bigcap_x in A E_x$ is empty.
The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_x$ if $x < y$. Hence $y notin bigcap_x epsilon A E_x$
I didn't understand the proof & also here is my understanding.
Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.
Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.
Please explain how rudin got this result ?
real-analysis set-theory
edited 3 hours ago
Lucas Corrêa
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago
add a comment |
$begingroup$
Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_x$ be the set of real numbers $y$ such that $0< y< x$. Then
$bigcap_x in A E_x$ is empty.
The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_x$ if $x < y$. Hence $y notin bigcap_x epsilon A E_x$
I didn't understand the proof & also here is my understanding.
Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.
Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.
Please explain how rudin got this result ?
real-analysis set-theory
edited 3 hours ago
Lucas Corrêa
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_x$ be the set of real numbers $y$ such that $0< y< x$. Then
$bigcap_x in A E_x$ is empty.
The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_x$ if $x < y$. Hence $y notin bigcap_x epsilon A E_x$
I didn't understand the proof & also here is my understanding.
Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.
Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.
Please explain how rudin got this result ?
real-analysis set-theory
real-analysis set-theory
edited 3 hours ago
Lucas Corrêa
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Lucas Corrêa
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Lucas Corrêa
1,6151421
edited 3 hours ago
Lucas Corrêa
1,6151421
edited 3 hours ago
Lucas Corrêa
1,6151421
1,6151421
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
asked 4 hours ago
Mahendra ReddyMahendra Reddy
212
212
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago
add a comment |
$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago
$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).
Regarding your comment, the nested interval property is about compact sets. These are open and not closed.
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
answered 3 hours ago
Andreé RíosAndreé Ríos
1
New contributor
Andreé Ríos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
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3 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).
Regarding your comment, the nested interval property is about compact sets. These are open and not closed.
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).
Regarding your comment, the nested interval property is about compact sets. These are open and not closed.
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).
Regarding your comment, the nested interval property is about compact sets. These are open and not closed.
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).
Regarding your comment, the nested interval property is about compact sets. These are open and not closed.
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
edited 3 hours ago
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
add a comment |
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
answered 3 hours ago
Andreé RíosAndreé Ríos
1
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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– dantopa
3 hours ago
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
answered 3 hours ago
Andreé RíosAndreé Ríos
1
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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$begingroup$
Note that $xnotin E_x$ for every $x$
answered 3 hours ago
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Note that $xnotin E_x$ for every $x$
answered 3 hours ago
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answered 3 hours ago
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answered 3 hours ago
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
3 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
3 hours ago