Rudin 2.10 (b) ExampleRudin 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
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
$begingroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 3 hours ago
Lucas Corrêa
1,6151421
1,6151421
New contributor
asked 3 hours ago
Mahendra ReddyMahendra Reddy
212
212
New contributor
New contributor
$begingroup$
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
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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.
$endgroup$
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
New contributor
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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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2 hours ago
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited 3 hours ago
answered 3 hours ago
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
add a comment |
add a comment |
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Note that $xnotin E_x$ for every $x$
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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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Note that $xnotin E_x$ for every $x$
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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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Note that $xnotin E_x$ for every $x$
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Note that $xnotin E_x$ for every $x$
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answered 2 hours ago
Andreé RíosAndreé Ríos
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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