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
What is the term for a person whose job is to place products on shelves in stores?
Creating a chemical industry from a medieval tech level without petroleum
Drawing a german abacus as in the books of Adam Ries
Why is the underscore command _ useful?
How bug prioritization works in agile projects vs non agile
Where was the County of Thurn und Taxis located?
"My boss was furious with me and I have been fired" vs. "My boss was furious with me and I was fired"
Are there moral objections to a life motivated purely by money? How to sway a person from this lifestyle?
"The cow" OR "a cow" OR "cows" in this context
Older movie/show about humans on derelict alien warship which refuels by passing through a star
Partitioning values in a sequence
How did Captain America manage to do this?
Is there really no use for MD5 anymore?
Is there metaphorical meaning of "aus der Haft entlassen"?
How to not starve gigantic beasts
Can I criticise the more senior developers around me for not writing clean code?
Was Dennis Ritchie being too modest in this quote about C and Pascal?
All ASCII characters with a given bit count
Co-worker works way more than he should
A Note on N!
Should the Product Owner dictate what info the UI needs to display?
Is Electric Central Heating worth it if using Solar Panels?
Why must Chinese maps be obfuscated?
A Paper Record is What I Hamper
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
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 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 4 hours ago
Lucas Corrêa
1,6151421
1,6151421
New contributor
asked 5 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
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 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
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 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
4 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
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 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.
$endgroup$
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
New contributor
$endgroup$
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Mahendra Reddy is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3202687%2frudin-2-10-b-example%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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.
$endgroup$
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 4 hours ago
answered 4 hours ago
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
add a comment |
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
New contributor
$endgroup$
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
New contributor
$endgroup$
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
add a comment |
$begingroup$
Note that $xnotin E_x$ for every $x$
New contributor
$endgroup$
Note that $xnotin E_x$ for every $x$
New contributor
New contributor
answered 4 hours ago
Andreé RíosAndreé Ríos
1
1
New contributor
New contributor
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
add a comment |
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
$begingroup$
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.
$endgroup$
– dantopa
3 hours ago
add a comment |
Mahendra Reddy is a new contributor. Be nice, and check out our Code of Conduct.
Mahendra Reddy is a new contributor. Be nice, and check out our Code of Conduct.
Mahendra Reddy is a new contributor. Be nice, and check out our Code of Conduct.
Mahendra Reddy is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3202687%2frudin-2-10-b-example%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
4 hours ago
$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 hours ago