Mixing
Determine $lambda^{2}({(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]})$
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
$endgroup$
add a comment |
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
$endgroup$
add a comment |
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
$endgroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
asked Jan 13 at 13:58
MinaThumaMinaThuma
1458
1458
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
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%2f3072026%2fdetermine-lambda2-x-y-in-mathbb-r2-x-in-mathbb-q-cap-0-1-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
$endgroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,72621652
8,72621652
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
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%2f3072026%2fdetermine-lambda2-x-y-in-mathbb-r2-x-in-mathbb-q-cap-0-1-y%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
