Mixing
Is there a dense set in $mathbb{R}$ with outer measure $1$? And what about with the Density Topology?
$begingroup$
I'm looking for a dense set in $mathbb{R}$ with outer measure $1$.
There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$..
Also I've tried with Bernstein sets but I don't get it.
Any hint is appreciated.
Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf
In the proof of theorem 4.13, he said:
...A set of outer measure $1$ is clearly dense...
In the Density Topology, $mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $mathbb{R}$ with outer measure $1$, I would appreciate it.
general-topology measure-theory lebesgue-measure descriptive-set-theory
asked Jan 9 at 8:13
guchiheguchihe
446
$endgroup$
|
show 1 more comment
$begingroup$
I'm looking for a dense set in $mathbb{R}$ with outer measure $1$.
There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$..
Also I've tried with Bernstein sets but I don't get it.
Any hint is appreciated.
Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf
In the proof of theorem 4.13, he said:
...A set of outer measure $1$ is clearly dense...
In the Density Topology, $mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $mathbb{R}$ with outer measure $1$, I would appreciate it.
general-topology measure-theory lebesgue-measure descriptive-set-theory
asked Jan 9 at 8:13
guchiheguchihe
446
$endgroup$
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17
|
show 1 more comment
$begingroup$
I'm looking for a dense set in $mathbb{R}$ with outer measure $1$.
There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$..
Also I've tried with Bernstein sets but I don't get it.
Any hint is appreciated.
Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf
In the proof of theorem 4.13, he said:
...A set of outer measure $1$ is clearly dense...
In the Density Topology, $mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $mathbb{R}$ with outer measure $1$, I would appreciate it.
general-topology measure-theory lebesgue-measure descriptive-set-theory
asked Jan 9 at 8:13
guchiheguchihe
446
$endgroup$
I'm looking for a dense set in $mathbb{R}$ with outer measure $1$.
There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$..
Also I've tried with Bernstein sets but I don't get it.
Any hint is appreciated.
Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf
In the proof of theorem 4.13, he said:
...A set of outer measure $1$ is clearly dense...
In the Density Topology, $mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $mathbb{R}$ with outer measure $1$, I would appreciate it.
general-topology measure-theory lebesgue-measure descriptive-set-theory
general-topology measure-theory lebesgue-measure descriptive-set-theory
asked Jan 9 at 8:13
guchiheguchihe
446
asked Jan 9 at 8:13
guchiheguchihe
446
edited Jan 9 at 8:59
guchihe
asked Jan 9 at 8:13
guchiheguchihe
446
asked Jan 9 at 8:13
guchiheguchihe
446
asked Jan 9 at 8:13
guchiheguchihe
446
446
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17
|
show 1 more comment
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
In the usual (Euclidean) topology, $Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$
Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$
For $nin Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:
$K(n)$ is bounded so $Bbb Q not subset K(n).$ Let $i(n)$ be the least $iin Bbb N$ such that $q_inot in K(n).$ Since $q_{i(n)} not in overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}in U(n)subset Bbb R setminus K(n).$ Since $K(n)cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$
Let $K(n+1)=K(n)cup U(n)cup V(n).$
Let $S=cup_{nin Bbb N}K(n).$
It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_nin S$ by induction on $n,$ as follows:
We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$
Now if $q_{n+1}in K(g(n))$ then $q_{n+1}in S.$
But if $q_{n+1}not in K(g(n)),$ then, since ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
add a comment |
$begingroup$
Take $mathbb{Q}cup[0,1]$, for instance.
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
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%2f3067208%2fis-there-a-dense-set-in-mathbbr-with-outer-measure-1-and-what-about-with%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
$begingroup$
In the usual (Euclidean) topology, $Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$
Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$
For $nin Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:
$K(n)$ is bounded so $Bbb Q not subset K(n).$ Let $i(n)$ be the least $iin Bbb N$ such that $q_inot in K(n).$ Since $q_{i(n)} not in overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}in U(n)subset Bbb R setminus K(n).$ Since $K(n)cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$
Let $K(n+1)=K(n)cup U(n)cup V(n).$
Let $S=cup_{nin Bbb N}K(n).$
It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_nin S$ by induction on $n,$ as follows:
We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$
Now if $q_{n+1}in K(g(n))$ then $q_{n+1}in S.$
But if $q_{n+1}not in K(g(n)),$ then, since ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
add a comment |
$begingroup$
In the usual (Euclidean) topology, $Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$
Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$
For $nin Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:
$K(n)$ is bounded so $Bbb Q not subset K(n).$ Let $i(n)$ be the least $iin Bbb N$ such that $q_inot in K(n).$ Since $q_{i(n)} not in overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}in U(n)subset Bbb R setminus K(n).$ Since $K(n)cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$
Let $K(n+1)=K(n)cup U(n)cup V(n).$
Let $S=cup_{nin Bbb N}K(n).$
It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_nin S$ by induction on $n,$ as follows:
We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$
Now if $q_{n+1}in K(g(n))$ then $q_{n+1}in S.$
But if $q_{n+1}not in K(g(n)),$ then, since ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
add a comment |
$begingroup$
In the usual (Euclidean) topology, $Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$
Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$
For $nin Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:
$K(n)$ is bounded so $Bbb Q not subset K(n).$ Let $i(n)$ be the least $iin Bbb N$ such that $q_inot in K(n).$ Since $q_{i(n)} not in overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}in U(n)subset Bbb R setminus K(n).$ Since $K(n)cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$
Let $K(n+1)=K(n)cup U(n)cup V(n).$
Let $S=cup_{nin Bbb N}K(n).$
It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_nin S$ by induction on $n,$ as follows:
We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$
Now if $q_{n+1}in K(g(n))$ then $q_{n+1}in S.$
But if $q_{n+1}not in K(g(n)),$ then, since ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
In the usual (Euclidean) topology, $Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$
Let $Bbb Q={q_i: iin Bbb N}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1in K(1).$
For $nin Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:
$K(n)$ is bounded so $Bbb Q not subset K(n).$ Let $i(n)$ be the least $iin Bbb N$ such that $q_inot in K(n).$ Since $q_{i(n)} not in overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}in U(n)subset Bbb R setminus K(n).$ Since $K(n)cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$
Let $K(n+1)=K(n)cup U(n)cup V(n).$
Let $S=cup_{nin Bbb N}K(n).$
It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_nin S$ by induction on $n,$ as follows:
We have $q_1in S.$ For $nin Bbb N,$ if ${q_i:ile n}subset S,$ then for each $ ile n$ let $f(i)$ be the least (or any) $j$ such that $q_iin K(j),$ and let $g(n)=max {f(i):ile n}.$
Now if $q_{n+1}in K(g(n))$ then $q_{n+1}in S.$
But if $q_{n+1}not in K(g(n)),$ then, since ${q_i:ile n}subset cup_{ile n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
edited Jan 9 at 22:40
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
answered Jan 9 at 22:20
DanielWainfleetDanielWainfleet
34.9k31648
34.9k31648
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
add a comment |
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
By a variation on the theme of a "fat Cantor set" we can also construct a closed nowhere-dense $Tsubset Bbb R$ such that $m(T)=1,$ and $ m(Tcap (n,n+1))ne 0$ for every $nin Bbb Z.$ Although this is not the kind of set in the Q.
$endgroup$
– DanielWainfleet
Jan 10 at 0:26
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
Maybe this can help me. I have to see wether $S$ is dense in the density topology or not. If it is, I just should to intersect an arbitrary dense with $S$ and i'll get a dense set with outer measure $1$ and cardinality of the dense, and I'll have the inequality... so thanks! If I get it, i'll put the answer.
$endgroup$
– guchihe
Jan 10 at 2:19
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
$begingroup$
By the Lebesgue Density Lemma, any measurable $Esubset Bbb R$ has a measurable subset $E^*$ with $m(E)=m(E^*)$ and such that, in the notation of F.D. Tall, $phi(E^*)supseteq E^*,$ which makes $E^*$ open in the Density Topology. And if $m(E)ne 0$ then $E^*$ is not empty because $m(E^*)=m(E)$.... In particular let $E=Bbb Rsetminus S.$
$endgroup$
– DanielWainfleet
Jan 10 at 9:14
1
1
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
$begingroup$
Oh it's true. In fact, just by that, I need a non-measurable set.
$endgroup$
– guchihe
Jan 10 at 19:57
add a comment |
$begingroup$
Take $mathbb{Q}cup[0,1]$, for instance.
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
add a comment |
$begingroup$
Take $mathbb{Q}cup[0,1]$, for instance.
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
add a comment |
$begingroup$
Take $mathbb{Q}cup[0,1]$, for instance.
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
Take $mathbb{Q}cup[0,1]$, for instance.
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 9 at 8:21
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
add a comment |
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
Yes, it works. Now I need a dense more complicated. I want a dense not ''composed'' by countable denses, nor nullsets.
$endgroup$
– guchihe
Jan 9 at 21:07
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
$begingroup$
I suggest that you post it as another answer. But explain better what you're after.
$endgroup$
– José Carlos Santos
Jan 9 at 21:48
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%2f3067208%2fis-there-a-dense-set-in-mathbbr-with-outer-measure-1-and-what-about-with%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
$begingroup$
$(0,1)$ works. Also $[0,1]$. Also $[0,1]setminusBbb Q$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:16
$begingroup$
But I need it to be dense in $mathbb{R}$.
$endgroup$
– guchihe
Jan 9 at 8:18
$begingroup$
I suspect that he's working in a measure space where the measure of the whole space is $1$ (e.g. the Cantor set with its associated Haar measure, or $[0,1]$).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:50
$begingroup$
No, the space is defined in $mathbb{R}$ and use the Lebesgue measure.
$endgroup$
– guchihe
Jan 9 at 8:55
$begingroup$
I also do not understand "... A set of outer measure $1$ is clearly dense..." in the paragraph following the assertions 4.13 and 4.14, because $[0,1]$ has outer measure $1$ and its complement is open in both the Euclidean and the Density topologies
$endgroup$
– DanielWainfleet
Jan 10 at 0:17