Is there a dense set in $mathbb{R}$ with outer measure $1$? And what about with the Density Topology?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $(0,1)$ 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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $(0,1)$ 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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 8:59







guchihe

















asked Jan 9 at 8:13









guchiheguchihe

1458




1458












  • $begingroup$
    $(0,1)$ 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










  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















7












$begingroup$

Take $mathbb{Q}cup[0,1]$, for instance.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$






share|cite|improve this answer











$endgroup$



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 $q_{n+1}in U(g(n))subset K(g(n)+1)subset S.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 22:40

























answered Jan 9 at 22:20









DanielWainfleetDanielWainfleet

35.8k31648




35.8k31648












  • $begingroup$
    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












  • $begingroup$
    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











7












$begingroup$

Take $mathbb{Q}cup[0,1]$, for instance.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















7












$begingroup$

Take $mathbb{Q}cup[0,1]$, for instance.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














7












7








7





$begingroup$

Take $mathbb{Q}cup[0,1]$, for instance.






share|cite|improve this answer









$endgroup$



Take $mathbb{Q}cup[0,1]$, for instance.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 8:21









José Carlos SantosJosé Carlos Santos

172k22132239




172k22132239












  • $begingroup$
    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












  • $begingroup$
    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


















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