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Given $min[0,1]$, can we find a dense subset of $[0,1]$ whose Lebesgue measure is exactly $m$?

Subset $Asubsetmathbb R$ such that for any interval $I$ of length $a$ the set $Acap I$ has Lebesgue measure $a/2$Lebesgue's Density Theorem - intuition and weaker formsRevisiting a Lebesgue measure question involving a dense subset of R, translates of a measurable set, etc.Lebesgue vs. Counting Measure on $[0,1]$Find an example of Lebesgue measurable subsets of $[0,1]$ these conditions:Nowhere dense subset of unit square of measure $1$Lebesgue measure on [0,1]A confusion over the lebesgue outer measure of $(0,1) cap mathbbQ^c$Proof that the complement of zero-measure set on $[0,1]$ is dense in $[0,1]$Take a set of points in a closed interval whose closure has 0 lebesgue measure. Can we cover them with finitely many intervals of small measure?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

1

$begingroup$

Consider the collection of subsets $A$ of the unit interval $[0,1]$ which are dense, meaning that for every $xin[0,1]$, for every $varepsilon>0$, there exists $ain A$ such that $|x-a|<varepsilon$. What are the Lebesgue measures of these sets?

Clearly these sets are bounded above by the unit interval itself, which is dense and has Lebesgue measure $1$. On the other hand, the set $Bbb Q cap [0,1]$ is dense and has Lebesgue measure null.

My question is this: for any $min[0,1]$, does there exist a dense subset $Asubseteq[0,1]$ with Lebesgue measure $m$?

---

Edit: I found out that if $A$ has measure $m$ and satisfies $|Acap I|=m|I|$ for every interval $Isubseteq[0,1]$ (a better, stronger condition) where $|cdot|$ denotes Lebesgue measure, then the density at a point $xin A$ is given by

$$ d(x) = lim_varepsilonrightarrow0
frac
= begincases |A|/2 & textif x=0text or 1 \
|A| & textif xin(0,1) endcases$$

Lebesgue's density theorem says that if $A$ is measurable then $d(x)=1$ for almost all $xin A$, and since we established $d(x)=|A|$ for $xin(0,1)$, which is almost all of $[0,1]$, this implies $|A|=1$.

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 13 at 17:31

M. Nestor

asked Aug 9 at 4:53

M. NestorM. Nestor

1,07611 silver badge1515 bronze badges

$endgroup$

add a comment | 

1

1

$begingroup$

Consider the collection of subsets $A$ of the unit interval $[0,1]$ which are dense, meaning that for every $xin[0,1]$, for every $varepsilon>0$, there exists $ain A$ such that $|x-a|<varepsilon$. What are the Lebesgue measures of these sets?

Clearly these sets are bounded above by the unit interval itself, which is dense and has Lebesgue measure $1$. On the other hand, the set $Bbb Q cap [0,1]$ is dense and has Lebesgue measure null.

My question is this: for any $min[0,1]$, does there exist a dense subset $Asubseteq[0,1]$ with Lebesgue measure $m$?

---

Edit: I found out that if $A$ has measure $m$ and satisfies $|Acap I|=m|I|$ for every interval $Isubseteq[0,1]$ (a better, stronger condition) where $|cdot|$ denotes Lebesgue measure, then the density at a point $xin A$ is given by

$$ d(x) = lim_varepsilonrightarrow0
frac
= begincases |A|/2 & textif x=0text or 1 \
|A| & textif xin(0,1) endcases$$

Lebesgue's density theorem says that if $A$ is measurable then $d(x)=1$ for almost all $xin A$, and since we established $d(x)=|A|$ for $xin(0,1)$, which is almost all of $[0,1]$, this implies $|A|=1$.

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 13 at 17:31

M. Nestor

asked Aug 9 at 4:53

M. NestorM. Nestor

1,07611 silver badge1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

Consider the collection of subsets $A$ of the unit interval $[0,1]$ which are dense, meaning that for every $xin[0,1]$, for every $varepsilon>0$, there exists $ain A$ such that $|x-a|<varepsilon$. What are the Lebesgue measures of these sets?

Clearly these sets are bounded above by the unit interval itself, which is dense and has Lebesgue measure $1$. On the other hand, the set $Bbb Q cap [0,1]$ is dense and has Lebesgue measure null.

My question is this: for any $min[0,1]$, does there exist a dense subset $Asubseteq[0,1]$ with Lebesgue measure $m$?

---

Edit: I found out that if $A$ has measure $m$ and satisfies $|Acap I|=m|I|$ for every interval $Isubseteq[0,1]$ (a better, stronger condition) where $|cdot|$ denotes Lebesgue measure, then the density at a point $xin A$ is given by

$$ d(x) = lim_varepsilonrightarrow0
frac
= begincases |A|/2 & textif x=0text or 1 \
|A| & textif xin(0,1) endcases$$

Lebesgue's density theorem says that if $A$ is measurable then $d(x)=1$ for almost all $xin A$, and since we established $d(x)=|A|$ for $xin(0,1)$, which is almost all of $[0,1]$, this implies $|A|=1$.

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 13 at 17:31

M. Nestor

asked Aug 9 at 4:53

M. NestorM. Nestor

1,07611 silver badge1515 bronze badges

$endgroup$

share|cite|improve this question

edited Aug 13 at 17:31

M. Nestor

asked Aug 9 at 4:53

M. NestorM. Nestor

1,07611 silver badge1515 bronze badges

share|cite|improve this question

edited Aug 13 at 17:31

M. Nestor

asked Aug 9 at 4:53

M. NestorM. Nestor

1,07611 silver badge1515 bronze badges

asked Aug 9 at 4:53

M. NestorM. Nestor

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asked Aug 9 at 4:53

M. NestorM. Nestor

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3 Answers
3

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oldest

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7

$begingroup$

The answer is yes. For $min [0,1]$ consider the set $A:=[0,m]cup(mathbbQcap [0,1])$.
This is clearly dense and has measure $m$.

share|cite|improve this answer

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Great! However, this set is quite "lop-sided". I am wondering if there is a more evenly distributed example.
  $endgroup$
  – M. Nestor
  Aug 9 at 4:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then I refer to the answer of saz. Take an evenly distributed set of Lebesgue measure $m$ and "add" $mathbbQcap [0,1]$.
  $endgroup$
  – Jonas Lenz
  Aug 9 at 5:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Nestor Depends on what you mean by evenly distributed. It would be cool if, given a constant $0 < c < 1$, it were possible to find a measurable set $A$ such that $m(A cap I) = c m(I)$ for every interval $I subseteq [0,1]$. This turns out to be impossible
  $endgroup$
  – Bungo
  Aug 9 at 5:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bungo After reading that post, it seems that what I am after is precisely Lebesgue's density theorem. Thanks!
  $endgroup$
  – M. Nestor
  Aug 9 at 5:17
  

  
  
  
  

add a comment | 

5

$begingroup$

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider

$$A := B cup (mathbbQ cap [0,1]).$$

The set has Lebesgue measure $m$ and it is dense in $[0,1]$.

It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_n in mathbbN$ of $mathbbQ cap [0,1]$ and $$A := bigcup_n in mathbbN (q_n-epsilon 2^-n,q_n+epsilon 2^-n)$$ for fixed $epsilon>0$. The set $A$ is open and has Lebesgue measure $leq epsilon$.

Remark: Note that there does not exist an open dense set with Lebesgue measure zero. In this sense, the best we can achieve is to have an open dense set of arbitrarily small Lebesgue measure, as above.

share|cite|improve this answer

edited Aug 9 at 5:08

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the 2nd example, the set must be less than $epsilon$ since there are some overlaps between those intervals, right?
  $endgroup$
  – efhvcjnfdbgefg
  Aug 9 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @efhvcjnfdbgefg Yes, that's correct. There are plenty of overlaps (... exactly because the rationals are dense).
  $endgroup$
  – saz
  Aug 9 at 5:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact $varepsilon$ must always be an overestimate, since any interval $(q_n-2^-nvarepsilon,q_n+2^-nvarepsilon)$ must contain another $q_m$, guaranteeing overlap at least $minvarepsilon2^-m,varepsilon2^-n$.
  $endgroup$
  – M. Nestor
  Aug 9 at 5:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My above comment doesn't change the validity, since the overestimate varies continuously, you can always find $varepsilon$ such that the true measure is precisely $m$.
  $endgroup$
  – M. Nestor
  Aug 13 at 17:47
  

  
  
  
  

add a comment | 

2

$begingroup$

Sure. Take
$$
[0,m]cup(mathbbQcap [0,1])=[0,m]oversetcdotcup(mathbbQcap(m,1]).
$$

share|cite|improve this answer

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

$endgroup$

add a comment | 

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7

$begingroup$

The answer is yes. For $min [0,1]$ consider the set $A:=[0,m]cup(mathbbQcap [0,1])$.
This is clearly dense and has measure $m$.

share|cite|improve this answer

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Great! However, this set is quite "lop-sided". I am wondering if there is a more evenly distributed example.
  $endgroup$
  – M. Nestor
  Aug 9 at 4:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then I refer to the answer of saz. Take an evenly distributed set of Lebesgue measure $m$ and "add" $mathbbQcap [0,1]$.
  $endgroup$
  – Jonas Lenz
  Aug 9 at 5:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Nestor Depends on what you mean by evenly distributed. It would be cool if, given a constant $0 < c < 1$, it were possible to find a measurable set $A$ such that $m(A cap I) = c m(I)$ for every interval $I subseteq [0,1]$. This turns out to be impossible
  $endgroup$
  – Bungo
  Aug 9 at 5:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bungo After reading that post, it seems that what I am after is precisely Lebesgue's density theorem. Thanks!
  $endgroup$
  – M. Nestor
  Aug 9 at 5:17
  

  
  
  
  

add a comment | 

7

$begingroup$

The answer is yes. For $min [0,1]$ consider the set $A:=[0,m]cup(mathbbQcap [0,1])$.
This is clearly dense and has measure $m$.

share|cite|improve this answer

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Great! However, this set is quite "lop-sided". I am wondering if there is a more evenly distributed example.
  $endgroup$
  – M. Nestor
  Aug 9 at 4:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then I refer to the answer of saz. Take an evenly distributed set of Lebesgue measure $m$ and "add" $mathbbQcap [0,1]$.
  $endgroup$
  – Jonas Lenz
  Aug 9 at 5:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Nestor Depends on what you mean by evenly distributed. It would be cool if, given a constant $0 < c < 1$, it were possible to find a measurable set $A$ such that $m(A cap I) = c m(I)$ for every interval $I subseteq [0,1]$. This turns out to be impossible
  $endgroup$
  – Bungo
  Aug 9 at 5:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bungo After reading that post, it seems that what I am after is precisely Lebesgue's density theorem. Thanks!
  $endgroup$
  – M. Nestor
  Aug 9 at 5:17
  

  
  
  
  

add a comment | 

$begingroup$

The answer is yes. For $min [0,1]$ consider the set $A:=[0,m]cup(mathbbQcap [0,1])$.
This is clearly dense and has measure $m$.

share|cite|improve this answer

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

answered Aug 9 at 4:58

Jonas LenzJonas Lenz

1,03411 gold badge44 silver badges1919 bronze badges

5

$begingroup$

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider

$$A := B cup (mathbbQ cap [0,1]).$$

The set has Lebesgue measure $m$ and it is dense in $[0,1]$.

It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_n in mathbbN$ of $mathbbQ cap [0,1]$ and $$A := bigcup_n in mathbbN (q_n-epsilon 2^-n,q_n+epsilon 2^-n)$$ for fixed $epsilon>0$. The set $A$ is open and has Lebesgue measure $leq epsilon$.

Remark: Note that there does not exist an open dense set with Lebesgue measure zero. In this sense, the best we can achieve is to have an open dense set of arbitrarily small Lebesgue measure, as above.

share|cite|improve this answer

edited Aug 9 at 5:08

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the 2nd example, the set must be less than $epsilon$ since there are some overlaps between those intervals, right?
  $endgroup$
  – efhvcjnfdbgefg
  Aug 9 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @efhvcjnfdbgefg Yes, that's correct. There are plenty of overlaps (... exactly because the rationals are dense).
  $endgroup$
  – saz
  Aug 9 at 5:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact $varepsilon$ must always be an overestimate, since any interval $(q_n-2^-nvarepsilon,q_n+2^-nvarepsilon)$ must contain another $q_m$, guaranteeing overlap at least $minvarepsilon2^-m,varepsilon2^-n$.
  $endgroup$
  – M. Nestor
  Aug 9 at 5:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My above comment doesn't change the validity, since the overestimate varies continuously, you can always find $varepsilon$ such that the true measure is precisely $m$.
  $endgroup$
  – M. Nestor
  Aug 13 at 17:47
  

  
  
  
  

add a comment | 

5

$begingroup$

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider

$$A := B cup (mathbbQ cap [0,1]).$$

The set has Lebesgue measure $m$ and it is dense in $[0,1]$.

It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_n in mathbbN$ of $mathbbQ cap [0,1]$ and $$A := bigcup_n in mathbbN (q_n-epsilon 2^-n,q_n+epsilon 2^-n)$$ for fixed $epsilon>0$. The set $A$ is open and has Lebesgue measure $leq epsilon$.

Remark: Note that there does not exist an open dense set with Lebesgue measure zero. In this sense, the best we can achieve is to have an open dense set of arbitrarily small Lebesgue measure, as above.

share|cite|improve this answer

edited Aug 9 at 5:08

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the 2nd example, the set must be less than $epsilon$ since there are some overlaps between those intervals, right?
  $endgroup$
  – efhvcjnfdbgefg
  Aug 9 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @efhvcjnfdbgefg Yes, that's correct. There are plenty of overlaps (... exactly because the rationals are dense).
  $endgroup$
  – saz
  Aug 9 at 5:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact $varepsilon$ must always be an overestimate, since any interval $(q_n-2^-nvarepsilon,q_n+2^-nvarepsilon)$ must contain another $q_m$, guaranteeing overlap at least $minvarepsilon2^-m,varepsilon2^-n$.
  $endgroup$
  – M. Nestor
  Aug 9 at 5:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My above comment doesn't change the validity, since the overestimate varies continuously, you can always find $varepsilon$ such that the true measure is precisely $m$.
  $endgroup$
  – M. Nestor
  Aug 13 at 17:47
  

  
  
  
  

add a comment | 

$begingroup$

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider

$$A := B cup (mathbbQ cap [0,1]).$$

The set has Lebesgue measure $m$ and it is dense in $[0,1]$.

It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_n in mathbbN$ of $mathbbQ cap [0,1]$ and $$A := bigcup_n in mathbbN (q_n-epsilon 2^-n,q_n+epsilon 2^-n)$$ for fixed $epsilon>0$. The set $A$ is open and has Lebesgue measure $leq epsilon$.

Remark: Note that there does not exist an open dense set with Lebesgue measure zero. In this sense, the best we can achieve is to have an open dense set of arbitrarily small Lebesgue measure, as above.

share|cite|improve this answer

edited Aug 9 at 5:08

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 9 at 5:08

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

answered Aug 9 at 4:58

sazsaz

87.3k88 gold badges7070 silver badges136136 bronze badges

2

$begingroup$

Sure. Take
$$
[0,m]cup(mathbbQcap [0,1])=[0,m]oversetcdotcup(mathbbQcap(m,1]).
$$

share|cite|improve this answer

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Sure. Take
$$
[0,m]cup(mathbbQcap [0,1])=[0,m]oversetcdotcup(mathbbQcap(m,1]).
$$

share|cite|improve this answer

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

$endgroup$

add a comment | 

$begingroup$

Sure. Take
$$
[0,m]cup(mathbbQcap [0,1])=[0,m]oversetcdotcup(mathbbQcap(m,1]).
$$

share|cite|improve this answer

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges

answered Aug 9 at 4:58

Nick PetersonNick Peterson

27.2k22 gold badges4040 silver badges6262 bronze badges