Mixing
A question about the Vitali set.
$begingroup$
Let $X=[0,1)$ and we define on $X$ the relation $sim$ in the following way:$$xsim yquadiffquad x-yinmathbb{Q}.$$ The relation $sim$ is an equivalence relation. For all $alphain[0,1)$ the equivalent class is $$E_{alpha}:={yin[0,1);|;y-alphainmathbb{Q}}.$$ The class $E_alpha$ is countable for all $alphain[0,1)$, indeed if $alphainmathbb{Q}cap[0,1)$, $E_alpha=mathbb{Q}cap [0,1)$, while if $alphainbig(mathbb{R}setminusmathbb{Q}big)cap [0,1)$, $yin E_alpha$ if and only if $y=(alpha+q)$, where $qinmathbb{Q}cap [0,1).$
Question 1. Is my reasoning correct to show that each class $E_alpha$ is countable?
Therefore $$[0,1)=bigcup_{alphain[0,1)} E_alpha.tag1$$ From $(1)$ it is clear that exists an uncountable infinity of classes $E_alpha.$
For the axiom of choice exists $Vsubseteq [0,1)$ such that $$Vcap E_alpha={x_alpha}quadtext{for all};alphain [0,1).$$ The set $V$ is called the Vitali set. Let ${q_n}_{ninmathbb{N}}$ an enumeration of $mathbb{Q}cap [0,1)$ and we define for all $ninmathbb{N}$ $$V_n:={x_alpha+q_n;text{mod};1;|;x_alphain V}.$$I must prove that $$[0,1)=bigcup_{ninmathbb{N}} V_n.$$ Let $xinbigcup_{n} V_n$ then $xin V_n$ for same $ninmathbb{N}$, therefore $x=x_{alpha}+ q_n;text{mod};1$, where $x_alphain V$, then $xin [0,1).$ Let's proceed with the other inclusion: let $xin [0,1)$ then $xin E_alpha$ for same $alphain [0,1)$, in particular, since the classes $E_alpha$ are disjointed $alpha$ is unique, then $$x-x_alphainmathbb{Q},$$ in particular, since $xin [0,1)$ and $x_alphain [0,1)$ we have that $$x-x_alphainmathbb{Q}cap (-1,1)tag2$$
Question. From $(2)$ how can I proceed to show inclusion?
Thanks!
measure-theory proof-verification elementary-set-theory lebesgue-measure
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
$endgroup$
add a comment |
$begingroup$
Let $X=[0,1)$ and we define on $X$ the relation $sim$ in the following way:$$xsim yquadiffquad x-yinmathbb{Q}.$$ The relation $sim$ is an equivalence relation. For all $alphain[0,1)$ the equivalent class is $$E_{alpha}:={yin[0,1);|;y-alphainmathbb{Q}}.$$ The class $E_alpha$ is countable for all $alphain[0,1)$, indeed if $alphainmathbb{Q}cap[0,1)$, $E_alpha=mathbb{Q}cap [0,1)$, while if $alphainbig(mathbb{R}setminusmathbb{Q}big)cap [0,1)$, $yin E_alpha$ if and only if $y=(alpha+q)$, where $qinmathbb{Q}cap [0,1).$
Question 1. Is my reasoning correct to show that each class $E_alpha$ is countable?
Therefore $$[0,1)=bigcup_{alphain[0,1)} E_alpha.tag1$$ From $(1)$ it is clear that exists an uncountable infinity of classes $E_alpha.$
For the axiom of choice exists $Vsubseteq [0,1)$ such that $$Vcap E_alpha={x_alpha}quadtext{for all};alphain [0,1).$$ The set $V$ is called the Vitali set. Let ${q_n}_{ninmathbb{N}}$ an enumeration of $mathbb{Q}cap [0,1)$ and we define for all $ninmathbb{N}$ $$V_n:={x_alpha+q_n;text{mod};1;|;x_alphain V}.$$I must prove that $$[0,1)=bigcup_{ninmathbb{N}} V_n.$$ Let $xinbigcup_{n} V_n$ then $xin V_n$ for same $ninmathbb{N}$, therefore $x=x_{alpha}+ q_n;text{mod};1$, where $x_alphain V$, then $xin [0,1).$ Let's proceed with the other inclusion: let $xin [0,1)$ then $xin E_alpha$ for same $alphain [0,1)$, in particular, since the classes $E_alpha$ are disjointed $alpha$ is unique, then $$x-x_alphainmathbb{Q},$$ in particular, since $xin [0,1)$ and $x_alphain [0,1)$ we have that $$x-x_alphainmathbb{Q}cap (-1,1)tag2$$
Question. From $(2)$ how can I proceed to show inclusion?
Thanks!
measure-theory proof-verification elementary-set-theory lebesgue-measure
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
$endgroup$
add a comment |
$begingroup$
Let $X=[0,1)$ and we define on $X$ the relation $sim$ in the following way:$$xsim yquadiffquad x-yinmathbb{Q}.$$ The relation $sim$ is an equivalence relation. For all $alphain[0,1)$ the equivalent class is $$E_{alpha}:={yin[0,1);|;y-alphainmathbb{Q}}.$$ The class $E_alpha$ is countable for all $alphain[0,1)$, indeed if $alphainmathbb{Q}cap[0,1)$, $E_alpha=mathbb{Q}cap [0,1)$, while if $alphainbig(mathbb{R}setminusmathbb{Q}big)cap [0,1)$, $yin E_alpha$ if and only if $y=(alpha+q)$, where $qinmathbb{Q}cap [0,1).$
Question 1. Is my reasoning correct to show that each class $E_alpha$ is countable?
Therefore $$[0,1)=bigcup_{alphain[0,1)} E_alpha.tag1$$ From $(1)$ it is clear that exists an uncountable infinity of classes $E_alpha.$
For the axiom of choice exists $Vsubseteq [0,1)$ such that $$Vcap E_alpha={x_alpha}quadtext{for all};alphain [0,1).$$ The set $V$ is called the Vitali set. Let ${q_n}_{ninmathbb{N}}$ an enumeration of $mathbb{Q}cap [0,1)$ and we define for all $ninmathbb{N}$ $$V_n:={x_alpha+q_n;text{mod};1;|;x_alphain V}.$$I must prove that $$[0,1)=bigcup_{ninmathbb{N}} V_n.$$ Let $xinbigcup_{n} V_n$ then $xin V_n$ for same $ninmathbb{N}$, therefore $x=x_{alpha}+ q_n;text{mod};1$, where $x_alphain V$, then $xin [0,1).$ Let's proceed with the other inclusion: let $xin [0,1)$ then $xin E_alpha$ for same $alphain [0,1)$, in particular, since the classes $E_alpha$ are disjointed $alpha$ is unique, then $$x-x_alphainmathbb{Q},$$ in particular, since $xin [0,1)$ and $x_alphain [0,1)$ we have that $$x-x_alphainmathbb{Q}cap (-1,1)tag2$$
Question. From $(2)$ how can I proceed to show inclusion?
Thanks!
measure-theory proof-verification elementary-set-theory lebesgue-measure
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
$endgroup$
Let $X=[0,1)$ and we define on $X$ the relation $sim$ in the following way:$$xsim yquadiffquad x-yinmathbb{Q}.$$ The relation $sim$ is an equivalence relation. For all $alphain[0,1)$ the equivalent class is $$E_{alpha}:={yin[0,1);|;y-alphainmathbb{Q}}.$$ The class $E_alpha$ is countable for all $alphain[0,1)$, indeed if $alphainmathbb{Q}cap[0,1)$, $E_alpha=mathbb{Q}cap [0,1)$, while if $alphainbig(mathbb{R}setminusmathbb{Q}big)cap [0,1)$, $yin E_alpha$ if and only if $y=(alpha+q)$, where $qinmathbb{Q}cap [0,1).$
Question 1. Is my reasoning correct to show that each class $E_alpha$ is countable?
Therefore $$[0,1)=bigcup_{alphain[0,1)} E_alpha.tag1$$ From $(1)$ it is clear that exists an uncountable infinity of classes $E_alpha.$
For the axiom of choice exists $Vsubseteq [0,1)$ such that $$Vcap E_alpha={x_alpha}quadtext{for all};alphain [0,1).$$ The set $V$ is called the Vitali set. Let ${q_n}_{ninmathbb{N}}$ an enumeration of $mathbb{Q}cap [0,1)$ and we define for all $ninmathbb{N}$ $$V_n:={x_alpha+q_n;text{mod};1;|;x_alphain V}.$$I must prove that $$[0,1)=bigcup_{ninmathbb{N}} V_n.$$ Let $xinbigcup_{n} V_n$ then $xin V_n$ for same $ninmathbb{N}$, therefore $x=x_{alpha}+ q_n;text{mod};1$, where $x_alphain V$, then $xin [0,1).$ Let's proceed with the other inclusion: let $xin [0,1)$ then $xin E_alpha$ for same $alphain [0,1)$, in particular, since the classes $E_alpha$ are disjointed $alpha$ is unique, then $$x-x_alphainmathbb{Q},$$ in particular, since $xin [0,1)$ and $x_alphain [0,1)$ we have that $$x-x_alphainmathbb{Q}cap (-1,1)tag2$$
Question. From $(2)$ how can I proceed to show inclusion?
Thanks!
measure-theory proof-verification elementary-set-theory lebesgue-measure
measure-theory proof-verification elementary-set-theory lebesgue-measure
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
edited Feb 1 at 16:09
José Carlos Santos
174k23133242
174k23133242
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
asked Feb 1 at 15:22
Jack J.Jack J.
3592419
3592419
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- It is not entirely correct. For instance, you claim that$$yin E_alphaiff y=alpha+qtext{, where }qinmathbb{Q}cap[0,1).$$What if $y<alpha$? Besides, there is no nedd to treat the cases $alphainmathbb Q$ and $alphanotinmathbb Q$ separately.
- Why induction? If $x-x_alpha=qinmathbb{Q}$, then $q-lfloor qrfloor=q_n$ for some $ninmathbb N$ (since $q-lfloor qrfloorinmathbb{Q}cap[0,1)$) and therefore $xin V_n$.
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
|
show 3 more comments
$begingroup$
One way to see countability of all classes is to note that the class of $xin mathbb{R}$ is just $[x]={x+q: q in mathbb{Q}}cap [0,1)$: every $x+q in [x]$ is equivalent to $x$, as $x+q-x = q in mathbb{Q}$ and if $x sim y$ then $q:=y-x in mathbb{Q}$ and $y = x+q in [x]$.
So every class is in bijective correspondence with $mathbb{Q}$ (via $q to x+q$ for the class of $x$).
$[0,1)$ is a partition of classes so (assuming choice) there are $mathfrak{c}=|mathbb{R}|$ many different classes; if there are $kappa ge aleph_0$ classes each of size $aleph_0$, $|[0,1)|$ has size $kappa cdot aleph_0=kappa$ as well. So $kappa=mathfrak{c}$.
There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).
To see that $[0,1) = bigcup_n (V+{q_n} pmod{1})$, let $x in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v in [x] cap V$ as a representative. So $x-v= q_k pmod{1}$ for some $k$ so $x in V+{q_k} pmod{1}$.
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- It is not entirely correct. For instance, you claim that$$yin E_alphaiff y=alpha+qtext{, where }qinmathbb{Q}cap[0,1).$$What if $y<alpha$? Besides, there is no nedd to treat the cases $alphainmathbb Q$ and $alphanotinmathbb Q$ separately.
- Why induction? If $x-x_alpha=qinmathbb{Q}$, then $q-lfloor qrfloor=q_n$ for some $ninmathbb N$ (since $q-lfloor qrfloorinmathbb{Q}cap[0,1)$) and therefore $xin V_n$.
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
|
show 3 more comments
$begingroup$
- It is not entirely correct. For instance, you claim that$$yin E_alphaiff y=alpha+qtext{, where }qinmathbb{Q}cap[0,1).$$What if $y<alpha$? Besides, there is no nedd to treat the cases $alphainmathbb Q$ and $alphanotinmathbb Q$ separately.
- Why induction? If $x-x_alpha=qinmathbb{Q}$, then $q-lfloor qrfloor=q_n$ for some $ninmathbb N$ (since $q-lfloor qrfloorinmathbb{Q}cap[0,1)$) and therefore $xin V_n$.
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
|
show 3 more comments
$begingroup$
- It is not entirely correct. For instance, you claim that$$yin E_alphaiff y=alpha+qtext{, where }qinmathbb{Q}cap[0,1).$$What if $y<alpha$? Besides, there is no nedd to treat the cases $alphainmathbb Q$ and $alphanotinmathbb Q$ separately.
- Why induction? If $x-x_alpha=qinmathbb{Q}$, then $q-lfloor qrfloor=q_n$ for some $ninmathbb N$ (since $q-lfloor qrfloorinmathbb{Q}cap[0,1)$) and therefore $xin V_n$.
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
- It is not entirely correct. For instance, you claim that$$yin E_alphaiff y=alpha+qtext{, where }qinmathbb{Q}cap[0,1).$$What if $y<alpha$? Besides, there is no nedd to treat the cases $alphainmathbb Q$ and $alphanotinmathbb Q$ separately.
- Why induction? If $x-x_alpha=qinmathbb{Q}$, then $q-lfloor qrfloor=q_n$ for some $ninmathbb N$ (since $q-lfloor qrfloorinmathbb{Q}cap[0,1)$) and therefore $xin V_n$.
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
edited Feb 1 at 16:02
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
answered Feb 1 at 15:31
José Carlos SantosJosé Carlos Santos
174k23133242
174k23133242
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
|
show 3 more comments
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
$begingroup$
Thanks for your answer. Why if $x-x_alpha=qinmathbb{Q}$, then $q=q_n$? $q_ninmathbb{Q}cap [0,1)$, while $x-x_alphain (-1,1).$ 1. At this point how can I show that every $E_alpha$ is countable?
$endgroup$
– Jack J.
Feb 1 at 15:58
1
1
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
I've edited my answer to the second doubt. What do you think now?
$endgroup$
– José Carlos Santos
Feb 1 at 16:02
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
So are we using the fact that there is a bjection between $mathbb{Q}$ and $mathbb{Q}cap [0,1)$?
$endgroup$
– Jack J.
Feb 1 at 16:44
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
$begingroup$
No, I did not use that.
$endgroup$
– José Carlos Santos
Feb 1 at 16:59
1
1
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Feb 1 at 18:50
|
show 3 more comments
$begingroup$
One way to see countability of all classes is to note that the class of $xin mathbb{R}$ is just $[x]={x+q: q in mathbb{Q}}cap [0,1)$: every $x+q in [x]$ is equivalent to $x$, as $x+q-x = q in mathbb{Q}$ and if $x sim y$ then $q:=y-x in mathbb{Q}$ and $y = x+q in [x]$.
So every class is in bijective correspondence with $mathbb{Q}$ (via $q to x+q$ for the class of $x$).
$[0,1)$ is a partition of classes so (assuming choice) there are $mathfrak{c}=|mathbb{R}|$ many different classes; if there are $kappa ge aleph_0$ classes each of size $aleph_0$, $|[0,1)|$ has size $kappa cdot aleph_0=kappa$ as well. So $kappa=mathfrak{c}$.
There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).
To see that $[0,1) = bigcup_n (V+{q_n} pmod{1})$, let $x in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v in [x] cap V$ as a representative. So $x-v= q_k pmod{1}$ for some $k$ so $x in V+{q_k} pmod{1}$.
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
add a comment |
$begingroup$
One way to see countability of all classes is to note that the class of $xin mathbb{R}$ is just $[x]={x+q: q in mathbb{Q}}cap [0,1)$: every $x+q in [x]$ is equivalent to $x$, as $x+q-x = q in mathbb{Q}$ and if $x sim y$ then $q:=y-x in mathbb{Q}$ and $y = x+q in [x]$.
So every class is in bijective correspondence with $mathbb{Q}$ (via $q to x+q$ for the class of $x$).
$[0,1)$ is a partition of classes so (assuming choice) there are $mathfrak{c}=|mathbb{R}|$ many different classes; if there are $kappa ge aleph_0$ classes each of size $aleph_0$, $|[0,1)|$ has size $kappa cdot aleph_0=kappa$ as well. So $kappa=mathfrak{c}$.
There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).
To see that $[0,1) = bigcup_n (V+{q_n} pmod{1})$, let $x in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v in [x] cap V$ as a representative. So $x-v= q_k pmod{1}$ for some $k$ so $x in V+{q_k} pmod{1}$.
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
add a comment |
$begingroup$
One way to see countability of all classes is to note that the class of $xin mathbb{R}$ is just $[x]={x+q: q in mathbb{Q}}cap [0,1)$: every $x+q in [x]$ is equivalent to $x$, as $x+q-x = q in mathbb{Q}$ and if $x sim y$ then $q:=y-x in mathbb{Q}$ and $y = x+q in [x]$.
So every class is in bijective correspondence with $mathbb{Q}$ (via $q to x+q$ for the class of $x$).
$[0,1)$ is a partition of classes so (assuming choice) there are $mathfrak{c}=|mathbb{R}|$ many different classes; if there are $kappa ge aleph_0$ classes each of size $aleph_0$, $|[0,1)|$ has size $kappa cdot aleph_0=kappa$ as well. So $kappa=mathfrak{c}$.
There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).
To see that $[0,1) = bigcup_n (V+{q_n} pmod{1})$, let $x in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v in [x] cap V$ as a representative. So $x-v= q_k pmod{1}$ for some $k$ so $x in V+{q_k} pmod{1}$.
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
One way to see countability of all classes is to note that the class of $xin mathbb{R}$ is just $[x]={x+q: q in mathbb{Q}}cap [0,1)$: every $x+q in [x]$ is equivalent to $x$, as $x+q-x = q in mathbb{Q}$ and if $x sim y$ then $q:=y-x in mathbb{Q}$ and $y = x+q in [x]$.
So every class is in bijective correspondence with $mathbb{Q}$ (via $q to x+q$ for the class of $x$).
$[0,1)$ is a partition of classes so (assuming choice) there are $mathfrak{c}=|mathbb{R}|$ many different classes; if there are $kappa ge aleph_0$ classes each of size $aleph_0$, $|[0,1)|$ has size $kappa cdot aleph_0=kappa$ as well. So $kappa=mathfrak{c}$.
There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).
To see that $[0,1) = bigcup_n (V+{q_n} pmod{1})$, let $x in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v in [x] cap V$ as a representative. So $x-v= q_k pmod{1}$ for some $k$ so $x in V+{q_k} pmod{1}$.
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:32
Henno BrandsmaHenno Brandsma
116k349127
116k349127
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
add a comment |
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
Let's see if I understand what you said in the first part: we fix $alphain [0,1)$ and we consider the map $fcolon mathbb{Q}to E_alpha$ defined as $qmapsto f(q):=q+alpha$. The map $f$ is well defined and injective. Now, let $yin E_alpha$, then $y-alphainmathbb{Q}$, therefore $q:=y-alpha$, then $y=q+alpha$. Since $q+alphasimalpha$, $y=f(q)in E_alpha$. Correct?
$endgroup$
– Jack J.
Feb 1 at 18:45
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
$begingroup$
@JackJ. Yes, that’s about it.
$endgroup$
– Henno Brandsma
Feb 1 at 19:20
add a comment |
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