Mixing
Linear order of the quotient generated from Vitali relation implies non-measurability of subset of reals
$begingroup$
Vitali relation, $a,binBbb R; asim biff a-binBbb Q$, is used to prove that there exists a non-measurable set of reals: we look at $A=Bbb R/sim$, from each $ain A$ we take $b_ain a$, then $bigcup_{ain A}{b_a}$ is the desired set.
This construction require choice function for the power set of $Bbb R$, and without some kind of axiom of choice there is no guarantee such function exists(it can be seen by, for example, Solovay's model).
I remember reading the following:
In $sf ZF$, $Bbb R/sim$ has a linear order implies the existence of of non-measurable subset of the reals.
This fact will also imply that compactness implies non-measurable set of reals, so my question is: what proof is there for the above theorem?
measure-theory set-theory lebesgue-measure axiom-of-choice
asked Jan 1 at 19:42
HoloHolo
5,60321030
$endgroup$
|
show 3 more comments
$begingroup$
Vitali relation, $a,binBbb R; asim biff a-binBbb Q$, is used to prove that there exists a non-measurable set of reals: we look at $A=Bbb R/sim$, from each $ain A$ we take $b_ain a$, then $bigcup_{ain A}{b_a}$ is the desired set.
This construction require choice function for the power set of $Bbb R$, and without some kind of axiom of choice there is no guarantee such function exists(it can be seen by, for example, Solovay's model).
I remember reading the following:
In $sf ZF$, $Bbb R/sim$ has a linear order implies the existence of of non-measurable subset of the reals.
This fact will also imply that compactness implies non-measurable set of reals, so my question is: what proof is there for the above theorem?
measure-theory set-theory lebesgue-measure axiom-of-choice
asked Jan 1 at 19:42
HoloHolo
5,60321030
$endgroup$
$begingroup$
For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:57
$begingroup$
mathoverflow.net/questions/26861/…
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:58
$begingroup$
@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
$endgroup$
– Holo
Jan 1 at 20:21
$begingroup$
It is a proof of the theorem you ask for.
$endgroup$
– Asaf Karagila♦
Jan 1 at 20:25
$begingroup$
@AsafKaragila oh, thanks! I will look at it soon!
$endgroup$
– Holo
Jan 1 at 20:28
|
show 3 more comments
$begingroup$
Vitali relation, $a,binBbb R; asim biff a-binBbb Q$, is used to prove that there exists a non-measurable set of reals: we look at $A=Bbb R/sim$, from each $ain A$ we take $b_ain a$, then $bigcup_{ain A}{b_a}$ is the desired set.
This construction require choice function for the power set of $Bbb R$, and without some kind of axiom of choice there is no guarantee such function exists(it can be seen by, for example, Solovay's model).
I remember reading the following:
In $sf ZF$, $Bbb R/sim$ has a linear order implies the existence of of non-measurable subset of the reals.
This fact will also imply that compactness implies non-measurable set of reals, so my question is: what proof is there for the above theorem?
measure-theory set-theory lebesgue-measure axiom-of-choice
asked Jan 1 at 19:42
HoloHolo
5,60321030
$endgroup$
Vitali relation, $a,binBbb R; asim biff a-binBbb Q$, is used to prove that there exists a non-measurable set of reals: we look at $A=Bbb R/sim$, from each $ain A$ we take $b_ain a$, then $bigcup_{ain A}{b_a}$ is the desired set.
This construction require choice function for the power set of $Bbb R$, and without some kind of axiom of choice there is no guarantee such function exists(it can be seen by, for example, Solovay's model).
I remember reading the following:
In $sf ZF$, $Bbb R/sim$ has a linear order implies the existence of of non-measurable subset of the reals.
This fact will also imply that compactness implies non-measurable set of reals, so my question is: what proof is there for the above theorem?
measure-theory set-theory lebesgue-measure axiom-of-choice
measure-theory set-theory lebesgue-measure axiom-of-choice
asked Jan 1 at 19:42
HoloHolo
5,60321030
asked Jan 1 at 19:42
HoloHolo
5,60321030
asked Jan 1 at 19:42
HoloHolo
5,60321030
asked Jan 1 at 19:42
HoloHolo
5,60321030
asked Jan 1 at 19:42
HoloHolo
5,60321030
5,60321030
$begingroup$
For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:57
$begingroup$
mathoverflow.net/questions/26861/…
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:58
$begingroup$
@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
$endgroup$
– Holo
Jan 1 at 20:21
$begingroup$
It is a proof of the theorem you ask for.
$endgroup$
– Asaf Karagila♦
Jan 1 at 20:25
$begingroup$
@AsafKaragila oh, thanks! I will look at it soon!
$endgroup$
– Holo
Jan 1 at 20:28
|
show 3 more comments
$begingroup$
For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:57
$begingroup$
mathoverflow.net/questions/26861/…
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:58
$begingroup$
@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
$endgroup$
– Holo
Jan 1 at 20:21
$begingroup$
It is a proof of the theorem you ask for.
$endgroup$
– Asaf Karagila♦
Jan 1 at 20:25
$begingroup$
@AsafKaragila oh, thanks! I will look at it soon!
$endgroup$
– Holo
Jan 1 at 20:28
$begingroup$
For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:57
$begingroup$
For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:57
$begingroup$
mathoverflow.net/questions/26861/…
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:58
$begingroup$
mathoverflow.net/questions/26861/…
$endgroup$
– Asaf Karagila♦
Jan 1 at 19:58
$begingroup$
@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
$endgroup$
– Holo
Jan 1 at 20:21
$begingroup$
@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
$endgroup$
– Holo
Jan 1 at 20:21
$begingroup$
It is a proof of the theorem you ask for.
$endgroup$
– Asaf Karagila♦
Jan 1 at 20:25
$begingroup$
It is a proof of the theorem you ask for.
$endgroup$
– Asaf Karagila♦
Jan 1 at 20:25
$begingroup$
@AsafKaragila oh, thanks! I will look at it soon!
$endgroup$
– Holo
Jan 1 at 20:28
$begingroup$
@AsafKaragila oh, thanks! I will look at it soon!
$endgroup$
– Holo
Jan 1 at 20:28
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Taking help from @Noah comment and from @Wojowu:
Let's assume the contrary, that is, $mathbb R/sim$ is linearly ordered but every subset $mathbb R^2$ is measurable we may use the ergodicity:
Let $prec$ be the linear order of $mathbb R/sim$, and define $xtriangleleft yiff [x]_simprec[y]_sim$. Then either $triangleleft$ has measure $0$ or $triangleleft^c={(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}$.
We may notice that ${(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}=bigcup_{qinmathbb Q}{(x,y)∈mathbb R^2mid x-y=q}$, which is countable union of lines, so has measure $0$, and ${(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}$ has the same measure as $triangleleft$, so in both cases we get that the measure of $mathbb R^2$ is $0$, which is contradiction.
answered Jan 2 at 10:48
HoloHolo
5,60321030
$endgroup$
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
add a comment |
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1 Answer
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$begingroup$
Taking help from @Noah comment and from @Wojowu:
Let's assume the contrary, that is, $mathbb R/sim$ is linearly ordered but every subset $mathbb R^2$ is measurable we may use the ergodicity:
Let $prec$ be the linear order of $mathbb R/sim$, and define $xtriangleleft yiff [x]_simprec[y]_sim$. Then either $triangleleft$ has measure $0$ or $triangleleft^c={(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}$.
We may notice that ${(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}=bigcup_{qinmathbb Q}{(x,y)∈mathbb R^2mid x-y=q}$, which is countable union of lines, so has measure $0$, and ${(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}$ has the same measure as $triangleleft$, so in both cases we get that the measure of $mathbb R^2$ is $0$, which is contradiction.
answered Jan 2 at 10:48
HoloHolo
5,60321030
$endgroup$
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
add a comment |
$begingroup$
Taking help from @Noah comment and from @Wojowu:
Let's assume the contrary, that is, $mathbb R/sim$ is linearly ordered but every subset $mathbb R^2$ is measurable we may use the ergodicity:
Let $prec$ be the linear order of $mathbb R/sim$, and define $xtriangleleft yiff [x]_simprec[y]_sim$. Then either $triangleleft$ has measure $0$ or $triangleleft^c={(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}$.
We may notice that ${(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}=bigcup_{qinmathbb Q}{(x,y)∈mathbb R^2mid x-y=q}$, which is countable union of lines, so has measure $0$, and ${(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}$ has the same measure as $triangleleft$, so in both cases we get that the measure of $mathbb R^2$ is $0$, which is contradiction.
answered Jan 2 at 10:48
HoloHolo
5,60321030
$endgroup$
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
add a comment |
$begingroup$
Taking help from @Noah comment and from @Wojowu:
Let's assume the contrary, that is, $mathbb R/sim$ is linearly ordered but every subset $mathbb R^2$ is measurable we may use the ergodicity:
Let $prec$ be the linear order of $mathbb R/sim$, and define $xtriangleleft yiff [x]_simprec[y]_sim$. Then either $triangleleft$ has measure $0$ or $triangleleft^c={(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}$.
We may notice that ${(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}=bigcup_{qinmathbb Q}{(x,y)∈mathbb R^2mid x-y=q}$, which is countable union of lines, so has measure $0$, and ${(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}$ has the same measure as $triangleleft$, so in both cases we get that the measure of $mathbb R^2$ is $0$, which is contradiction.
answered Jan 2 at 10:48
HoloHolo
5,60321030
$endgroup$
Taking help from @Noah comment and from @Wojowu:
Let's assume the contrary, that is, $mathbb R/sim$ is linearly ordered but every subset $mathbb R^2$ is measurable we may use the ergodicity:
Let $prec$ be the linear order of $mathbb R/sim$, and define $xtriangleleft yiff [x]_simprec[y]_sim$. Then either $triangleleft$ has measure $0$ or $triangleleft^c={(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}$.
We may notice that ${(a,b)∈mathbb R^2mid [a]_sim=[b]_sim}=bigcup_{qinmathbb Q}{(x,y)∈mathbb R^2mid x-y=q}$, which is countable union of lines, so has measure $0$, and ${(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}$ has the same measure as $triangleleft$, so in both cases we get that the measure of $mathbb R^2$ is $0$, which is contradiction.
answered Jan 2 at 10:48
HoloHolo
5,60321030
edited Jan 9 at 8:17
answered Jan 2 at 10:48
HoloHolo
5,60321030
answered Jan 2 at 10:48
HoloHolo
5,60321030
answered Jan 2 at 10:48
HoloHolo
5,60321030
5,60321030
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
add a comment |
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
My measure theory is a bit rusty, but why does it have to be the case that the preorder you define is either null or co-null? Moreover, just because $prec$ and $succ$ have the same measure doesn't imply contradiction since the measure space is not finite, but $sigma$-finite.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:12
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@AsafKaragila it is either null or co-null because it is closed under $Bbb Q^2$-translation, and by assumption it is measurable. And if $triangleleft$ has measure 0 then $μ(triangleleft∪triangleleft^c)=μ(triangleleft)+μ(triangleleft^c)=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim}cup{(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_simsucc[b]_sim})+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})=μ(triangleleft)+μ({(a,b)∈mathbb R^2mid [a]_sim=[b]_sim})$, those 2 has the same measure and at least one of them is $0$
$endgroup$
– Holo
Jan 2 at 11:39
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
$begingroup$
@Asaf the same measure alone is not enough, the ergodicity is crucial
$endgroup$
– Holo
Jan 2 at 11:45
add a comment |
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For compactness there is a direct route. Or rather, more direct. It implies the Ultrafilter Theorem. Then show that any free ultrafilter must violate the 0-1 Law in $2^Bbb N$ with its standard measure. Since all Polish spaces are Borel isomorphic, this implies there is a non-measurable set of reals. Or, if for you "reals" is just $2^Bbb N$ (because all Polish spaces are Borel isomorphic), then it implies that immediately.
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– Asaf Karagila♦
Jan 1 at 19:57
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mathoverflow.net/questions/26861/…
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– Asaf Karagila♦
Jan 1 at 19:58
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@AsafKaragila thanks for giving the reference, I will read it soon. But I still wish for proof for the above theorem if possible(as it is very interesting by itself)
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– Holo
Jan 1 at 20:21
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It is a proof of the theorem you ask for.
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– Asaf Karagila♦
Jan 1 at 20:25
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@AsafKaragila oh, thanks! I will look at it soon!
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– Holo
Jan 1 at 20:28