Equivalence relation on $mathbbR$: $xsim y iff x-yinmathbbQ$ - disjoint equivalence classes

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0
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Consider the equivalence relation on $mathbbR$ given by $xsim y iff x-yinmathbbQ$.



Several questions about this have been asked before, but I could not find anywhere what the disjoint equivalence classes of $mathbbR/sim$ are.



I know that the equivalence classes are of the form
$$[a]=a+q:qinmathbbQ$$
where $a$ is irrational.



But how would I go about writing down $mathbbR$ as a union of disjoint equivalence classes? (I believe this is possible for any equivalence relation on any set).



I want to write $$mathbbR=mathbbQsqcup bigsqcup_textsome condition on $a$[a],.$$



I know that the condition cannot be $anotinmathbbQ$, because then, for example, $[pi+1]=[pi]$, so the union is not disjoint.



What condition on $a$ makes this work? Is it $ain[0,1):anotinmathbbQ$? If so, can anyone give a hint for how to prove that these are all disjoint?




elementary-set-theory equivalence-relations




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  • 2




    Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
    – quasi
    2 days ago











  • It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
    – Dog_69
    2 days ago















up vote
0
down vote

favorite












Consider the equivalence relation on $mathbbR$ given by $xsim y iff x-yinmathbbQ$.



Several questions about this have been asked before, but I could not find anywhere what the disjoint equivalence classes of $mathbbR/sim$ are.



I know that the equivalence classes are of the form
$$[a]=a+q:qinmathbbQ$$
where $a$ is irrational.



But how would I go about writing down $mathbbR$ as a union of disjoint equivalence classes? (I believe this is possible for any equivalence relation on any set).



I want to write $$mathbbR=mathbbQsqcup bigsqcup_textsome condition on $a$[a],.$$



I know that the condition cannot be $anotinmathbbQ$, because then, for example, $[pi+1]=[pi]$, so the union is not disjoint.



What condition on $a$ makes this work? Is it $ain[0,1):anotinmathbbQ$? If so, can anyone give a hint for how to prove that these are all disjoint?




elementary-set-theory equivalence-relations




share|cite|improve this question















  • 2




    Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
    – quasi
    2 days ago











  • It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
    – Dog_69
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider the equivalence relation on $mathbbR$ given by $xsim y iff x-yinmathbbQ$.



Several questions about this have been asked before, but I could not find anywhere what the disjoint equivalence classes of $mathbbR/sim$ are.



I know that the equivalence classes are of the form
$$[a]=a+q:qinmathbbQ$$
where $a$ is irrational.



But how would I go about writing down $mathbbR$ as a union of disjoint equivalence classes? (I believe this is possible for any equivalence relation on any set).



I want to write $$mathbbR=mathbbQsqcup bigsqcup_textsome condition on $a$[a],.$$



I know that the condition cannot be $anotinmathbbQ$, because then, for example, $[pi+1]=[pi]$, so the union is not disjoint.



What condition on $a$ makes this work? Is it $ain[0,1):anotinmathbbQ$? If so, can anyone give a hint for how to prove that these are all disjoint?




elementary-set-theory equivalence-relations




share|cite|improve this question











Consider the equivalence relation on $mathbbR$ given by $xsim y iff x-yinmathbbQ$.



Several questions about this have been asked before, but I could not find anywhere what the disjoint equivalence classes of $mathbbR/sim$ are.



I know that the equivalence classes are of the form
$$[a]=a+q:qinmathbbQ$$
where $a$ is irrational.



But how would I go about writing down $mathbbR$ as a union of disjoint equivalence classes? (I believe this is possible for any equivalence relation on any set).



I want to write $$mathbbR=mathbbQsqcup bigsqcup_textsome condition on $a$[a],.$$



I know that the condition cannot be $anotinmathbbQ$, because then, for example, $[pi+1]=[pi]$, so the union is not disjoint.



What condition on $a$ makes this work? Is it $ain[0,1):anotinmathbbQ$? If so, can anyone give a hint for how to prove that these are all disjoint?





elementary-set-theory equivalence-relations





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share|cite|improve this question




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asked 2 days ago









rbird

87912




87912







  • 2




    Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
    – quasi
    2 days ago











  • It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
    – Dog_69
    2 days ago













  • 2




    Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
    – quasi
    2 days ago











  • It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
    – Dog_69
    2 days ago








2




2




Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
– quasi
2 days ago





Note: If $ain(0,1)$ with $anotin mathbbQ$, then $asim a-largefrac1n$, for all positive integers $n$, and $a-largefrac1nin(0,1)$, for all sufficiently large positive integers $n$.
– quasi
2 days ago













It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
– Dog_69
2 days ago





It is quite tautological but if $I$ denotes the set of irrational numbers and $sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/sim$ you could get your goal.
– Dog_69
2 days ago











1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.



Unfortunately, that's not possible.



Suppose $A$ is a set of unique representatives for the set of equivalence classes.



If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.



But one can show that for any such set $A$, the set $Acap mathbbR$ is non-measurable.



Since the assertion that there exist non-measurable subsets of $mathbbR$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.






share|cite|improve this answer























  • I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
    – Gastón Burrull
    2 days ago










  • Well, I said "Suppose $A$ is ..."
    – quasi
    2 days ago











  • I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
    – Gastón Burrull
    2 days ago










  • I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
    – quasi
    2 days ago







  • 1




    Ok, I see your point. I'll edit my answer . . .
    – quasi
    2 days ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.



Unfortunately, that's not possible.



Suppose $A$ is a set of unique representatives for the set of equivalence classes.



If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.



But one can show that for any such set $A$, the set $Acap mathbbR$ is non-measurable.



Since the assertion that there exist non-measurable subsets of $mathbbR$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.






share|cite|improve this answer























  • I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
    – Gastón Burrull
    2 days ago










  • Well, I said "Suppose $A$ is ..."
    – quasi
    2 days ago











  • I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
    – Gastón Burrull
    2 days ago










  • I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
    – quasi
    2 days ago







  • 1




    Ok, I see your point. I'll edit my answer . . .
    – quasi
    2 days ago















up vote
3
down vote



accepted










It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.



Unfortunately, that's not possible.



Suppose $A$ is a set of unique representatives for the set of equivalence classes.



If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.



But one can show that for any such set $A$, the set $Acap mathbbR$ is non-measurable.



Since the assertion that there exist non-measurable subsets of $mathbbR$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.






share|cite|improve this answer























  • I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
    – Gastón Burrull
    2 days ago










  • Well, I said "Suppose $A$ is ..."
    – quasi
    2 days ago











  • I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
    – Gastón Burrull
    2 days ago










  • I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
    – quasi
    2 days ago







  • 1




    Ok, I see your point. I'll edit my answer . . .
    – quasi
    2 days ago













up vote
3
down vote



accepted







up vote
3
down vote



accepted






It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.



Unfortunately, that's not possible.



Suppose $A$ is a set of unique representatives for the set of equivalence classes.



If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.



But one can show that for any such set $A$, the set $Acap mathbbR$ is non-measurable.



Since the assertion that there exist non-measurable subsets of $mathbbR$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.






share|cite|improve this answer















It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.



Unfortunately, that's not possible.



Suppose $A$ is a set of unique representatives for the set of equivalence classes.



If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.



But one can show that for any such set $A$, the set $Acap mathbbR$ is non-measurable.



Since the assertion that there exist non-measurable subsets of $mathbbR$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago


























answered 2 days ago









quasi

32.9k22359




32.9k22359











  • I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
    – Gastón Burrull
    2 days ago










  • Well, I said "Suppose $A$ is ..."
    – quasi
    2 days ago











  • I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
    – Gastón Burrull
    2 days ago










  • I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
    – quasi
    2 days ago







  • 1




    Ok, I see your point. I'll edit my answer . . .
    – quasi
    2 days ago

















  • I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
    – Gastón Burrull
    2 days ago










  • Well, I said "Suppose $A$ is ..."
    – quasi
    2 days ago











  • I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
    – Gastón Burrull
    2 days ago










  • I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
    – quasi
    2 days ago







  • 1




    Ok, I see your point. I'll edit my answer . . .
    – quasi
    2 days ago
















I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
– Gastón Burrull
2 days ago




I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives."
– Gastón Burrull
2 days ago












Well, I said "Suppose $A$ is ..."
– quasi
2 days ago





Well, I said "Suppose $A$ is ..."
– quasi
2 days ago













I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
– Gastón Burrull
2 days ago




I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true.
– Gastón Burrull
2 days ago












I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
– quasi
2 days ago





I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice.
– quasi
2 days ago





1




1




Ok, I see your point. I'll edit my answer . . .
– quasi
2 days ago





Ok, I see your point. I'll edit my answer . . .
– quasi
2 days ago













 

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