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Equivalence relation on $mathbbR$: $xsim y iff x-yinmathbbQ$ - disjoint equivalence classes
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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
asked 2 days ago
rbird
87912
add a comment |Â
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
asked 2 days ago
rbird
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
add a comment |Â
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
asked 2 days ago
rbird
87912
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
asked 2 days ago
rbird
87912
asked 2 days ago
rbird
87912
asked 2 days ago
rbird
87912
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
add a comment |Â
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
add a comment |Â
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 $A$.
answered 2 days ago
quasi
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
 |Â
show 2 more comments
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 $A$.
answered 2 days ago
quasi
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
 |Â
show 2 more comments
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 $A$.
answered 2 days ago
quasi
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
 |Â
show 2 more comments
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 $A$.
answered 2 days ago
quasi
32.9k22359
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 $A$.
answered 2 days ago
quasi
32.9k22359
edited 2 days ago
answered 2 days ago
quasi
32.9k22359
answered 2 days ago
quasi
32.9k22359
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
 |Â
show 2 more comments
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
 |Â
show 2 more comments
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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