Equivalence relation by the symmetric difference of setsWhy isn't reflexivity redundant in the definition of equivalence relation?Show that the restriction of an equivalence relation is an equivalence relation.Equivalence-relations question.Equivalence relation $gsim h :Longleftrightarrow h in g,g^-1$Equivalence relation question with functionsProving something is an equivalence relationProving that rational equivalence is an equivalence relation on any set.Elementary set theory - challenging problem from relations and equivalence classesProve that the union of relations is an equivalence relationEquivalence Relation Requiring Set of all Sets
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Equivalence relation by the symmetric difference of sets
Why isn't reflexivity redundant in the definition of equivalence relation?Show that the restriction of an equivalence relation is an equivalence relation.Equivalence-relations question.Equivalence relation $gsim h :Longleftrightarrow h in g,g^-1$Equivalence relation question with functionsProving something is an equivalence relationProving that rational equivalence is an equivalence relation on any set.Elementary set theory - challenging problem from relations and equivalence classesProve that the union of relations is an equivalence relationEquivalence Relation Requiring Set of all Sets
$begingroup$
Let $A, B$ subsets of $X$ and $mathbb P(X)$ the power set,
we define the following equivalence relation on $mathbb P(X)$:
Let $ Ssubseteq X$ a fixed subset of $X$ and $A$~$B$ $iff A△B subseteq S$
Prove that is is an equivalence relation and find the class of $X$ and $S$
My work:
I have already shown that the relationship satisfies reflexivity and symmetry, all this is justified respectively by the fact that the empty set is a subset of any set and the symmetric difference is commutative.
My problem is with transitivity, I do not know how to do it, that is when I try to use it for the definition of symmetric difference I fall in many cases. There is some way to test transitivity using only operations between sets. And with respect to the equivalence class of $S$, I showed that they are all subsets of $X$ contained in $S$. But with respect to the equivalence class of $X4 I do not see what it is.
Any help would be useful. Thank you!
elementary-set-theory
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
$endgroup$
add a comment |
$begingroup$
Let $A, B$ subsets of $X$ and $mathbb P(X)$ the power set,
we define the following equivalence relation on $mathbb P(X)$:
Let $ Ssubseteq X$ a fixed subset of $X$ and $A$~$B$ $iff A△B subseteq S$
Prove that is is an equivalence relation and find the class of $X$ and $S$
My work:
I have already shown that the relationship satisfies reflexivity and symmetry, all this is justified respectively by the fact that the empty set is a subset of any set and the symmetric difference is commutative.
My problem is with transitivity, I do not know how to do it, that is when I try to use it for the definition of symmetric difference I fall in many cases. There is some way to test transitivity using only operations between sets. And with respect to the equivalence class of $S$, I showed that they are all subsets of $X$ contained in $S$. But with respect to the equivalence class of $X4 I do not see what it is.
Any help would be useful. Thank you!
elementary-set-theory
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
$endgroup$
$begingroup$
@EthanBolker You're right, my mistake!
$endgroup$
– Hendrik Matamoros
May 23 at 21:54
add a comment |
$begingroup$
Let $A, B$ subsets of $X$ and $mathbb P(X)$ the power set,
we define the following equivalence relation on $mathbb P(X)$:
Let $ Ssubseteq X$ a fixed subset of $X$ and $A$~$B$ $iff A△B subseteq S$
Prove that is is an equivalence relation and find the class of $X$ and $S$
My work:
I have already shown that the relationship satisfies reflexivity and symmetry, all this is justified respectively by the fact that the empty set is a subset of any set and the symmetric difference is commutative.
My problem is with transitivity, I do not know how to do it, that is when I try to use it for the definition of symmetric difference I fall in many cases. There is some way to test transitivity using only operations between sets. And with respect to the equivalence class of $S$, I showed that they are all subsets of $X$ contained in $S$. But with respect to the equivalence class of $X4 I do not see what it is.
Any help would be useful. Thank you!
elementary-set-theory
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
$endgroup$
Let $A, B$ subsets of $X$ and $mathbb P(X)$ the power set,
we define the following equivalence relation on $mathbb P(X)$:
Let $ Ssubseteq X$ a fixed subset of $X$ and $A$~$B$ $iff A△B subseteq S$
Prove that is is an equivalence relation and find the class of $X$ and $S$
My work:
I have already shown that the relationship satisfies reflexivity and symmetry, all this is justified respectively by the fact that the empty set is a subset of any set and the symmetric difference is commutative.
My problem is with transitivity, I do not know how to do it, that is when I try to use it for the definition of symmetric difference I fall in many cases. There is some way to test transitivity using only operations between sets. And with respect to the equivalence class of $S$, I showed that they are all subsets of $X$ contained in $S$. But with respect to the equivalence class of $X4 I do not see what it is.
Any help would be useful. Thank you!
elementary-set-theory
elementary-set-theory
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
edited May 23 at 21:54
Hendrik Matamoros
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
asked May 23 at 21:43
Hendrik MatamorosHendrik Matamoros
460310
460310
$begingroup$
@EthanBolker You're right, my mistake!
$endgroup$
– Hendrik Matamoros
May 23 at 21:54
add a comment |
$begingroup$
@EthanBolker You're right, my mistake!
$endgroup$
– Hendrik Matamoros
May 23 at 21:54
$begingroup$
@EthanBolker You're right, my mistake!
$endgroup$
– Hendrik Matamoros
May 23 at 21:54
$begingroup$
@EthanBolker You're right, my mistake!
$endgroup$
– Hendrik Matamoros
May 23 at 21:54
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Two hints about ways to go.
- Draw a Venn diagram for $A, B, C, X$ showing $A Delta B$ and so on.
- Know or show that $Delta$ is associative. That and the fact that $B
Delta B$ is empty leads to an algebraic proof.
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
$endgroup$
add a comment |
$begingroup$
Hint:
By definition, $Asim B$ means $A-B$ and $B-Asubset S$. So you have to show that, if $A-B, B-A, B-C, C-Bsubset S$, then both $A-C$ and $C-A$ are subsets of $S$.
Consider first an element $xin A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?
answered May 23 at 22:00
BernardBernard
127k743120
$endgroup$
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
add a comment |
$begingroup$
We need to show theat $Asim B$ and $B sim C$ give $A sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!)
To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.
edited May 23 at 22:09
Bernard
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $A Delta B subseteq S$ and $B Delta C subseteq S$, then
$A Delta C= (A Delta B) Delta (B Delta C) subseteq S$ as well.
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
add a comment |
$begingroup$
Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive
Below, please read " we'll " instead of " will".
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Two hints about ways to go.
- Draw a Venn diagram for $A, B, C, X$ showing $A Delta B$ and so on.
- Know or show that $Delta$ is associative. That and the fact that $B
Delta B$ is empty leads to an algebraic proof.
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
$endgroup$
add a comment |
$begingroup$
Two hints about ways to go.
- Draw a Venn diagram for $A, B, C, X$ showing $A Delta B$ and so on.
- Know or show that $Delta$ is associative. That and the fact that $B
Delta B$ is empty leads to an algebraic proof.
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
$endgroup$
add a comment |
$begingroup$
Two hints about ways to go.
- Draw a Venn diagram for $A, B, C, X$ showing $A Delta B$ and so on.
- Know or show that $Delta$ is associative. That and the fact that $B
Delta B$ is empty leads to an algebraic proof.
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
$endgroup$
Two hints about ways to go.
- Draw a Venn diagram for $A, B, C, X$ showing $A Delta B$ and so on.
- Know or show that $Delta$ is associative. That and the fact that $B
Delta B$ is empty leads to an algebraic proof.
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
answered May 23 at 21:53
Ethan BolkerEthan Bolker
49.4k556126
49.4k556126
add a comment |
add a comment |
$begingroup$
Hint:
By definition, $Asim B$ means $A-B$ and $B-Asubset S$. So you have to show that, if $A-B, B-A, B-C, C-Bsubset S$, then both $A-C$ and $C-A$ are subsets of $S$.
Consider first an element $xin A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?
answered May 23 at 22:00
BernardBernard
127k743120
$endgroup$
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
add a comment |
$begingroup$
Hint:
By definition, $Asim B$ means $A-B$ and $B-Asubset S$. So you have to show that, if $A-B, B-A, B-C, C-Bsubset S$, then both $A-C$ and $C-A$ are subsets of $S$.
Consider first an element $xin A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?
answered May 23 at 22:00
BernardBernard
127k743120
$endgroup$
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
add a comment |
$begingroup$
Hint:
By definition, $Asim B$ means $A-B$ and $B-Asubset S$. So you have to show that, if $A-B, B-A, B-C, C-Bsubset S$, then both $A-C$ and $C-A$ are subsets of $S$.
Consider first an element $xin A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?
answered May 23 at 22:00
BernardBernard
127k743120
$endgroup$
Hint:
By definition, $Asim B$ means $A-B$ and $B-Asubset S$. So you have to show that, if $A-B, B-A, B-C, C-Bsubset S$, then both $A-C$ and $C-A$ are subsets of $S$.
Consider first an element $xin A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?
answered May 23 at 22:00
BernardBernard
127k743120
answered May 23 at 22:00
BernardBernard
127k743120
answered May 23 at 22:00
BernardBernard
127k743120
answered May 23 at 22:00
BernardBernard
127k743120
127k743120
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
add a comment |
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
1
1
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
$begingroup$
Thanks! My problem is now, what is the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 22:01
2
2
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
$begingroup$
Well, it seems to be made up of the subsets of $X$ which contain $X-S$.
$endgroup$
– Bernard
May 23 at 22:16
add a comment |
$begingroup$
We need to show theat $Asim B$ and $B sim C$ give $A sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!)
To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.
edited May 23 at 22:09
Bernard
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We need to show theat $Asim B$ and $B sim C$ give $A sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!)
To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.
edited May 23 at 22:09
Bernard
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We need to show theat $Asim B$ and $B sim C$ give $A sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!)
To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.
edited May 23 at 22:09
Bernard
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We need to show theat $Asim B$ and $B sim C$ give $A sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!)
To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.
edited May 23 at 22:09
Bernard
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 23 at 22:09
Bernard
127k743120
edited May 23 at 22:09
Bernard
127k743120
edited May 23 at 22:09
Bernard
127k743120
127k743120
answered May 23 at 21:56
auscryptauscrypt
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered May 23 at 21:56
auscryptauscrypt
3,501110
answered May 23 at 21:56
auscryptauscrypt
3,501110
3,501110
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
auscrypt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
If $A Delta B subseteq S$ and $B Delta C subseteq S$, then
$A Delta C= (A Delta B) Delta (B Delta C) subseteq S$ as well.
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
add a comment |
$begingroup$
If $A Delta B subseteq S$ and $B Delta C subseteq S$, then
$A Delta C= (A Delta B) Delta (B Delta C) subseteq S$ as well.
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
add a comment |
$begingroup$
If $A Delta B subseteq S$ and $B Delta C subseteq S$, then
$A Delta C= (A Delta B) Delta (B Delta C) subseteq S$ as well.
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
If $A Delta B subseteq S$ and $B Delta C subseteq S$, then
$A Delta C= (A Delta B) Delta (B Delta C) subseteq S$ as well.
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
edited May 24 at 4:02
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
answered May 23 at 21:55
Henno BrandsmaHenno Brandsma
121k351134
121k351134
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
add a comment |
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
$begingroup$
Must be, subset of $S$. Any help about the equivalence class of $X$?
$endgroup$
– Hendrik Matamoros
May 23 at 21:59
1
1
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
$begingroup$
@HendrikMatamoros: if $A$ is a subset of $X$, then $ADelta X=A^c$, the complement of $A$ in $X$. This should help
$endgroup$
– Taladris
May 24 at 12:13
add a comment |
$begingroup$
Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive
Below, please read " we'll " instead of " will".
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
$endgroup$
add a comment |
$begingroup$
Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive
Below, please read " we'll " instead of " will".
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
$endgroup$
add a comment |
$begingroup$
Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive
Below, please read " we'll " instead of " will".
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
$endgroup$
Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive
Below, please read " we'll " instead of " will".
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
edited May 26 at 11:33
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
answered May 25 at 11:46
Eleonore Saint JamesEleonore Saint James
1,267118
1,267118
add a comment |
add a comment |
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@EthanBolker You're right, my mistake!
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– Hendrik Matamoros
May 23 at 21:54