Proving equality of 3 sets
$begingroup$
$A, B, C$ are sets. I have to prove equality of this:
$$A cap C = B cap C land A cup C = B cup C ⇒ A = B$$
I did this, but I don't know what to do next and whether I even did the right thing:
$A cap C = x in A land x in C$
$B cap C ∧ A cup C = (x in B land x in C) land (x in A lor x in C)$
$B cup C implies A = x in B lor x in C ⇒ A $
$B$
Any ideas what to do next? Or maybe how to solve this somehow else?
elementary-set-theory proof-writing
edited Dec 5 '18 at 21:14
platty
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
$endgroup$
add a comment |
$begingroup$
$A, B, C$ are sets. I have to prove equality of this:
$$A cap C = B cap C land A cup C = B cup C ⇒ A = B$$
I did this, but I don't know what to do next and whether I even did the right thing:
$A cap C = x in A land x in C$
$B cap C ∧ A cup C = (x in B land x in C) land (x in A lor x in C)$
$B cup C implies A = x in B lor x in C ⇒ A $
$B$
Any ideas what to do next? Or maybe how to solve this somehow else?
elementary-set-theory proof-writing
edited Dec 5 '18 at 21:14
platty
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
$endgroup$
$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15
add a comment |
$begingroup$
$A, B, C$ are sets. I have to prove equality of this:
$$A cap C = B cap C land A cup C = B cup C ⇒ A = B$$
I did this, but I don't know what to do next and whether I even did the right thing:
$A cap C = x in A land x in C$
$B cap C ∧ A cup C = (x in B land x in C) land (x in A lor x in C)$
$B cup C implies A = x in B lor x in C ⇒ A $
$B$
Any ideas what to do next? Or maybe how to solve this somehow else?
elementary-set-theory proof-writing
edited Dec 5 '18 at 21:14
platty
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
$endgroup$
$A, B, C$ are sets. I have to prove equality of this:
$$A cap C = B cap C land A cup C = B cup C ⇒ A = B$$
I did this, but I don't know what to do next and whether I even did the right thing:
$A cap C = x in A land x in C$
$B cap C ∧ A cup C = (x in B land x in C) land (x in A lor x in C)$
$B cup C implies A = x in B lor x in C ⇒ A $
$B$
Any ideas what to do next? Or maybe how to solve this somehow else?
elementary-set-theory proof-writing
elementary-set-theory proof-writing
edited Dec 5 '18 at 21:14
platty
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
edited Dec 5 '18 at 21:14
platty
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
edited Dec 5 '18 at 21:14
platty
3,370320
edited Dec 5 '18 at 21:14
platty
3,370320
edited Dec 5 '18 at 21:14
platty
3,370320
3,370320
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
asked Dec 5 '18 at 21:09
qwerty1qwerty1
11
11
$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15
add a comment |
$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15
$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Partition $A$ into the two sets $A'=Acap C$ and $A''=Abackslash (Acap C)$, and partition $B$ into the two sets $B'=Bcap C$ and $B''=Bbackslash (Bcap C)$. The first equality
$$Acap C=Bcap C$$
tells us that $A'=B'$. Since $(Acup C)backslash C=A''$ and $(Bcup C)backslash C=B''$, the second equality
$$Acup C=Bcup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
$endgroup$
add a comment |
$begingroup$
Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x in A$ and $x in C$.
2) $x in A$ and $x not in C$.
3) $x not in A$ and $x in C$.
4) $x not in A$ and $ xnot in C$
For each of those cases we ask: Is $xin B$?
If we discover that $x in B iff x in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A ne B$ or that $A$ need not equal $B$
So
1) $x in A$ and $x in C$.
So $x in Acap C = Bcap C$ so $xin B$.
2) $x in A$ and $x not in C$.
So $xin Acup C= Bcup C$ so $x$ is in either $B$ or $C$. But $x not in C$ so $x in B$.
3) $x not in A$ and $x in C$.
So $x not in Acap C = Bcap C$. So $x$ is not in both $B$ and $C$. But $x in C$ so $x$ can't be in $B$.
4) $x not in A$ and $ xnot in C$
$xnot in Acup C = Bcup C$ so $x$ is in neither $B$ nor $C$. So $xnot in B$.
So $xin B$ precisely when $x in A$ and $x not in B$ precisely what $x not in A$. So $A$ an $B$ have precisely the same elements and $A = B$.
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
Partition $A$ into the two sets $A'=Acap C$ and $A''=Abackslash (Acap C)$, and partition $B$ into the two sets $B'=Bcap C$ and $B''=Bbackslash (Bcap C)$. The first equality
$$Acap C=Bcap C$$
tells us that $A'=B'$. Since $(Acup C)backslash C=A''$ and $(Bcup C)backslash C=B''$, the second equality
$$Acup C=Bcup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
$endgroup$
add a comment |
$begingroup$
Partition $A$ into the two sets $A'=Acap C$ and $A''=Abackslash (Acap C)$, and partition $B$ into the two sets $B'=Bcap C$ and $B''=Bbackslash (Bcap C)$. The first equality
$$Acap C=Bcap C$$
tells us that $A'=B'$. Since $(Acup C)backslash C=A''$ and $(Bcup C)backslash C=B''$, the second equality
$$Acup C=Bcup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
$endgroup$
add a comment |
$begingroup$
Partition $A$ into the two sets $A'=Acap C$ and $A''=Abackslash (Acap C)$, and partition $B$ into the two sets $B'=Bcap C$ and $B''=Bbackslash (Bcap C)$. The first equality
$$Acap C=Bcap C$$
tells us that $A'=B'$. Since $(Acup C)backslash C=A''$ and $(Bcup C)backslash C=B''$, the second equality
$$Acup C=Bcup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
$endgroup$
Partition $A$ into the two sets $A'=Acap C$ and $A''=Abackslash (Acap C)$, and partition $B$ into the two sets $B'=Bcap C$ and $B''=Bbackslash (Bcap C)$. The first equality
$$Acap C=Bcap C$$
tells us that $A'=B'$. Since $(Acup C)backslash C=A''$ and $(Bcup C)backslash C=B''$, the second equality
$$Acup C=Bcup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
answered Dec 5 '18 at 21:14
FrpzzdFrpzzd
22.7k840108
22.7k840108
add a comment |
add a comment |
$begingroup$
Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x in A$ and $x in C$.
2) $x in A$ and $x not in C$.
3) $x not in A$ and $x in C$.
4) $x not in A$ and $ xnot in C$
For each of those cases we ask: Is $xin B$?
If we discover that $x in B iff x in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A ne B$ or that $A$ need not equal $B$
So
1) $x in A$ and $x in C$.
So $x in Acap C = Bcap C$ so $xin B$.
2) $x in A$ and $x not in C$.
So $xin Acup C= Bcup C$ so $x$ is in either $B$ or $C$. But $x not in C$ so $x in B$.
3) $x not in A$ and $x in C$.
So $x not in Acap C = Bcap C$. So $x$ is not in both $B$ and $C$. But $x in C$ so $x$ can't be in $B$.
4) $x not in A$ and $ xnot in C$
$xnot in Acup C = Bcup C$ so $x$ is in neither $B$ nor $C$. So $xnot in B$.
So $xin B$ precisely when $x in A$ and $x not in B$ precisely what $x not in A$. So $A$ an $B$ have precisely the same elements and $A = B$.
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x in A$ and $x in C$.
2) $x in A$ and $x not in C$.
3) $x not in A$ and $x in C$.
4) $x not in A$ and $ xnot in C$
For each of those cases we ask: Is $xin B$?
If we discover that $x in B iff x in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A ne B$ or that $A$ need not equal $B$
So
1) $x in A$ and $x in C$.
So $x in Acap C = Bcap C$ so $xin B$.
2) $x in A$ and $x not in C$.
So $xin Acup C= Bcup C$ so $x$ is in either $B$ or $C$. But $x not in C$ so $x in B$.
3) $x not in A$ and $x in C$.
So $x not in Acap C = Bcap C$. So $x$ is not in both $B$ and $C$. But $x in C$ so $x$ can't be in $B$.
4) $x not in A$ and $ xnot in C$
$xnot in Acup C = Bcup C$ so $x$ is in neither $B$ nor $C$. So $xnot in B$.
So $xin B$ precisely when $x in A$ and $x not in B$ precisely what $x not in A$. So $A$ an $B$ have precisely the same elements and $A = B$.
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x in A$ and $x in C$.
2) $x in A$ and $x not in C$.
3) $x not in A$ and $x in C$.
4) $x not in A$ and $ xnot in C$
For each of those cases we ask: Is $xin B$?
If we discover that $x in B iff x in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A ne B$ or that $A$ need not equal $B$
So
1) $x in A$ and $x in C$.
So $x in Acap C = Bcap C$ so $xin B$.
2) $x in A$ and $x not in C$.
So $xin Acup C= Bcup C$ so $x$ is in either $B$ or $C$. But $x not in C$ so $x in B$.
3) $x not in A$ and $x in C$.
So $x not in Acap C = Bcap C$. So $x$ is not in both $B$ and $C$. But $x in C$ so $x$ can't be in $B$.
4) $x not in A$ and $ xnot in C$
$xnot in Acup C = Bcup C$ so $x$ is in neither $B$ nor $C$. So $xnot in B$.
So $xin B$ precisely when $x in A$ and $x not in B$ precisely what $x not in A$. So $A$ an $B$ have precisely the same elements and $A = B$.
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
$endgroup$
Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x in A$ and $x in C$.
2) $x in A$ and $x not in C$.
3) $x not in A$ and $x in C$.
4) $x not in A$ and $ xnot in C$
For each of those cases we ask: Is $xin B$?
If we discover that $x in B iff x in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A ne B$ or that $A$ need not equal $B$
So
1) $x in A$ and $x in C$.
So $x in Acap C = Bcap C$ so $xin B$.
2) $x in A$ and $x not in C$.
So $xin Acup C= Bcup C$ so $x$ is in either $B$ or $C$. But $x not in C$ so $x in B$.
3) $x not in A$ and $x in C$.
So $x not in Acap C = Bcap C$. So $x$ is not in both $B$ and $C$. But $x in C$ so $x$ can't be in $B$.
4) $x not in A$ and $ xnot in C$
$xnot in Acup C = Bcup C$ so $x$ is in neither $B$ nor $C$. So $xnot in B$.
So $xin B$ precisely when $x in A$ and $x not in B$ precisely what $x not in A$. So $A$ an $B$ have precisely the same elements and $A = B$.
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
answered Dec 5 '18 at 21:47
fleabloodfleablood
69.3k22685
69.3k22685
add a comment |
add a comment |
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$begingroup$
You are mixing logical statements and expressions in ways they should not be. Do try to be more consistent. For example your fourth line should read $xin Acap Ciff xin Awedge xin C$ or it should read $Acap C = {x~:~xin Awedge xin C}$.
$endgroup$
– JMoravitz
Dec 5 '18 at 21:12
$begingroup$
I recommend approaching via contrapositive. Suppose that $Aneq B$. Without loss of generality, then there is some $ain A$ such that $anotin B$. There are two possibilities to continue, either $anotin C$ or $ain C$. In the first case then can you see and explain why $Acup Cneq Bcup C$? In the second case can you see and explain why $Acap Cneq Bcap C$? Do you see why showing this proves what you want?
$endgroup$
– JMoravitz
Dec 5 '18 at 21:15