Proof about sets: $(A cap C = B cap C land A cup C = B cup C) implies A=B$ [closed]
$begingroup$
If I have undefined sets $A, B, C$.
How can I prove that this statement is true?
$$ (A cap C = B cap C land A cup C = B cup C) implies A= B $$
I absolutely do not know how to start or what to use. Please try to advise me or how to proceed. Thanks
elementary-set-theory
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
$endgroup$
closed as off-topic by Andrés E. Caicedo, Saad, Martin Sleziak, Chinnapparaj R, Siong Thye Goh Dec 12 '18 at 2:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chinnapparaj R, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If I have undefined sets $A, B, C$.
How can I prove that this statement is true?
$$ (A cap C = B cap C land A cup C = B cup C) implies A= B $$
I absolutely do not know how to start or what to use. Please try to advise me or how to proceed. Thanks
elementary-set-theory
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
$endgroup$
closed as off-topic by Andrés E. Caicedo, Saad, Martin Sleziak, Chinnapparaj R, Siong Thye Goh Dec 12 '18 at 2:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chinnapparaj R, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
1
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
1
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25
add a comment |
$begingroup$
If I have undefined sets $A, B, C$.
How can I prove that this statement is true?
$$ (A cap C = B cap C land A cup C = B cup C) implies A= B $$
I absolutely do not know how to start or what to use. Please try to advise me or how to proceed. Thanks
elementary-set-theory
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
$endgroup$
If I have undefined sets $A, B, C$.
How can I prove that this statement is true?
$$ (A cap C = B cap C land A cup C = B cup C) implies A= B $$
I absolutely do not know how to start or what to use. Please try to advise me or how to proceed. Thanks
elementary-set-theory
elementary-set-theory
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
edited Dec 12 '18 at 1:24
Martin Sleziak
44.8k9118272
44.8k9118272
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
asked Dec 10 '18 at 12:42
Marek MarkeMarek Marke
266
266
closed as off-topic by Andrés E. Caicedo, Saad, Martin Sleziak, Chinnapparaj R, Siong Thye Goh Dec 12 '18 at 2:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chinnapparaj R, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Andrés E. Caicedo, Saad, Martin Sleziak, Chinnapparaj R, Siong Thye Goh Dec 12 '18 at 2:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Chinnapparaj R, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
1
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
1
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25
add a comment |
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
1
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
1
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
1
1
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
1
1
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Using a contradiction. Suppose there is $x in A$ which is not in $B$ ($x notin B$). As $A neq B$ and $Acup C = B cup C$, $xin C$. Therefore, $x in A cap C$. As we know $Acap C = B cap C$, $x in B cap C$. Hence, $xin B$! which this contradicts by the first assumption. and we can proof that all $x$ in $A$ are in $B$. In the same way we can say that all $x$ in $B$ are in $A$. Therefore, $A = B$.
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
$endgroup$
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
add a comment |
$begingroup$
Let $x in A$.
Case 1: $x in C$. Then $x in B cap C$, thus $x in B$.
Case 2: $x notin C$. Then $x in B cup C$, thus $x in B$.
Therefore we have shown that $A subseteq B.$
The reversed implication can be shown by similar arguments.
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
$endgroup$
add a comment |
$begingroup$
$$A = A cup (A cap C) = A cup (B cap C) = (A cup B) cap (A cup C) = (A cup B) cap (B cup C) = (B cup A) cap (B cup C) = B cup (A cap C) = B cup (B cap C) = B$$
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using a contradiction. Suppose there is $x in A$ which is not in $B$ ($x notin B$). As $A neq B$ and $Acup C = B cup C$, $xin C$. Therefore, $x in A cap C$. As we know $Acap C = B cap C$, $x in B cap C$. Hence, $xin B$! which this contradicts by the first assumption. and we can proof that all $x$ in $A$ are in $B$. In the same way we can say that all $x$ in $B$ are in $A$. Therefore, $A = B$.
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
$endgroup$
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
add a comment |
$begingroup$
Using a contradiction. Suppose there is $x in A$ which is not in $B$ ($x notin B$). As $A neq B$ and $Acup C = B cup C$, $xin C$. Therefore, $x in A cap C$. As we know $Acap C = B cap C$, $x in B cap C$. Hence, $xin B$! which this contradicts by the first assumption. and we can proof that all $x$ in $A$ are in $B$. In the same way we can say that all $x$ in $B$ are in $A$. Therefore, $A = B$.
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
$endgroup$
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
add a comment |
$begingroup$
Using a contradiction. Suppose there is $x in A$ which is not in $B$ ($x notin B$). As $A neq B$ and $Acup C = B cup C$, $xin C$. Therefore, $x in A cap C$. As we know $Acap C = B cap C$, $x in B cap C$. Hence, $xin B$! which this contradicts by the first assumption. and we can proof that all $x$ in $A$ are in $B$. In the same way we can say that all $x$ in $B$ are in $A$. Therefore, $A = B$.
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
$endgroup$
Using a contradiction. Suppose there is $x in A$ which is not in $B$ ($x notin B$). As $A neq B$ and $Acup C = B cup C$, $xin C$. Therefore, $x in A cap C$. As we know $Acap C = B cap C$, $x in B cap C$. Hence, $xin B$! which this contradicts by the first assumption. and we can proof that all $x$ in $A$ are in $B$. In the same way we can say that all $x$ in $B$ are in $A$. Therefore, $A = B$.
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
edited Dec 10 '18 at 12:50
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
answered Dec 10 '18 at 12:48
OmGOmG
2,512722
2,512722
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
add a comment |
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
$begingroup$
$$ forall x in A subseteq B wedge forall x in B subseteq A $$ Is correct when i write it like this?
$endgroup$
– Marek Marke
Dec 10 '18 at 16:16
add a comment |
$begingroup$
Let $x in A$.
Case 1: $x in C$. Then $x in B cap C$, thus $x in B$.
Case 2: $x notin C$. Then $x in B cup C$, thus $x in B$.
Therefore we have shown that $A subseteq B.$
The reversed implication can be shown by similar arguments.
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
$endgroup$
add a comment |
$begingroup$
Let $x in A$.
Case 1: $x in C$. Then $x in B cap C$, thus $x in B$.
Case 2: $x notin C$. Then $x in B cup C$, thus $x in B$.
Therefore we have shown that $A subseteq B.$
The reversed implication can be shown by similar arguments.
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
$endgroup$
add a comment |
$begingroup$
Let $x in A$.
Case 1: $x in C$. Then $x in B cap C$, thus $x in B$.
Case 2: $x notin C$. Then $x in B cup C$, thus $x in B$.
Therefore we have shown that $A subseteq B.$
The reversed implication can be shown by similar arguments.
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
$endgroup$
Let $x in A$.
Case 1: $x in C$. Then $x in B cap C$, thus $x in B$.
Case 2: $x notin C$. Then $x in B cup C$, thus $x in B$.
Therefore we have shown that $A subseteq B.$
The reversed implication can be shown by similar arguments.
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
answered Dec 10 '18 at 12:53
FredFred
45.6k1848
45.6k1848
add a comment |
add a comment |
$begingroup$
$$A = A cup (A cap C) = A cup (B cap C) = (A cup B) cap (A cup C) = (A cup B) cap (B cup C) = (B cup A) cap (B cup C) = B cup (A cap C) = B cup (B cap C) = B$$
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
$endgroup$
add a comment |
$begingroup$
$$A = A cup (A cap C) = A cup (B cap C) = (A cup B) cap (A cup C) = (A cup B) cap (B cup C) = (B cup A) cap (B cup C) = B cup (A cap C) = B cup (B cap C) = B$$
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
$endgroup$
add a comment |
$begingroup$
$$A = A cup (A cap C) = A cup (B cap C) = (A cup B) cap (A cup C) = (A cup B) cap (B cup C) = (B cup A) cap (B cup C) = B cup (A cap C) = B cup (B cap C) = B$$
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
$endgroup$
$$A = A cup (A cap C) = A cup (B cap C) = (A cup B) cap (A cup C) = (A cup B) cap (B cup C) = (B cup A) cap (B cup C) = B cup (A cap C) = B cup (B cap C) = B$$
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
answered Dec 10 '18 at 18:36
Bram28Bram28
61.8k44793
61.8k44793
add a comment |
add a comment |
$begingroup$
You say you absolutely don't know how to start or what to use, I say I don't believe you. I think that if you just think back to all the other times you've been asked to prove that two sets are equal (surely this isn't your first such problem), you can get a few ideas for things you can try. Maybe it works, maybe it doesn't.
$endgroup$
– Arthur
Dec 10 '18 at 12:48
$begingroup$
@Arthur, i don't know how to proceed in evidence or how to verify their correctness. I try to understand but I do not know where I'm making a mistake
$endgroup$
– Marek Marke
Dec 10 '18 at 12:53
1
$begingroup$
karagila.org/2015/how-to-solve-your-problems
$endgroup$
– Asaf Karagila♦
Dec 10 '18 at 13:00
1
$begingroup$
Possible duplicate of Unions and intersections: $(A cup B = A ∪ C) land (A cap B = A ∩ C) implies B = C.$
$endgroup$
– Martin Sleziak
Dec 12 '18 at 1:25