Prove that for any sets $A$ and $B $ with $(A - B) ∪ (B - A) = A ∪ B$ , then $A ∩ B = ∅$ ...
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Prove that for any sets $A$ and $B $ with $(A - B) ∪ (B - A) = A ∪ B$ , then $A ∩ B = ∅$
The Next CEO of Stack OverflowProving that the sets $A$ and $Asetminus B$ have the same cardinality if $A$ is uncoutnable and $B$ is countableUnion and Intersection of setsShow that if A, B, and C are sets thenProve that sets $A$ and $B$ are disjoint iff $A cup B = A bigtriangleup B$Prove that if sets $A$ and $B$ are countable, then their union $Acup B$ is countableProve A = $(A cap B) cup (A cap B^c)$ using the distributive and associative law.Let A be a denumerable set. Prove that the set ${B:Bsubset A}$ and cardinality of B=1 of all 1-element subsets of A is denumerable.Show that if $A$ and $C$ are disjoint sets, then $(f cup g): A cup C to B cup D$Show that the intersection of any two intervals is an intervalFor any two finite sets $X$ and $Y$, prove that $#(Y^X)=#(Y)^{#(X)}$ by induction
$begingroup$
So this is an assignment question. I'm not sure how to do it so I started off assuming that intersection of $A$ and $B$ is not empty so if $ p$ is an element in it, then $p$ is an element of $A cup{ B } $ too.
Is this right and if so where do I go from there?
Thanks in advance!
proof-verification elementary-set-theory
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
$endgroup$
add a comment |
$begingroup$
So this is an assignment question. I'm not sure how to do it so I started off assuming that intersection of $A$ and $B$ is not empty so if $ p$ is an element in it, then $p$ is an element of $A cup{ B } $ too.
Is this right and if so where do I go from there?
Thanks in advance!
proof-verification elementary-set-theory
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
$endgroup$
add a comment |
$begingroup$
So this is an assignment question. I'm not sure how to do it so I started off assuming that intersection of $A$ and $B$ is not empty so if $ p$ is an element in it, then $p$ is an element of $A cup{ B } $ too.
Is this right and if so where do I go from there?
Thanks in advance!
proof-verification elementary-set-theory
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
$endgroup$
So this is an assignment question. I'm not sure how to do it so I started off assuming that intersection of $A$ and $B$ is not empty so if $ p$ is an element in it, then $p$ is an element of $A cup{ B } $ too.
Is this right and if so where do I go from there?
Thanks in advance!
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
edited Mar 17 at 2:58
YuiTo Cheng
2,1862937
2,1862937
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
asked Nov 7 '18 at 14:33
AnoUser1AnoUser1
836
836
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume $exists xin A cap B$. Then $x notin A $ $B $ and $x notin B$ $A$ (do you see why?)
Hence $x notin (A $ $B)cup (B$ $A)$. However $xin Acup B$ hence a contradiction.
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
$endgroup$
add a comment |
$begingroup$
Yes, it's a good starting point. Now, since $pnotin Asetminus B$ and $pnotin Bsetminus A$, $pnotin(Asetminus B)cup(Bsetminus A)$. This proves that $Acup Bneq(Asetminus B)cup(Bsetminus A)$.
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Assume $exists xin A cap B$. Then $x notin A $ $B $ and $x notin B$ $A$ (do you see why?)
Hence $x notin (A $ $B)cup (B$ $A)$. However $xin Acup B$ hence a contradiction.
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
$endgroup$
add a comment |
$begingroup$
Assume $exists xin A cap B$. Then $x notin A $ $B $ and $x notin B$ $A$ (do you see why?)
Hence $x notin (A $ $B)cup (B$ $A)$. However $xin Acup B$ hence a contradiction.
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
$endgroup$
add a comment |
$begingroup$
Assume $exists xin A cap B$. Then $x notin A $ $B $ and $x notin B$ $A$ (do you see why?)
Hence $x notin (A $ $B)cup (B$ $A)$. However $xin Acup B$ hence a contradiction.
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
$endgroup$
Assume $exists xin A cap B$. Then $x notin A $ $B $ and $x notin B$ $A$ (do you see why?)
Hence $x notin (A $ $B)cup (B$ $A)$. However $xin Acup B$ hence a contradiction.
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
answered Nov 7 '18 at 14:39
asdfasdf
3,701519
3,701519
add a comment |
add a comment |
$begingroup$
Yes, it's a good starting point. Now, since $pnotin Asetminus B$ and $pnotin Bsetminus A$, $pnotin(Asetminus B)cup(Bsetminus A)$. This proves that $Acup Bneq(Asetminus B)cup(Bsetminus A)$.
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Yes, it's a good starting point. Now, since $pnotin Asetminus B$ and $pnotin Bsetminus A$, $pnotin(Asetminus B)cup(Bsetminus A)$. This proves that $Acup Bneq(Asetminus B)cup(Bsetminus A)$.
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Yes, it's a good starting point. Now, since $pnotin Asetminus B$ and $pnotin Bsetminus A$, $pnotin(Asetminus B)cup(Bsetminus A)$. This proves that $Acup Bneq(Asetminus B)cup(Bsetminus A)$.
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Yes, it's a good starting point. Now, since $pnotin Asetminus B$ and $pnotin Bsetminus A$, $pnotin(Asetminus B)cup(Bsetminus A)$. This proves that $Acup Bneq(Asetminus B)cup(Bsetminus A)$.
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
edited Nov 7 '18 at 15:32
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
answered Nov 7 '18 at 14:40
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
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