How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
add a comment |
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
edited Dec 9 '18 at 22:32
Shaun
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
edited Dec 9 '18 at 22:32
Shaun
9,226113684
edited Dec 9 '18 at 22:32
Shaun
9,226113684
edited Dec 9 '18 at 22:32
Shaun
9,226113684
9,226113684
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
62
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
1
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
$endgroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
answered Dec 9 '18 at 22:38
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
$endgroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,586722
4,586722
add a comment |
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
$endgroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
answered Dec 10 '18 at 0:13
YiFanYiFan
4,0101627
4,0101627
add a comment |
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
$endgroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
answered Dec 9 '18 at 23:10
fleabloodfleablood
71k22686
71k22686
add a comment |
add a comment |
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1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09