Mixing
For all sets A,B, and C, If B ∩ C ⊆ A, then (A-B) ∩ (A-C) ≠∅
Clash Royale CLAN TAG#URR8PPP
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I’m having trouble working in the following subset proof:
For all sets A,B, and C, If $varnothingne B ∩ C ⊆ A$, then (A-B) ∩ (A-C) ≠∅
My train of thought of creating a proof of via the following. Am I headed towards the right direction or am I making a mistake in my proofs?
Assuming B ∩ C ⊆ A
X exists in B and X exists in C
If x doesn’t exist in b or C then x doesn’t exist in A
Hence x doesnt in B(universal) and x doesn’t exist in C universal and does not exist in A(universal)
Equivalentelty x doesn’t exist in A-B and x doesn’t exist in A-C
Or equivalently, (A-B) ∩ (A-C) ) ≠∅
elementary-set-theory
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
asked Jul 23 at 15:09
Jlee
542
add a comment |Â
up vote
0
down vote
favorite
I’m having trouble working in the following subset proof:
For all sets A,B, and C, If $varnothingne B ∩ C ⊆ A$, then (A-B) ∩ (A-C) ≠∅
My train of thought of creating a proof of via the following. Am I headed towards the right direction or am I making a mistake in my proofs?
Assuming B ∩ C ⊆ A
X exists in B and X exists in C
If x doesn’t exist in b or C then x doesn’t exist in A
Hence x doesnt in B(universal) and x doesn’t exist in C universal and does not exist in A(universal)
Equivalentelty x doesn’t exist in A-B and x doesn’t exist in A-C
Or equivalently, (A-B) ∩ (A-C) ) ≠∅
elementary-set-theory
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
asked Jul 23 at 15:09
Jlee
542
That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
1
You might want to learnsetminus,cap,neq,emptyset.
– Arnaud Mortier
Jul 23 at 15:20
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I’m having trouble working in the following subset proof:
For all sets A,B, and C, If $varnothingne B ∩ C ⊆ A$, then (A-B) ∩ (A-C) ≠∅
My train of thought of creating a proof of via the following. Am I headed towards the right direction or am I making a mistake in my proofs?
Assuming B ∩ C ⊆ A
X exists in B and X exists in C
If x doesn’t exist in b or C then x doesn’t exist in A
Hence x doesnt in B(universal) and x doesn’t exist in C universal and does not exist in A(universal)
Equivalentelty x doesn’t exist in A-B and x doesn’t exist in A-C
Or equivalently, (A-B) ∩ (A-C) ) ≠∅
elementary-set-theory
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
asked Jul 23 at 15:09
Jlee
542
I’m having trouble working in the following subset proof:
For all sets A,B, and C, If $varnothingne B ∩ C ⊆ A$, then (A-B) ∩ (A-C) ≠∅
My train of thought of creating a proof of via the following. Am I headed towards the right direction or am I making a mistake in my proofs?
Assuming B ∩ C ⊆ A
X exists in B and X exists in C
If x doesn’t exist in b or C then x doesn’t exist in A
Hence x doesnt in B(universal) and x doesn’t exist in C universal and does not exist in A(universal)
Equivalentelty x doesn’t exist in A-B and x doesn’t exist in A-C
Or equivalently, (A-B) ∩ (A-C) ) ≠∅
elementary-set-theory
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
asked Jul 23 at 15:09
Jlee
542
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
edited Jul 23 at 15:27
Parcly Taxel
33.5k136588
33.5k136588
asked Jul 23 at 15:09
Jlee
542
asked Jul 23 at 15:09
Jlee
542
asked Jul 23 at 15:09
Jlee
542
542
That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
1
You might want to learnsetminus,cap,neq,emptyset.
– Arnaud Mortier
Jul 23 at 15:20
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27
add a comment |Â
That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
1
You might want to learnsetminus,cap,neq,emptyset.
– Arnaud Mortier
Jul 23 at 15:20
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27
That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
1
1
You might want to learn
setminus, cap, neq, emptyset.– Arnaud Mortier
Jul 23 at 15:20
You might want to learn
setminus, cap, neq, emptyset.– Arnaud Mortier
Jul 23 at 15:20
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27
add a comment |Â
3 Answers
3
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up vote
1
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That is wrong. Take $A=B=C=emptyset$.
It becomes true if you replace the hypothesis by $$Bcup Csubsetneq A$$ because then $$(Asetminus B)cap (Asetminus C)=Asetminus(Bcup C)neq emptyset$$
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
add a comment |Â
up vote
0
down vote
This is wrong. Let $$A=Bcap C$$ or even $$A=B$$ an example is $$B=1,2\C=2,3\A=2$$
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
add a comment |Â
up vote
0
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Let $A_alone = A- (Bcup C)$ which consists of all the elements of $A$ but none of the elements of $B$ or $C$.
$A_in B = Acap Bcap C^c$ which consists of all the elements of $A$ that are in $B$ but none of the elements of $C$.
$A_in C = Acap Ccap B^c$ which consists of all the elements of $A$ that are in $C$ but none of the elements of $B$.
$A_inallthree = Acap B cap C$.
Let $B_alone = B- (Acup C)$ which consists of all the elements of $B$ but none of the elements of $A$ or $C$.
$B_A = A_in B = Acap Bcap C^c$ which consists of all the elements of $B$ that are in $A$ but none of the elements of $C$.
$B_in C = Bcap Ccap A^c$ which consists of all the elements of $B$ that are in $C$ but none of the elements of $A$.
$B_inallthree = A_inallthree = Acap B cap C$.
Let $C_alone = C- (Acup B)$ which consists of all the elements of $C$ but none of the elements of $A$ or $B$.
$C_A = A_in C = Acap Ccap B^c$ which consists of all the elements of $C$ that are in $A$ but none of the elements of $B$.
$C_in B =B_in C = Bcap Ccap A^c$ which consists of all the elements of $C$ that are in $B$ but none of the elements of $A$.
$C_inallthree =B_inallthree= A_inallthree = Acap B cap C$.
These seven distinct sets are disjoint and partition the universal $Acup B cup C$.[1]
=====
So you are given:
$emptysetne B ∩ C ⊆ A$
$emptyset ne (B_alone cup B_inAcup B_in C cup B_inallthree)cap (C_alone cup C_inAcup C_in B cup C_inallthree)subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
$emptyset ne ( B_in C cup B_inallthree) subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
Which means $cup B_in C = emptyset$ and $Acap Bcap C ne emptyset$.
That's .... all we can conclude.
Now: $(A-B)cap(A-C)=$
$(A_alonecup A_inC)cap (A_alonecup A_in B)=$
$A_alone$.
There is simply no reason $A_alone$ can't be empty. And there is also no reason it should. It could go either way.
So as a counter example let:
$A_alone = emptyset; A_in B=ab; A_in C = ac, A_inallthree = abc; B_alone=b; B_inC = emptyset; C_alone = c$.
i.e. $A =ab,ac,abc; B= b,ab,abc; C= c,ac,abc$.
Then $emptyset ne Bcap C = abc subset A$.
And $(A-B)cap (A-C) = accap ab = emptyset$.
=======
[1] A Venn diagram of the three sets will cut that universe into $2^3 = 8$ distinct disjoint subsets. The eighth one, that I did not mention would be $N_none = (Acup Bcup C)^c$. It's not relevant.
answered Jul 23 at 17:16
fleablood
60.3k22575
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
That is wrong. Take $A=B=C=emptyset$.
It becomes true if you replace the hypothesis by $$Bcup Csubsetneq A$$ because then $$(Asetminus B)cap (Asetminus C)=Asetminus(Bcup C)neq emptyset$$
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
add a comment |Â
up vote
1
down vote
That is wrong. Take $A=B=C=emptyset$.
It becomes true if you replace the hypothesis by $$Bcup Csubsetneq A$$ because then $$(Asetminus B)cap (Asetminus C)=Asetminus(Bcup C)neq emptyset$$
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
That is wrong. Take $A=B=C=emptyset$.
It becomes true if you replace the hypothesis by $$Bcup Csubsetneq A$$ because then $$(Asetminus B)cap (Asetminus C)=Asetminus(Bcup C)neq emptyset$$
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
That is wrong. Take $A=B=C=emptyset$.
It becomes true if you replace the hypothesis by $$Bcup Csubsetneq A$$ because then $$(Asetminus B)cap (Asetminus C)=Asetminus(Bcup C)neq emptyset$$
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
edited Jul 23 at 16:39
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
answered Jul 23 at 15:17
Arnaud Mortier
18.9k22159
18.9k22159
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
add a comment |Â
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
No, even with $subsetneq$ in the hypothesis it's still wrong. Take $A=1,2$, $B=1$, and $C=2$.
– Andreas Blass
Jul 23 at 16:28
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass, if you take $A=1,2,3, B=1,3$ and $C=2,3$, you even get $B cap C neq emptyset$, which the questioner assumed.
– Babelfish
Jul 23 at 16:35
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
@AndreasBlass Thanks. Corrected (typoed $cap$ instead of $cup$).
– Arnaud Mortier
Jul 23 at 16:40
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
Ah. So I think I understand it now. Here’s my attempt at a proof by contradiction. If B ∩ C ⊆A, then (A-B) ∩ (A-C) ≠∅. Suppose not, so let (A-B) ∩ (A-C). Then x exists in A and x does not in exist in B and x does not exist in C. But because assume x exists B ∩ C( as B ∩ C ⊆ A). We have a contradiction. Thus B ∩ C ≠∅.
– Jlee
Jul 23 at 16:51
1
1
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
@Jlee First of all, you say "$x$ lies in $A$" for $xin A$, not "$x$ exists $A$". Also, your last argument doesn't make sense. If you assumed that $xin Asetminus (B cup C)$, then there is no reason for $x$ to be in $B cap X$. In principle, you will have a hard day to show something if someone already proved that the claim is wrong.
– Babelfish
Jul 23 at 17:57
add a comment |Â
up vote
0
down vote
This is wrong. Let $$A=Bcap C$$ or even $$A=B$$ an example is $$B=1,2\C=2,3\A=2$$
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
add a comment |Â
up vote
0
down vote
This is wrong. Let $$A=Bcap C$$ or even $$A=B$$ an example is $$B=1,2\C=2,3\A=2$$
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is wrong. Let $$A=Bcap C$$ or even $$A=B$$ an example is $$B=1,2\C=2,3\A=2$$
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
This is wrong. Let $$A=Bcap C$$ or even $$A=B$$ an example is $$B=1,2\C=2,3\A=2$$
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
answered Jul 23 at 15:55
Mostafa Ayaz
8,5823630
8,5823630
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
add a comment |Â
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
How would I find the solution from A = B ∩ C? The numerical set made sense but I’m not sure where to start my proof
– Jlee
Jul 23 at 16:22
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
The point is, that if $A=B cap C$, then your statement is wrong (if $Aneq emptyset$). If $A=Bcap C$, then $Asubseteq B$ and $Asubseteq C$, so $Asetminus B=emptyset=Asetminus C$, so the intersection of both is empty aswell.
– Babelfish
Jul 23 at 16:37
add a comment |Â
up vote
0
down vote
Let $A_alone = A- (Bcup C)$ which consists of all the elements of $A$ but none of the elements of $B$ or $C$.
$A_in B = Acap Bcap C^c$ which consists of all the elements of $A$ that are in $B$ but none of the elements of $C$.
$A_in C = Acap Ccap B^c$ which consists of all the elements of $A$ that are in $C$ but none of the elements of $B$.
$A_inallthree = Acap B cap C$.
Let $B_alone = B- (Acup C)$ which consists of all the elements of $B$ but none of the elements of $A$ or $C$.
$B_A = A_in B = Acap Bcap C^c$ which consists of all the elements of $B$ that are in $A$ but none of the elements of $C$.
$B_in C = Bcap Ccap A^c$ which consists of all the elements of $B$ that are in $C$ but none of the elements of $A$.
$B_inallthree = A_inallthree = Acap B cap C$.
Let $C_alone = C- (Acup B)$ which consists of all the elements of $C$ but none of the elements of $A$ or $B$.
$C_A = A_in C = Acap Ccap B^c$ which consists of all the elements of $C$ that are in $A$ but none of the elements of $B$.
$C_in B =B_in C = Bcap Ccap A^c$ which consists of all the elements of $C$ that are in $B$ but none of the elements of $A$.
$C_inallthree =B_inallthree= A_inallthree = Acap B cap C$.
These seven distinct sets are disjoint and partition the universal $Acup B cup C$.[1]
=====
So you are given:
$emptysetne B ∩ C ⊆ A$
$emptyset ne (B_alone cup B_inAcup B_in C cup B_inallthree)cap (C_alone cup C_inAcup C_in B cup C_inallthree)subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
$emptyset ne ( B_in C cup B_inallthree) subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
Which means $cup B_in C = emptyset$ and $Acap Bcap C ne emptyset$.
That's .... all we can conclude.
Now: $(A-B)cap(A-C)=$
$(A_alonecup A_inC)cap (A_alonecup A_in B)=$
$A_alone$.
There is simply no reason $A_alone$ can't be empty. And there is also no reason it should. It could go either way.
So as a counter example let:
$A_alone = emptyset; A_in B=ab; A_in C = ac, A_inallthree = abc; B_alone=b; B_inC = emptyset; C_alone = c$.
i.e. $A =ab,ac,abc; B= b,ab,abc; C= c,ac,abc$.
Then $emptyset ne Bcap C = abc subset A$.
And $(A-B)cap (A-C) = accap ab = emptyset$.
=======
[1] A Venn diagram of the three sets will cut that universe into $2^3 = 8$ distinct disjoint subsets. The eighth one, that I did not mention would be $N_none = (Acup Bcup C)^c$. It's not relevant.
answered Jul 23 at 17:16
fleablood
60.3k22575
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Let $A_alone = A- (Bcup C)$ which consists of all the elements of $A$ but none of the elements of $B$ or $C$.
$A_in B = Acap Bcap C^c$ which consists of all the elements of $A$ that are in $B$ but none of the elements of $C$.
$A_in C = Acap Ccap B^c$ which consists of all the elements of $A$ that are in $C$ but none of the elements of $B$.
$A_inallthree = Acap B cap C$.
Let $B_alone = B- (Acup C)$ which consists of all the elements of $B$ but none of the elements of $A$ or $C$.
$B_A = A_in B = Acap Bcap C^c$ which consists of all the elements of $B$ that are in $A$ but none of the elements of $C$.
$B_in C = Bcap Ccap A^c$ which consists of all the elements of $B$ that are in $C$ but none of the elements of $A$.
$B_inallthree = A_inallthree = Acap B cap C$.
Let $C_alone = C- (Acup B)$ which consists of all the elements of $C$ but none of the elements of $A$ or $B$.
$C_A = A_in C = Acap Ccap B^c$ which consists of all the elements of $C$ that are in $A$ but none of the elements of $B$.
$C_in B =B_in C = Bcap Ccap A^c$ which consists of all the elements of $C$ that are in $B$ but none of the elements of $A$.
$C_inallthree =B_inallthree= A_inallthree = Acap B cap C$.
These seven distinct sets are disjoint and partition the universal $Acup B cup C$.[1]
=====
So you are given:
$emptysetne B ∩ C ⊆ A$
$emptyset ne (B_alone cup B_inAcup B_in C cup B_inallthree)cap (C_alone cup C_inAcup C_in B cup C_inallthree)subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
$emptyset ne ( B_in C cup B_inallthree) subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
Which means $cup B_in C = emptyset$ and $Acap Bcap C ne emptyset$.
That's .... all we can conclude.
Now: $(A-B)cap(A-C)=$
$(A_alonecup A_inC)cap (A_alonecup A_in B)=$
$A_alone$.
There is simply no reason $A_alone$ can't be empty. And there is also no reason it should. It could go either way.
So as a counter example let:
$A_alone = emptyset; A_in B=ab; A_in C = ac, A_inallthree = abc; B_alone=b; B_inC = emptyset; C_alone = c$.
i.e. $A =ab,ac,abc; B= b,ab,abc; C= c,ac,abc$.
Then $emptyset ne Bcap C = abc subset A$.
And $(A-B)cap (A-C) = accap ab = emptyset$.
=======
[1] A Venn diagram of the three sets will cut that universe into $2^3 = 8$ distinct disjoint subsets. The eighth one, that I did not mention would be $N_none = (Acup Bcup C)^c$. It's not relevant.
answered Jul 23 at 17:16
fleablood
60.3k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $A_alone = A- (Bcup C)$ which consists of all the elements of $A$ but none of the elements of $B$ or $C$.
$A_in B = Acap Bcap C^c$ which consists of all the elements of $A$ that are in $B$ but none of the elements of $C$.
$A_in C = Acap Ccap B^c$ which consists of all the elements of $A$ that are in $C$ but none of the elements of $B$.
$A_inallthree = Acap B cap C$.
Let $B_alone = B- (Acup C)$ which consists of all the elements of $B$ but none of the elements of $A$ or $C$.
$B_A = A_in B = Acap Bcap C^c$ which consists of all the elements of $B$ that are in $A$ but none of the elements of $C$.
$B_in C = Bcap Ccap A^c$ which consists of all the elements of $B$ that are in $C$ but none of the elements of $A$.
$B_inallthree = A_inallthree = Acap B cap C$.
Let $C_alone = C- (Acup B)$ which consists of all the elements of $C$ but none of the elements of $A$ or $B$.
$C_A = A_in C = Acap Ccap B^c$ which consists of all the elements of $C$ that are in $A$ but none of the elements of $B$.
$C_in B =B_in C = Bcap Ccap A^c$ which consists of all the elements of $C$ that are in $B$ but none of the elements of $A$.
$C_inallthree =B_inallthree= A_inallthree = Acap B cap C$.
These seven distinct sets are disjoint and partition the universal $Acup B cup C$.[1]
=====
So you are given:
$emptysetne B ∩ C ⊆ A$
$emptyset ne (B_alone cup B_inAcup B_in C cup B_inallthree)cap (C_alone cup C_inAcup C_in B cup C_inallthree)subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
$emptyset ne ( B_in C cup B_inallthree) subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
Which means $cup B_in C = emptyset$ and $Acap Bcap C ne emptyset$.
That's .... all we can conclude.
Now: $(A-B)cap(A-C)=$
$(A_alonecup A_inC)cap (A_alonecup A_in B)=$
$A_alone$.
There is simply no reason $A_alone$ can't be empty. And there is also no reason it should. It could go either way.
So as a counter example let:
$A_alone = emptyset; A_in B=ab; A_in C = ac, A_inallthree = abc; B_alone=b; B_inC = emptyset; C_alone = c$.
i.e. $A =ab,ac,abc; B= b,ab,abc; C= c,ac,abc$.
Then $emptyset ne Bcap C = abc subset A$.
And $(A-B)cap (A-C) = accap ab = emptyset$.
=======
[1] A Venn diagram of the three sets will cut that universe into $2^3 = 8$ distinct disjoint subsets. The eighth one, that I did not mention would be $N_none = (Acup Bcup C)^c$. It's not relevant.
answered Jul 23 at 17:16
fleablood
60.3k22575
Let $A_alone = A- (Bcup C)$ which consists of all the elements of $A$ but none of the elements of $B$ or $C$.
$A_in B = Acap Bcap C^c$ which consists of all the elements of $A$ that are in $B$ but none of the elements of $C$.
$A_in C = Acap Ccap B^c$ which consists of all the elements of $A$ that are in $C$ but none of the elements of $B$.
$A_inallthree = Acap B cap C$.
Let $B_alone = B- (Acup C)$ which consists of all the elements of $B$ but none of the elements of $A$ or $C$.
$B_A = A_in B = Acap Bcap C^c$ which consists of all the elements of $B$ that are in $A$ but none of the elements of $C$.
$B_in C = Bcap Ccap A^c$ which consists of all the elements of $B$ that are in $C$ but none of the elements of $A$.
$B_inallthree = A_inallthree = Acap B cap C$.
Let $C_alone = C- (Acup B)$ which consists of all the elements of $C$ but none of the elements of $A$ or $B$.
$C_A = A_in C = Acap Ccap B^c$ which consists of all the elements of $C$ that are in $A$ but none of the elements of $B$.
$C_in B =B_in C = Bcap Ccap A^c$ which consists of all the elements of $C$ that are in $B$ but none of the elements of $A$.
$C_inallthree =B_inallthree= A_inallthree = Acap B cap C$.
These seven distinct sets are disjoint and partition the universal $Acup B cup C$.[1]
=====
So you are given:
$emptysetne B ∩ C ⊆ A$
$emptyset ne (B_alone cup B_inAcup B_in C cup B_inallthree)cap (C_alone cup C_inAcup C_in B cup C_inallthree)subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
$emptyset ne ( B_in C cup B_inallthree) subset (A_alone cup A_inBcup A_in B cup A_inallthree)$
Which means $cup B_in C = emptyset$ and $Acap Bcap C ne emptyset$.
That's .... all we can conclude.
Now: $(A-B)cap(A-C)=$
$(A_alonecup A_inC)cap (A_alonecup A_in B)=$
$A_alone$.
There is simply no reason $A_alone$ can't be empty. And there is also no reason it should. It could go either way.
So as a counter example let:
$A_alone = emptyset; A_in B=ab; A_in C = ac, A_inallthree = abc; B_alone=b; B_inC = emptyset; C_alone = c$.
i.e. $A =ab,ac,abc; B= b,ab,abc; C= c,ac,abc$.
Then $emptyset ne Bcap C = abc subset A$.
And $(A-B)cap (A-C) = accap ab = emptyset$.
=======
[1] A Venn diagram of the three sets will cut that universe into $2^3 = 8$ distinct disjoint subsets. The eighth one, that I did not mention would be $N_none = (Acup Bcup C)^c$. It's not relevant.
answered Jul 23 at 17:16
fleablood
60.3k22575
edited Jul 23 at 17:22
answered Jul 23 at 17:16
fleablood
60.3k22575
answered Jul 23 at 17:16
fleablood
60.3k22575
answered Jul 23 at 17:16
fleablood
60.3k22575
60.3k22575
add a comment |Â
add a comment |Â
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That is wrong. Take $A=B=C=emptyset$.
– Arnaud Mortier
Jul 23 at 15:15
1
You might want to learn
setminus,cap,neq,emptyset.– Arnaud Mortier
Jul 23 at 15:20
@Parcly: Really? [set-theory]?
– Asaf Karagila
Jul 23 at 15:27