Mixing
How do I prove that A∩B∩C can be zero?
$begingroup$
I have a situation where,
n(A)=a
n(B)=b
n(C)=c
n(A∩B)=x
n(B∩C)=y
n(C∩A)=z
where a,b,c,x,y,z>0
Although I do understand that A∩B∩C can be {0,min(x,y,z)} but how do I prove that it can attain the value 0?
elementary-set-theory
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
$endgroup$
add a comment |
$begingroup$
I have a situation where,
n(A)=a
n(B)=b
n(C)=c
n(A∩B)=x
n(B∩C)=y
n(C∩A)=z
where a,b,c,x,y,z>0
Although I do understand that A∩B∩C can be {0,min(x,y,z)} but how do I prove that it can attain the value 0?
elementary-set-theory
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
$endgroup$
1
$begingroup$
If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
$endgroup$
– fleablood
Jan 8 at 0:16
add a comment |
$begingroup$
I have a situation where,
n(A)=a
n(B)=b
n(C)=c
n(A∩B)=x
n(B∩C)=y
n(C∩A)=z
where a,b,c,x,y,z>0
Although I do understand that A∩B∩C can be {0,min(x,y,z)} but how do I prove that it can attain the value 0?
elementary-set-theory
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
$endgroup$
I have a situation where,
n(A)=a
n(B)=b
n(C)=c
n(A∩B)=x
n(B∩C)=y
n(C∩A)=z
where a,b,c,x,y,z>0
Although I do understand that A∩B∩C can be {0,min(x,y,z)} but how do I prove that it can attain the value 0?
elementary-set-theory
elementary-set-theory
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
edited Jan 8 at 1:15
Andrés E. Caicedo
65.3k8158247
65.3k8158247
asked Jan 8 at 0:01
asds_asdsasds_asds
61
asked Jan 8 at 0:01
asds_asdsasds_asds
61
asked Jan 8 at 0:01
asds_asdsasds_asds
61
61
1
$begingroup$
If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
$endgroup$
– fleablood
Jan 8 at 0:16
add a comment |
1
$begingroup$
If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
$endgroup$
– fleablood
Jan 8 at 0:16
1
1
$begingroup$
If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
$endgroup$
– fleablood
Jan 8 at 0:16
$begingroup$
If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
$endgroup$
– fleablood
Jan 8 at 0:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Just find three sets with this property: you're just trying to show that it can happen, so one example is sufficient.
answered Jan 8 at 0:05
user3482749user3482749
4,196919
$endgroup$
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can provemin(x,y,z)for the upper bound I'd like a proof for the lower bound.
$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
add a comment |
$begingroup$
$min(x,y,z)$ was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be $0$ is a proof of what can be and the only thing nescessary for that is a simple example.
There are eight DISJOINT sets that combine to make the whole enchilada. They are (using the notation $M'$ to mean the compliment of $M$)
$A cap B cap C$ and $Acap B cap C'$ and $A cap B' cap C$ and $A cap B' cap C'$ and $A' cap B cap C$ and $A'cap B cap C'$ and $A' cap B' cap C$ and $A' cap B' cap C'$.
Any of these sets can contain any elements you desire so long as they are disjoint. Simply leave $A cap Bcap C$ empty.
You will have
$A cap B = (Acap B cap C) cup (Acap Bcap C')$ so just make sure $Acap Bcap C$ has no elements and $Acap B cap C'$ has $x$ elements.
$Bcap C = (Acap B cap C)cup (A'cap Bcap C)$ so make sure $(A'cap Bcap C)$ has $y$ elements.
$Acap C = (Acap Bcap C) cup (Acap B' cap C)$ so make sure $(Acap B' cap C)$ has $z$ elements.
Then $A = (Acap Bcap C) cup (Acap Bcap C') cup (Acap B' cap C)cup (Acap B'cap C')$. You can fill $Acap B'cap C'$ with as many elements or no elements as you wish. Fill it with $k$ elements and you have $|A| = a = x+z + k$.
Then $B = (Acap Bcap C) cup (Acap Bcap C') cup (A'cap B cap C)cup (Acap Bcap C')$. You can fill $A'cap Bcap C'$ with as many elements or no elements as you wish. Fill it with $j$ elements and you have $|B| = b = x+y + j$.
Then $C = (Acap Bcap C) cup (Acap B'cap C) cup (A'cap B cap C)cup (A'cap B'cap C)$. You can fill $A'cap B'cap C$ with as many elements or no elements as you wish. Fill it with $m$ elements and you have $|C| = c = y+z + m$.
And that's it. That's a proof.
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
$endgroup$
add a comment |
$begingroup$
"How do I prove that $A cap B cap C$ can be zero?"
Sometimes a figure is worth a thousand words:
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper boundmin(x,y,z)is proved.
$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just find three sets with this property: you're just trying to show that it can happen, so one example is sufficient.
answered Jan 8 at 0:05
user3482749user3482749
4,196919
$endgroup$
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can provemin(x,y,z)for the upper bound I'd like a proof for the lower bound.
$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
add a comment |
$begingroup$
Just find three sets with this property: you're just trying to show that it can happen, so one example is sufficient.
answered Jan 8 at 0:05
user3482749user3482749
4,196919
$endgroup$
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can provemin(x,y,z)for the upper bound I'd like a proof for the lower bound.
$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
add a comment |
$begingroup$
Just find three sets with this property: you're just trying to show that it can happen, so one example is sufficient.
answered Jan 8 at 0:05
user3482749user3482749
4,196919
$endgroup$
Just find three sets with this property: you're just trying to show that it can happen, so one example is sufficient.
answered Jan 8 at 0:05
user3482749user3482749
4,196919
answered Jan 8 at 0:05
user3482749user3482749
4,196919
answered Jan 8 at 0:05
user3482749user3482749
4,196919
answered Jan 8 at 0:05
user3482749user3482749
4,196919
4,196919
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can provemin(x,y,z)for the upper bound I'd like a proof for the lower bound.
$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
add a comment |
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can provemin(x,y,z)for the upper bound I'd like a proof for the lower bound.
$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can prove
min(x,y,z) for the upper bound I'd like a proof for the lower bound.$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
Yes that'd definitely work, I was just looking for some general proof , if there is one. Just the way I can prove
min(x,y,z) for the upper bound I'd like a proof for the lower bound.$endgroup$
– asds_asds
Jan 8 at 0:06
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
I can't conceive of any useful way to define "general proof" that doesn't include an example in this case.
$endgroup$
– user3482749
Jan 8 at 0:19
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:02
add a comment |
$begingroup$
$min(x,y,z)$ was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be $0$ is a proof of what can be and the only thing nescessary for that is a simple example.
There are eight DISJOINT sets that combine to make the whole enchilada. They are (using the notation $M'$ to mean the compliment of $M$)
$A cap B cap C$ and $Acap B cap C'$ and $A cap B' cap C$ and $A cap B' cap C'$ and $A' cap B cap C$ and $A'cap B cap C'$ and $A' cap B' cap C$ and $A' cap B' cap C'$.
Any of these sets can contain any elements you desire so long as they are disjoint. Simply leave $A cap Bcap C$ empty.
You will have
$A cap B = (Acap B cap C) cup (Acap Bcap C')$ so just make sure $Acap Bcap C$ has no elements and $Acap B cap C'$ has $x$ elements.
$Bcap C = (Acap B cap C)cup (A'cap Bcap C)$ so make sure $(A'cap Bcap C)$ has $y$ elements.
$Acap C = (Acap Bcap C) cup (Acap B' cap C)$ so make sure $(Acap B' cap C)$ has $z$ elements.
Then $A = (Acap Bcap C) cup (Acap Bcap C') cup (Acap B' cap C)cup (Acap B'cap C')$. You can fill $Acap B'cap C'$ with as many elements or no elements as you wish. Fill it with $k$ elements and you have $|A| = a = x+z + k$.
Then $B = (Acap Bcap C) cup (Acap Bcap C') cup (A'cap B cap C)cup (Acap Bcap C')$. You can fill $A'cap Bcap C'$ with as many elements or no elements as you wish. Fill it with $j$ elements and you have $|B| = b = x+y + j$.
Then $C = (Acap Bcap C) cup (Acap B'cap C) cup (A'cap B cap C)cup (A'cap B'cap C)$. You can fill $A'cap B'cap C$ with as many elements or no elements as you wish. Fill it with $m$ elements and you have $|C| = c = y+z + m$.
And that's it. That's a proof.
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
$endgroup$
add a comment |
$begingroup$
$min(x,y,z)$ was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be $0$ is a proof of what can be and the only thing nescessary for that is a simple example.
There are eight DISJOINT sets that combine to make the whole enchilada. They are (using the notation $M'$ to mean the compliment of $M$)
$A cap B cap C$ and $Acap B cap C'$ and $A cap B' cap C$ and $A cap B' cap C'$ and $A' cap B cap C$ and $A'cap B cap C'$ and $A' cap B' cap C$ and $A' cap B' cap C'$.
Any of these sets can contain any elements you desire so long as they are disjoint. Simply leave $A cap Bcap C$ empty.
You will have
$A cap B = (Acap B cap C) cup (Acap Bcap C')$ so just make sure $Acap Bcap C$ has no elements and $Acap B cap C'$ has $x$ elements.
$Bcap C = (Acap B cap C)cup (A'cap Bcap C)$ so make sure $(A'cap Bcap C)$ has $y$ elements.
$Acap C = (Acap Bcap C) cup (Acap B' cap C)$ so make sure $(Acap B' cap C)$ has $z$ elements.
Then $A = (Acap Bcap C) cup (Acap Bcap C') cup (Acap B' cap C)cup (Acap B'cap C')$. You can fill $Acap B'cap C'$ with as many elements or no elements as you wish. Fill it with $k$ elements and you have $|A| = a = x+z + k$.
Then $B = (Acap Bcap C) cup (Acap Bcap C') cup (A'cap B cap C)cup (Acap Bcap C')$. You can fill $A'cap Bcap C'$ with as many elements or no elements as you wish. Fill it with $j$ elements and you have $|B| = b = x+y + j$.
Then $C = (Acap Bcap C) cup (Acap B'cap C) cup (A'cap B cap C)cup (A'cap B'cap C)$. You can fill $A'cap B'cap C$ with as many elements or no elements as you wish. Fill it with $m$ elements and you have $|C| = c = y+z + m$.
And that's it. That's a proof.
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
$endgroup$
add a comment |
$begingroup$
$min(x,y,z)$ was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be $0$ is a proof of what can be and the only thing nescessary for that is a simple example.
There are eight DISJOINT sets that combine to make the whole enchilada. They are (using the notation $M'$ to mean the compliment of $M$)
$A cap B cap C$ and $Acap B cap C'$ and $A cap B' cap C$ and $A cap B' cap C'$ and $A' cap B cap C$ and $A'cap B cap C'$ and $A' cap B' cap C$ and $A' cap B' cap C'$.
Any of these sets can contain any elements you desire so long as they are disjoint. Simply leave $A cap Bcap C$ empty.
You will have
$A cap B = (Acap B cap C) cup (Acap Bcap C')$ so just make sure $Acap Bcap C$ has no elements and $Acap B cap C'$ has $x$ elements.
$Bcap C = (Acap B cap C)cup (A'cap Bcap C)$ so make sure $(A'cap Bcap C)$ has $y$ elements.
$Acap C = (Acap Bcap C) cup (Acap B' cap C)$ so make sure $(Acap B' cap C)$ has $z$ elements.
Then $A = (Acap Bcap C) cup (Acap Bcap C') cup (Acap B' cap C)cup (Acap B'cap C')$. You can fill $Acap B'cap C'$ with as many elements or no elements as you wish. Fill it with $k$ elements and you have $|A| = a = x+z + k$.
Then $B = (Acap Bcap C) cup (Acap Bcap C') cup (A'cap B cap C)cup (Acap Bcap C')$. You can fill $A'cap Bcap C'$ with as many elements or no elements as you wish. Fill it with $j$ elements and you have $|B| = b = x+y + j$.
Then $C = (Acap Bcap C) cup (Acap B'cap C) cup (A'cap B cap C)cup (A'cap B'cap C)$. You can fill $A'cap B'cap C$ with as many elements or no elements as you wish. Fill it with $m$ elements and you have $|C| = c = y+z + m$.
And that's it. That's a proof.
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
$endgroup$
$min(x,y,z)$ was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be $0$ is a proof of what can be and the only thing nescessary for that is a simple example.
There are eight DISJOINT sets that combine to make the whole enchilada. They are (using the notation $M'$ to mean the compliment of $M$)
$A cap B cap C$ and $Acap B cap C'$ and $A cap B' cap C$ and $A cap B' cap C'$ and $A' cap B cap C$ and $A'cap B cap C'$ and $A' cap B' cap C$ and $A' cap B' cap C'$.
Any of these sets can contain any elements you desire so long as they are disjoint. Simply leave $A cap Bcap C$ empty.
You will have
$A cap B = (Acap B cap C) cup (Acap Bcap C')$ so just make sure $Acap Bcap C$ has no elements and $Acap B cap C'$ has $x$ elements.
$Bcap C = (Acap B cap C)cup (A'cap Bcap C)$ so make sure $(A'cap Bcap C)$ has $y$ elements.
$Acap C = (Acap Bcap C) cup (Acap B' cap C)$ so make sure $(Acap B' cap C)$ has $z$ elements.
Then $A = (Acap Bcap C) cup (Acap Bcap C') cup (Acap B' cap C)cup (Acap B'cap C')$. You can fill $Acap B'cap C'$ with as many elements or no elements as you wish. Fill it with $k$ elements and you have $|A| = a = x+z + k$.
Then $B = (Acap Bcap C) cup (Acap Bcap C') cup (A'cap B cap C)cup (Acap Bcap C')$. You can fill $A'cap Bcap C'$ with as many elements or no elements as you wish. Fill it with $j$ elements and you have $|B| = b = x+y + j$.
Then $C = (Acap Bcap C) cup (Acap B'cap C) cup (A'cap B cap C)cup (A'cap B'cap C)$. You can fill $A'cap B'cap C$ with as many elements or no elements as you wish. Fill it with $m$ elements and you have $|C| = c = y+z + m$.
And that's it. That's a proof.
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
edited Jan 8 at 1:00
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
answered Jan 8 at 0:42
fleabloodfleablood
69.5k22685
69.5k22685
add a comment |
add a comment |
$begingroup$
"How do I prove that $A cap B cap C$ can be zero?"
Sometimes a figure is worth a thousand words:
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper boundmin(x,y,z)is proved.
$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
add a comment |
$begingroup$
"How do I prove that $A cap B cap C$ can be zero?"
Sometimes a figure is worth a thousand words:
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
$begingroup$
Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper boundmin(x,y,z)is proved.
$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
add a comment |
$begingroup$
"How do I prove that $A cap B cap C$ can be zero?"
Sometimes a figure is worth a thousand words:
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
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"How do I prove that $A cap B cap C$ can be zero?"
Sometimes a figure is worth a thousand words:
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
edited Jan 8 at 1:07
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
answered Jan 8 at 0:18
David G. StorkDavid G. Stork
10.9k31432
10.9k31432
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Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper boundmin(x,y,z)is proved.
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– asds_asds
Jan 8 at 0:25
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min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
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– fleablood
Jan 8 at 1:01
add a comment |
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Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper boundmin(x,y,z)is proved.
$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
$begingroup$
Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper bound
min(x,y,z) is proved.$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
Indeed it is. Thanks for the figure, I had the same figure in mind I was just wondering if there was a way I could prove it just the way the upper bound
min(x,y,z) is proved.$endgroup$
– asds_asds
Jan 8 at 0:25
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
$begingroup$
min(x,y,z) was a proof of what couldn't be. You couldn't have the intersection be larger. To prove it can be 0 is a proof of what can be and the only thing nescessary for that is a simple example.
$endgroup$
– fleablood
Jan 8 at 1:01
add a comment |
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If you understand why $Acap Bcap C$ can be anything between $0$ and $min(x,y,z)$ then you know how it can be $0$.
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– fleablood
Jan 8 at 0:16