Element is not in intersection of A and BExistence of infinite intersectionIntersection of collection of sets not in the collection of sets?Prove $A cup A' = U$ and$ A cap A' = emptyset$Logic and set theory proof helpWhy is the intersection of the empty set the universe?The distributive property of the intersection of sets applied to the intersection of two grouped unions of setsCountable intersection of open sets results in a closed setHow to prove that intersection of events is a partitioningIntersection of bounded decreasing sets.Why is A intersection A' null?
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Element is not in intersection of A and B
Existence of infinite intersectionIntersection of collection of sets not in the collection of sets?Prove $A cup A' = U$ and$ A cap A' = emptyset$Logic and set theory proof helpWhy is the intersection of the empty set the universe?The distributive property of the intersection of sets applied to the intersection of two grouped unions of setsCountable intersection of open sets results in a closed setHow to prove that intersection of events is a partitioningIntersection of bounded decreasing sets.Why is A intersection A' null?
$begingroup$
I have a statement that $x notin (A cap B)$. I know that this implies $(x notin A$ or $x notin B)$, as it makes sense intuitively.
My question is, by definition of intersection of sets, wouldn't it become $(xnotin A $ and $x notin B)$?? I know it's wrong, but just going by definitions, why is this incorrect?
elementary-set-theory
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
$endgroup$
add a comment |
$begingroup$
I have a statement that $x notin (A cap B)$. I know that this implies $(x notin A$ or $x notin B)$, as it makes sense intuitively.
My question is, by definition of intersection of sets, wouldn't it become $(xnotin A $ and $x notin B)$?? I know it's wrong, but just going by definitions, why is this incorrect?
elementary-set-theory
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
$endgroup$
$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51
add a comment |
$begingroup$
I have a statement that $x notin (A cap B)$. I know that this implies $(x notin A$ or $x notin B)$, as it makes sense intuitively.
My question is, by definition of intersection of sets, wouldn't it become $(xnotin A $ and $x notin B)$?? I know it's wrong, but just going by definitions, why is this incorrect?
elementary-set-theory
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
$endgroup$
I have a statement that $x notin (A cap B)$. I know that this implies $(x notin A$ or $x notin B)$, as it makes sense intuitively.
My question is, by definition of intersection of sets, wouldn't it become $(xnotin A $ and $x notin B)$?? I know it's wrong, but just going by definitions, why is this incorrect?
elementary-set-theory
elementary-set-theory
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
edited Mar 14 at 21:35
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
asked Mar 14 at 16:41
MinYoung KimMinYoung Kim
847
847
$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51
add a comment |
$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51
$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An element $x$ can be in one of the sets, but not the other:
If $A =a,b,c, B = a,b$, $c notin (Acap B)$ and $c notin B$, but $c in A$.
answered Mar 14 at 16:46
user458276user458276
7401314
$endgroup$
add a comment |
$begingroup$
The easiest way to see this is to use truth tables.
$beginarrayc
x in A & x in B & x in Acap B \ hline
T&T&T\
T&F&F\
F&T&F\
F&F&F\
endarray
$
answered Mar 14 at 16:48
nammienammie
3279
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
An element $x$ can be in one of the sets, but not the other:
If $A =a,b,c, B = a,b$, $c notin (Acap B)$ and $c notin B$, but $c in A$.
answered Mar 14 at 16:46
user458276user458276
7401314
$endgroup$
add a comment |
$begingroup$
An element $x$ can be in one of the sets, but not the other:
If $A =a,b,c, B = a,b$, $c notin (Acap B)$ and $c notin B$, but $c in A$.
answered Mar 14 at 16:46
user458276user458276
7401314
$endgroup$
add a comment |
$begingroup$
An element $x$ can be in one of the sets, but not the other:
If $A =a,b,c, B = a,b$, $c notin (Acap B)$ and $c notin B$, but $c in A$.
answered Mar 14 at 16:46
user458276user458276
7401314
$endgroup$
An element $x$ can be in one of the sets, but not the other:
If $A =a,b,c, B = a,b$, $c notin (Acap B)$ and $c notin B$, but $c in A$.
answered Mar 14 at 16:46
user458276user458276
7401314
answered Mar 14 at 16:46
user458276user458276
7401314
answered Mar 14 at 16:46
user458276user458276
7401314
answered Mar 14 at 16:46
user458276user458276
7401314
7401314
add a comment |
add a comment |
$begingroup$
The easiest way to see this is to use truth tables.
$beginarrayc
x in A & x in B & x in Acap B \ hline
T&T&T\
T&F&F\
F&T&F\
F&F&F\
endarray
$
answered Mar 14 at 16:48
nammienammie
3279
$endgroup$
add a comment |
$begingroup$
The easiest way to see this is to use truth tables.
$beginarrayc
x in A & x in B & x in Acap B \ hline
T&T&T\
T&F&F\
F&T&F\
F&F&F\
endarray
$
answered Mar 14 at 16:48
nammienammie
3279
$endgroup$
add a comment |
$begingroup$
The easiest way to see this is to use truth tables.
$beginarrayc
x in A & x in B & x in Acap B \ hline
T&T&T\
T&F&F\
F&T&F\
F&F&F\
endarray
$
answered Mar 14 at 16:48
nammienammie
3279
$endgroup$
The easiest way to see this is to use truth tables.
$beginarrayc
x in A & x in B & x in Acap B \ hline
T&T&T\
T&F&F\
F&T&F\
F&F&F\
endarray
$
answered Mar 14 at 16:48
nammienammie
3279
answered Mar 14 at 16:48
nammienammie
3279
answered Mar 14 at 16:48
nammienammie
3279
answered Mar 14 at 16:48
nammienammie
3279
3279
add a comment |
add a comment |
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$begingroup$
The intersection is amde of the elements that are in both $A$ and $B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:45
$begingroup$
Thus, if $x notin (A cap B)$ means that is not in one of them.
$endgroup$
– Mauro ALLEGRANZA
Mar 14 at 16:46
$begingroup$
This is closely related to De Morgan's Laws, one of which states the complement of an intersection is the union of two complements.
$endgroup$
– hardmath
Mar 14 at 16:51
$begingroup$
If I am not rich and famous then I am either not rich or not famous. I don't have to be both not rich and not famous.
$endgroup$
– John Douma
Mar 14 at 17:13
$begingroup$
I guess my question is, I know what the correct expression is, and it makes sense, but using the DEFINITION of intersection $x in A$ and $x in B$, wouldn't $x notin A cap B rightarrow x notin A and x notin B$? What is the error in my thinking here? should $x notin (A cap B)$ be viewed as $x in NOT(A cap B)$?
$endgroup$
– MinYoung Kim
Mar 14 at 18:51