Mixing
Why are set operations always showed with a Venn Diagram where both A and B are intersecting?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
So there is this question involving the pictorial representations of set operations (i.e. $A - B$).
In each Venn Diagram, the A and B circles (sets) are always shown overlapping. For example, $A - B$:
$A - B$">
What if the circles $A$ and $B$ were separate? How would defining the set operation be showed? Thanks.
elementary-set-theory proof-verification proof-writing
asked 4 hours ago
Strikers
16710
add a comment |Â
up vote
3
down vote
favorite
So there is this question involving the pictorial representations of set operations (i.e. $A - B$).
In each Venn Diagram, the A and B circles (sets) are always shown overlapping. For example, $A - B$:
$A - B$">
What if the circles $A$ and $B$ were separate? How would defining the set operation be showed? Thanks.
elementary-set-theory proof-verification proof-writing
asked 4 hours ago
Strikers
16710
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
1
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So there is this question involving the pictorial representations of set operations (i.e. $A - B$).
In each Venn Diagram, the A and B circles (sets) are always shown overlapping. For example, $A - B$:
$A - B$">
What if the circles $A$ and $B$ were separate? How would defining the set operation be showed? Thanks.
elementary-set-theory proof-verification proof-writing
asked 4 hours ago
Strikers
16710
So there is this question involving the pictorial representations of set operations (i.e. $A - B$).
In each Venn Diagram, the A and B circles (sets) are always shown overlapping. For example, $A - B$:
$A - B$">
What if the circles $A$ and $B$ were separate? How would defining the set operation be showed? Thanks.
elementary-set-theory proof-verification proof-writing
elementary-set-theory proof-verification proof-writing
asked 4 hours ago
Strikers
16710
asked 4 hours ago
Strikers
16710
edited 3 hours ago
asked 4 hours ago
Strikers
16710
asked 4 hours ago
Strikers
16710
asked 4 hours ago
Strikers
16710
16710
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
1
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago
add a comment |Â
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
1
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
1
1
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
In Venn diagrams, the circles are shown to intersect each other to provide for all possible scenarios. Now dots (representing elements) that lie in both circles in such diagrams represent elements that belong to both sets, and if these two sets are disjoint (have no elements in common), that just means that there are really no dots lying in both circles. If the circles were drawn as non-intersecting, there would be no way to represent elements that belong to both sets.
answered 3 hours ago
Will Hunting
35516
Okay, this makes senses, thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
1
down vote
I am assuming that you are using $A-B$ to denote $Asetminus B$, and that the right-hand circle is $A$, while the other is $B$. If the circles were not overlapping, then $Acap B=emptyset$. In that case:$$A-B=Asetminus B=A$$
answered 3 hours ago
clathratus
28210
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In Venn diagrams, the circles are shown to intersect each other to provide for all possible scenarios. Now dots (representing elements) that lie in both circles in such diagrams represent elements that belong to both sets, and if these two sets are disjoint (have no elements in common), that just means that there are really no dots lying in both circles. If the circles were drawn as non-intersecting, there would be no way to represent elements that belong to both sets.
answered 3 hours ago
Will Hunting
35516
Okay, this makes senses, thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
In Venn diagrams, the circles are shown to intersect each other to provide for all possible scenarios. Now dots (representing elements) that lie in both circles in such diagrams represent elements that belong to both sets, and if these two sets are disjoint (have no elements in common), that just means that there are really no dots lying in both circles. If the circles were drawn as non-intersecting, there would be no way to represent elements that belong to both sets.
answered 3 hours ago
Will Hunting
35516
Okay, this makes senses, thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In Venn diagrams, the circles are shown to intersect each other to provide for all possible scenarios. Now dots (representing elements) that lie in both circles in such diagrams represent elements that belong to both sets, and if these two sets are disjoint (have no elements in common), that just means that there are really no dots lying in both circles. If the circles were drawn as non-intersecting, there would be no way to represent elements that belong to both sets.
answered 3 hours ago
Will Hunting
35516
In Venn diagrams, the circles are shown to intersect each other to provide for all possible scenarios. Now dots (representing elements) that lie in both circles in such diagrams represent elements that belong to both sets, and if these two sets are disjoint (have no elements in common), that just means that there are really no dots lying in both circles. If the circles were drawn as non-intersecting, there would be no way to represent elements that belong to both sets.
answered 3 hours ago
Will Hunting
35516
answered 3 hours ago
Will Hunting
35516
answered 3 hours ago
Will Hunting
35516
answered 3 hours ago
Will Hunting
35516
35516
Okay, this makes senses, thanks.
– Strikers
3 hours ago
add a comment |Â
Okay, this makes senses, thanks.
– Strikers
3 hours ago
Okay, this makes senses, thanks.
– Strikers
3 hours ago
Okay, this makes senses, thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
1
down vote
I am assuming that you are using $A-B$ to denote $Asetminus B$, and that the right-hand circle is $A$, while the other is $B$. If the circles were not overlapping, then $Acap B=emptyset$. In that case:$$A-B=Asetminus B=A$$
answered 3 hours ago
clathratus
28210
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
1
down vote
I am assuming that you are using $A-B$ to denote $Asetminus B$, and that the right-hand circle is $A$, while the other is $B$. If the circles were not overlapping, then $Acap B=emptyset$. In that case:$$A-B=Asetminus B=A$$
answered 3 hours ago
clathratus
28210
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I am assuming that you are using $A-B$ to denote $Asetminus B$, and that the right-hand circle is $A$, while the other is $B$. If the circles were not overlapping, then $Acap B=emptyset$. In that case:$$A-B=Asetminus B=A$$
answered 3 hours ago
clathratus
28210
I am assuming that you are using $A-B$ to denote $Asetminus B$, and that the right-hand circle is $A$, while the other is $B$. If the circles were not overlapping, then $Acap B=emptyset$. In that case:$$A-B=Asetminus B=A$$
answered 3 hours ago
clathratus
28210
answered 3 hours ago
clathratus
28210
answered 3 hours ago
clathratus
28210
answered 3 hours ago
clathratus
28210
28210
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
add a comment |Â
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
Okay, this is good additional information to the question. Thanks.
– Strikers
3 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2941546%2fwhy-are-set-operations-always-showed-with-a-venn-diagram-where-both-a-and-b-are%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If the circles do not overlap, then this depicts the sets being disjoint, i.e. $Acap B=emptyset.$ The overlap represents the intersection: the elements that are in both sets. Beyond that, I'm not sure I understand your question.
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Hi, okay so if the sets are disjoint, why isn't that the case when describing i.e. $A - B$. Why is it that they are always joint when showing the different set operations?
– Strikers
3 hours ago
1
They are shading the region that represents the set. The red region is $A-B$ (if $A$ is the right circle and $B$ is the left circle). If the circles didn't overlap, all of $A$ would be red. This means that if $Acap B=emptyset,$ then $A-B = A.$
– spaceisdarkgreen
3 hours ago
@spaceisdarkgreen Okay, thanks.
– Strikers
3 hours ago