Why are set operations always showed with a Venn Diagram where both A and B are intersecting?
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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
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
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
Oct 4 at 2:10
@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?
– user595054
Oct 4 at 2:17
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
Oct 4 at 2:20
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21
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
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
edited Oct 4 at 2:15
asked Oct 4 at 2:05
user595054
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
Oct 4 at 2:10
@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?
– user595054
Oct 4 at 2:17
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
Oct 4 at 2:20
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21
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
Oct 4 at 2:10
@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?
– user595054
Oct 4 at 2:17
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
Oct 4 at 2:20
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21
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
Oct 4 at 2:10
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
Oct 4 at 2:10
@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?
– user595054
Oct 4 at 2:17
@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?
– user595054
Oct 4 at 2:17
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
Oct 4 at 2:20
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
Oct 4 at 2:20
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
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 Oct 4 at 2:17
Will Hunting
42016
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
up vote
0
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 Oct 4 at 2:24
clathratus
48512
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
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 Oct 4 at 2:17
Will Hunting
42016
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
up vote
2
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 Oct 4 at 2:17
Will Hunting
42016
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
up vote
2
down vote
accepted
up vote
2
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 Oct 4 at 2:17
Will Hunting
42016
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 Oct 4 at 2:17
Will Hunting
42016
answered Oct 4 at 2:17
Will Hunting
42016
answered Oct 4 at 2:17
Will Hunting
42016
answered Oct 4 at 2:17
Will Hunting
42016
42016
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
Okay, this makes senses, thanks.
– user595054
Oct 4 at 2:21
add a comment |Â
up vote
0
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 Oct 4 at 2:24
clathratus
48512
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
add a comment |Â
up vote
0
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 Oct 4 at 2:24
clathratus
48512
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
add a comment |Â
up vote
0
down vote
up vote
0
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 Oct 4 at 2:24
clathratus
48512
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 Oct 4 at 2:24
clathratus
48512
answered Oct 4 at 2:24
clathratus
48512
answered Oct 4 at 2:24
clathratus
48512
answered Oct 4 at 2:24
clathratus
48512
48512
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
add a comment |Â
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
Okay, this is good additional information to the question. Thanks.
– user595054
Oct 4 at 2:27
add a comment |Â
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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
Oct 4 at 2:10
@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?
– user595054
Oct 4 at 2:17
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
Oct 4 at 2:20
@spaceisdarkgreen Okay, thanks.
– user595054
Oct 4 at 2:21