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What do these sets mean? [on hold]
The Next CEO of Stack OverflowWhat does the following mean for sets?Equinumerosity of infinite setsWhat is the difference between these two simple sets?Find the cardinality of these setsNotation for a subset of a powersetWhat does $^(L)K$ mean?What do these sets mean? Relational AlgebraWhat Does the Logical Equivalence Symbol Mean Between Two Sets?Finite families of finite sets closed under union and intersectionWhat is the name of this operation between sets?
$begingroup$
What is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
In C, 1 is repeated...
Thank you so much
elementary-set-theory
edited 2 days ago
StackTD
24.3k2254
asked 2 days ago
WwwWww
112
$endgroup$
put on hold as off-topic by José Carlos Santos, Shailesh, user21820, Saad, RRL yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, user21820, Saad, RRL
add a comment |
$begingroup$
What is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
In C, 1 is repeated...
Thank you so much
elementary-set-theory
edited 2 days ago
StackTD
24.3k2254
asked 2 days ago
WwwWww
112
$endgroup$
put on hold as off-topic by José Carlos Santos, Shailesh, user21820, Saad, RRL yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, user21820, Saad, RRL
2
$begingroup$
1 is a set containing 1; it is not the same as 1 itself
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
$endgroup$
– drhab
2 days ago
add a comment |
$begingroup$
What is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
In C, 1 is repeated...
Thank you so much
elementary-set-theory
edited 2 days ago
StackTD
24.3k2254
asked 2 days ago
WwwWww
112
$endgroup$
What is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
In C, 1 is repeated...
Thank you so much
elementary-set-theory
elementary-set-theory
edited 2 days ago
StackTD
24.3k2254
asked 2 days ago
WwwWww
112
edited 2 days ago
StackTD
24.3k2254
asked 2 days ago
WwwWww
112
edited 2 days ago
StackTD
24.3k2254
edited 2 days ago
StackTD
24.3k2254
edited 2 days ago
StackTD
24.3k2254
24.3k2254
asked 2 days ago
WwwWww
112
asked 2 days ago
WwwWww
112
asked 2 days ago
WwwWww
112
112
put on hold as off-topic by José Carlos Santos, Shailesh, user21820, Saad, RRL yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, user21820, Saad, RRL
put on hold as off-topic by José Carlos Santos, Shailesh, user21820, Saad, RRL yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, user21820, Saad, RRL
2
$begingroup$
1 is a set containing 1; it is not the same as 1 itself
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
$endgroup$
– drhab
2 days ago
add a comment |
2
$begingroup$
1 is a set containing 1; it is not the same as 1 itself
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
$endgroup$
– drhab
2 days ago
2
2
$begingroup$
1 is a set containing 1; it is not the same as 1 itself
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
1 is a set containing 1; it is not the same as 1 itself
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
$endgroup$
– drhab
2 days ago
$begingroup$
The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
$endgroup$
– drhab
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In a nutshell: "the number x" and "the set with $x$ as an element", written $leftxright$, are not the same.
what is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
$B$ is a set with four elements, those elements are the numbers $1$, $2$, $3$ and $4$;
$A$ is a set with four elements: those elements are the three numbers $2$, $3$ and $4$ and the set containing $1$ (written $left1right$);
$C=leftcolorblueleft1,left2rightright,colorgreen3,colorpurpleleft1rightright$ is a set with three elements:- the number $3$ (in green);
- the set containing $1$ (in purple);
- the set containing: the number $1$ and the set $left2right$ (in blue).
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
|
show 2 more comments
$begingroup$
A contains one set; B contains none.
C contains two sets, and one of these sets contains a set.
answered 2 days ago
Jossie CalderonJossie Calderon
291111
$endgroup$
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
add a comment |
$begingroup$
A lot of the answers here just answer the question in a way that I'm sure the OP has seen or read before. So I would like to explain in a way that would explain sets to say a 10-year-old. Say you have three cases, the first where you have $2$ bags, the second case where you have $1$ bag and the third case where you have $4$. Now in the first case say you put all of your objects into one bag and then you decide that some of the objects in the bag have common properties and you can group them together, but they still belong to the larger bag, so you take your second bag, put all the common items in and then put that inside the bigger bag.
Now take your second case where you just dump all the objects into one bag and don't bother with any inter-grouping. This is the second set in your question.
Third, you take the $4$ bags and put all the items in one giant bag and then take one item out that does not share any properties with the other but you still feel is important enough to have its own category. Then you find that you can have another one where multiple items can be grouped together and you put these in another bag, inside which you see one thing is important enough once again to have its own place but still belong to this larger group. Now you place the two bags, one with only one special item inside and the other with some items and one special item kept inside that bag in its own bag.
I know, got a little confusing there but if you understood what I was trying to do then you'll know what I'm talking about, This is analogous to sets and just common elements. The common elements are like all things just placed inside a big bag and the ones that have set brackets around them within the sets are the ones that are placed inside smaller bags and put inside the larger bag along with the rest of the items.
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In a nutshell: "the number x" and "the set with $x$ as an element", written $leftxright$, are not the same.
what is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
$B$ is a set with four elements, those elements are the numbers $1$, $2$, $3$ and $4$;
$A$ is a set with four elements: those elements are the three numbers $2$, $3$ and $4$ and the set containing $1$ (written $left1right$);
$C=leftcolorblueleft1,left2rightright,colorgreen3,colorpurpleleft1rightright$ is a set with three elements:- the number $3$ (in green);
- the set containing $1$ (in purple);
- the set containing: the number $1$ and the set $left2right$ (in blue).
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
|
show 2 more comments
$begingroup$
In a nutshell: "the number x" and "the set with $x$ as an element", written $leftxright$, are not the same.
what is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
$B$ is a set with four elements, those elements are the numbers $1$, $2$, $3$ and $4$;
$A$ is a set with four elements: those elements are the three numbers $2$, $3$ and $4$ and the set containing $1$ (written $left1right$);
$C=leftcolorblueleft1,left2rightright,colorgreen3,colorpurpleleft1rightright$ is a set with three elements:- the number $3$ (in green);
- the set containing $1$ (in purple);
- the set containing: the number $1$ and the set $left2right$ (in blue).
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
|
show 2 more comments
$begingroup$
In a nutshell: "the number x" and "the set with $x$ as an element", written $leftxright$, are not the same.
what is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
$B$ is a set with four elements, those elements are the numbers $1$, $2$, $3$ and $4$;
$A$ is a set with four elements: those elements are the three numbers $2$, $3$ and $4$ and the set containing $1$ (written $left1right$);
$C=leftcolorblueleft1,left2rightright,colorgreen3,colorpurpleleft1rightright$ is a set with three elements:- the number $3$ (in green);
- the set containing $1$ (in purple);
- the set containing: the number $1$ and the set $left2right$ (in blue).
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
In a nutshell: "the number x" and "the set with $x$ as an element", written $leftxright$, are not the same.
what is the difference between A=1,2,3,4 and B=1,2,3,4?
why C=1,2,3,1 exists? what does these sets mean ?
$B$ is a set with four elements, those elements are the numbers $1$, $2$, $3$ and $4$;
$A$ is a set with four elements: those elements are the three numbers $2$, $3$ and $4$ and the set containing $1$ (written $left1right$);
$C=leftcolorblueleft1,left2rightright,colorgreen3,colorpurpleleft1rightright$ is a set with three elements:- the number $3$ (in green);
- the set containing $1$ (in purple);
- the set containing: the number $1$ and the set $left2right$ (in blue).
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
24.3k2254
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
|
show 2 more comments
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
one more question: C=1,2,3,2 then 2 should not exists two times right ? thank you btw, very clear
$endgroup$
– Www
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
What do you mean with "exist"? The set containing (only) 2, written 2, is an element of C. Another element of C, the set 1,2, has that same set containing (only) 2, written 2, as one of its (two) elements.
$endgroup$
– StackTD
2 days ago
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
I mean Can I delete one of the 2 in set C=1,2,3,2
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
Since 2 is repeated 2 times . thank you
$endgroup$
– Www
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
$begingroup$
In your question above that last number is a 1, not a 2...? But in both cases: no, you can't. In 1,2,2, you can delete a 2; in 1,2,2 you can't because there's only one element which is 2, since it's not the same as the element 2. See the first line of my answer.
$endgroup$
– StackTD
yesterday
|
show 2 more comments
$begingroup$
A contains one set; B contains none.
C contains two sets, and one of these sets contains a set.
answered 2 days ago
Jossie CalderonJossie Calderon
291111
$endgroup$
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
add a comment |
$begingroup$
A contains one set; B contains none.
C contains two sets, and one of these sets contains a set.
answered 2 days ago
Jossie CalderonJossie Calderon
291111
$endgroup$
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
add a comment |
$begingroup$
A contains one set; B contains none.
C contains two sets, and one of these sets contains a set.
answered 2 days ago
Jossie CalderonJossie Calderon
291111
$endgroup$
A contains one set; B contains none.
C contains two sets, and one of these sets contains a set.
answered 2 days ago
Jossie CalderonJossie Calderon
291111
answered 2 days ago
Jossie CalderonJossie Calderon
291111
answered 2 days ago
Jossie CalderonJossie Calderon
291111
answered 2 days ago
Jossie CalderonJossie Calderon
291111
291111
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
add a comment |
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
What if $2,3,4$ are also sets (which is quite common in set-theory)?
$endgroup$
– drhab
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
@drhab then you argue with the fact that the sets are distinct
$endgroup$
– B.Swan
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
$begingroup$
What I mean is that - if $2,3,4$ are sets - then e.g. $A$ contains $4$ sets as elements (not just one as you state). From the question on its own it cannot be deduced that $A$ contains one set as element (or that $B$ contains none).
$endgroup$
– drhab
2 days ago
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A lot of the answers here just answer the question in a way that I'm sure the OP has seen or read before. So I would like to explain in a way that would explain sets to say a 10-year-old. Say you have three cases, the first where you have $2$ bags, the second case where you have $1$ bag and the third case where you have $4$. Now in the first case say you put all of your objects into one bag and then you decide that some of the objects in the bag have common properties and you can group them together, but they still belong to the larger bag, so you take your second bag, put all the common items in and then put that inside the bigger bag.
Now take your second case where you just dump all the objects into one bag and don't bother with any inter-grouping. This is the second set in your question.
Third, you take the $4$ bags and put all the items in one giant bag and then take one item out that does not share any properties with the other but you still feel is important enough to have its own category. Then you find that you can have another one where multiple items can be grouped together and you put these in another bag, inside which you see one thing is important enough once again to have its own place but still belong to this larger group. Now you place the two bags, one with only one special item inside and the other with some items and one special item kept inside that bag in its own bag.
I know, got a little confusing there but if you understood what I was trying to do then you'll know what I'm talking about, This is analogous to sets and just common elements. The common elements are like all things just placed inside a big bag and the ones that have set brackets around them within the sets are the ones that are placed inside smaller bags and put inside the larger bag along with the rest of the items.
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
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add a comment |
$begingroup$
A lot of the answers here just answer the question in a way that I'm sure the OP has seen or read before. So I would like to explain in a way that would explain sets to say a 10-year-old. Say you have three cases, the first where you have $2$ bags, the second case where you have $1$ bag and the third case where you have $4$. Now in the first case say you put all of your objects into one bag and then you decide that some of the objects in the bag have common properties and you can group them together, but they still belong to the larger bag, so you take your second bag, put all the common items in and then put that inside the bigger bag.
Now take your second case where you just dump all the objects into one bag and don't bother with any inter-grouping. This is the second set in your question.
Third, you take the $4$ bags and put all the items in one giant bag and then take one item out that does not share any properties with the other but you still feel is important enough to have its own category. Then you find that you can have another one where multiple items can be grouped together and you put these in another bag, inside which you see one thing is important enough once again to have its own place but still belong to this larger group. Now you place the two bags, one with only one special item inside and the other with some items and one special item kept inside that bag in its own bag.
I know, got a little confusing there but if you understood what I was trying to do then you'll know what I'm talking about, This is analogous to sets and just common elements. The common elements are like all things just placed inside a big bag and the ones that have set brackets around them within the sets are the ones that are placed inside smaller bags and put inside the larger bag along with the rest of the items.
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
A lot of the answers here just answer the question in a way that I'm sure the OP has seen or read before. So I would like to explain in a way that would explain sets to say a 10-year-old. Say you have three cases, the first where you have $2$ bags, the second case where you have $1$ bag and the third case where you have $4$. Now in the first case say you put all of your objects into one bag and then you decide that some of the objects in the bag have common properties and you can group them together, but they still belong to the larger bag, so you take your second bag, put all the common items in and then put that inside the bigger bag.
Now take your second case where you just dump all the objects into one bag and don't bother with any inter-grouping. This is the second set in your question.
Third, you take the $4$ bags and put all the items in one giant bag and then take one item out that does not share any properties with the other but you still feel is important enough to have its own category. Then you find that you can have another one where multiple items can be grouped together and you put these in another bag, inside which you see one thing is important enough once again to have its own place but still belong to this larger group. Now you place the two bags, one with only one special item inside and the other with some items and one special item kept inside that bag in its own bag.
I know, got a little confusing there but if you understood what I was trying to do then you'll know what I'm talking about, This is analogous to sets and just common elements. The common elements are like all things just placed inside a big bag and the ones that have set brackets around them within the sets are the ones that are placed inside smaller bags and put inside the larger bag along with the rest of the items.
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
A lot of the answers here just answer the question in a way that I'm sure the OP has seen or read before. So I would like to explain in a way that would explain sets to say a 10-year-old. Say you have three cases, the first where you have $2$ bags, the second case where you have $1$ bag and the third case where you have $4$. Now in the first case say you put all of your objects into one bag and then you decide that some of the objects in the bag have common properties and you can group them together, but they still belong to the larger bag, so you take your second bag, put all the common items in and then put that inside the bigger bag.
Now take your second case where you just dump all the objects into one bag and don't bother with any inter-grouping. This is the second set in your question.
Third, you take the $4$ bags and put all the items in one giant bag and then take one item out that does not share any properties with the other but you still feel is important enough to have its own category. Then you find that you can have another one where multiple items can be grouped together and you put these in another bag, inside which you see one thing is important enough once again to have its own place but still belong to this larger group. Now you place the two bags, one with only one special item inside and the other with some items and one special item kept inside that bag in its own bag.
I know, got a little confusing there but if you understood what I was trying to do then you'll know what I'm talking about, This is analogous to sets and just common elements. The common elements are like all things just placed inside a big bag and the ones that have set brackets around them within the sets are the ones that are placed inside smaller bags and put inside the larger bag along with the rest of the items.
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
answered 2 days ago
Prakhar NagpalPrakhar Nagpal
752318
752318
add a comment |
add a comment |
2
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1 is a set containing 1; it is not the same as 1 itself
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– J. W. Tanner
2 days ago
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The set $C$ has $3$ elements: $1,2$, $3$ and $1$. None of them is mentioned twice. Two of the elements contain $1$ as an element.
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– drhab
2 days ago