Mixing
Can sets A, B and C be transitive or does transitivity hold just for individual relations. Also does missing...
$begingroup$
Can sets A, B and C be transitive or can we describe only individual relations as being transitive. Also does missing a single relation between two sets mean we can't describe the relation as transitive?
So, the following is transitive where the operator is <= (less than or equals) between A and B and B and C
A = {1,2} B = {2,3} C = {5,6}
A <= B
Ra 1 <= 2
Rb 2 <= 2
Rc 1 <= 3
Rd 2 <= 3
B <= C
Re 2 <= 5
Rf 2 <= 6
Rg 3 <= 5
Rh 3 <= 6
because all relations between A <= B and B <= C
we can say these sets have transitivity and therefore
A <= C
however...
if we change set C to {2,5,6} (I added an extra 2)
additional relations between B <= C are as follows
2 <= 2
// 3 <= 2 does not hold so this is not a relation in regards to the operator <= between 3 and 2
So, could we originally describe the relationship of the sets ARB -> BRC as being transitive but now we cannot?
(apologies, please correct any incorrect terminology, I'm self learning this stuff and I have never studied this stuff before! I may have got some of the fundamentals incorrect too)
relations equivalence-relations
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
$endgroup$
add a comment |
$begingroup$
Can sets A, B and C be transitive or can we describe only individual relations as being transitive. Also does missing a single relation between two sets mean we can't describe the relation as transitive?
So, the following is transitive where the operator is <= (less than or equals) between A and B and B and C
A = {1,2} B = {2,3} C = {5,6}
A <= B
Ra 1 <= 2
Rb 2 <= 2
Rc 1 <= 3
Rd 2 <= 3
B <= C
Re 2 <= 5
Rf 2 <= 6
Rg 3 <= 5
Rh 3 <= 6
because all relations between A <= B and B <= C
we can say these sets have transitivity and therefore
A <= C
however...
if we change set C to {2,5,6} (I added an extra 2)
additional relations between B <= C are as follows
2 <= 2
// 3 <= 2 does not hold so this is not a relation in regards to the operator <= between 3 and 2
So, could we originally describe the relationship of the sets ARB -> BRC as being transitive but now we cannot?
(apologies, please correct any incorrect terminology, I'm self learning this stuff and I have never studied this stuff before! I may have got some of the fundamentals incorrect too)
relations equivalence-relations
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
$endgroup$
add a comment |
$begingroup$
Can sets A, B and C be transitive or can we describe only individual relations as being transitive. Also does missing a single relation between two sets mean we can't describe the relation as transitive?
So, the following is transitive where the operator is <= (less than or equals) between A and B and B and C
A = {1,2} B = {2,3} C = {5,6}
A <= B
Ra 1 <= 2
Rb 2 <= 2
Rc 1 <= 3
Rd 2 <= 3
B <= C
Re 2 <= 5
Rf 2 <= 6
Rg 3 <= 5
Rh 3 <= 6
because all relations between A <= B and B <= C
we can say these sets have transitivity and therefore
A <= C
however...
if we change set C to {2,5,6} (I added an extra 2)
additional relations between B <= C are as follows
2 <= 2
// 3 <= 2 does not hold so this is not a relation in regards to the operator <= between 3 and 2
So, could we originally describe the relationship of the sets ARB -> BRC as being transitive but now we cannot?
(apologies, please correct any incorrect terminology, I'm self learning this stuff and I have never studied this stuff before! I may have got some of the fundamentals incorrect too)
relations equivalence-relations
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
$endgroup$
Can sets A, B and C be transitive or can we describe only individual relations as being transitive. Also does missing a single relation between two sets mean we can't describe the relation as transitive?
So, the following is transitive where the operator is <= (less than or equals) between A and B and B and C
A = {1,2} B = {2,3} C = {5,6}
A <= B
Ra 1 <= 2
Rb 2 <= 2
Rc 1 <= 3
Rd 2 <= 3
B <= C
Re 2 <= 5
Rf 2 <= 6
Rg 3 <= 5
Rh 3 <= 6
because all relations between A <= B and B <= C
we can say these sets have transitivity and therefore
A <= C
however...
if we change set C to {2,5,6} (I added an extra 2)
additional relations between B <= C are as follows
2 <= 2
// 3 <= 2 does not hold so this is not a relation in regards to the operator <= between 3 and 2
So, could we originally describe the relationship of the sets ARB -> BRC as being transitive but now we cannot?
(apologies, please correct any incorrect terminology, I'm self learning this stuff and I have never studied this stuff before! I may have got some of the fundamentals incorrect too)
relations equivalence-relations
relations equivalence-relations
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
asked Jan 9 at 16:40
orangesherbertorangesherbert
1033
1033
add a comment |
add a comment |
1 Answer
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$begingroup$
You need to define what you mean by one set being less than or equal to another. It seems your definition is that all elements of the first set are less than or equal to all the elements of the second, so $A le B$ is the same as $max(A) le min(B)$ for finite sets. That is a fine definition of a relation between sets of numbers.
Now we can ask if this relation is transitive across all sets of sets. Your example does not show that it is not transitive across all sets of sets because with the changed $C$ you do not have $B le C$. In fact it is, because given $Ale B$ and $B le C$ we can show that $A le C$ by invoking the definition. We are given $max (A) le min (B)$ and $max (B) le min (C)$. Then $max (A) le min (B)le max (B) le min (C)$ and we have shown that $A le C$
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
add a comment |
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1 Answer
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$begingroup$
You need to define what you mean by one set being less than or equal to another. It seems your definition is that all elements of the first set are less than or equal to all the elements of the second, so $A le B$ is the same as $max(A) le min(B)$ for finite sets. That is a fine definition of a relation between sets of numbers.
Now we can ask if this relation is transitive across all sets of sets. Your example does not show that it is not transitive across all sets of sets because with the changed $C$ you do not have $B le C$. In fact it is, because given $Ale B$ and $B le C$ we can show that $A le C$ by invoking the definition. We are given $max (A) le min (B)$ and $max (B) le min (C)$. Then $max (A) le min (B)le max (B) le min (C)$ and we have shown that $A le C$
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
add a comment |
$begingroup$
You need to define what you mean by one set being less than or equal to another. It seems your definition is that all elements of the first set are less than or equal to all the elements of the second, so $A le B$ is the same as $max(A) le min(B)$ for finite sets. That is a fine definition of a relation between sets of numbers.
Now we can ask if this relation is transitive across all sets of sets. Your example does not show that it is not transitive across all sets of sets because with the changed $C$ you do not have $B le C$. In fact it is, because given $Ale B$ and $B le C$ we can show that $A le C$ by invoking the definition. We are given $max (A) le min (B)$ and $max (B) le min (C)$. Then $max (A) le min (B)le max (B) le min (C)$ and we have shown that $A le C$
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
add a comment |
$begingroup$
You need to define what you mean by one set being less than or equal to another. It seems your definition is that all elements of the first set are less than or equal to all the elements of the second, so $A le B$ is the same as $max(A) le min(B)$ for finite sets. That is a fine definition of a relation between sets of numbers.
Now we can ask if this relation is transitive across all sets of sets. Your example does not show that it is not transitive across all sets of sets because with the changed $C$ you do not have $B le C$. In fact it is, because given $Ale B$ and $B le C$ we can show that $A le C$ by invoking the definition. We are given $max (A) le min (B)$ and $max (B) le min (C)$. Then $max (A) le min (B)le max (B) le min (C)$ and we have shown that $A le C$
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
You need to define what you mean by one set being less than or equal to another. It seems your definition is that all elements of the first set are less than or equal to all the elements of the second, so $A le B$ is the same as $max(A) le min(B)$ for finite sets. That is a fine definition of a relation between sets of numbers.
Now we can ask if this relation is transitive across all sets of sets. Your example does not show that it is not transitive across all sets of sets because with the changed $C$ you do not have $B le C$. In fact it is, because given $Ale B$ and $B le C$ we can show that $A le C$ by invoking the definition. We are given $max (A) le min (B)$ and $max (B) le min (C)$. Then $max (A) le min (B)le max (B) le min (C)$ and we have shown that $A le C$
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 17:02
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
add a comment |
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
Is it transitive due to the definition of the relationship between the sets or/and due to the individual elements and the operator? I don't follow how it is transitive
$endgroup$
– orangesherbert
Jan 9 at 17:35
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
We defined the relationship between sets based on the relationships between the elements. It is a partial order because there are sets that cannot be compared, like ${1,5}$ and ${2,4}$. We need the definition of the relationship between sets and the transitivity of $le$ on numbers to make the proof work. My last two sentences are the proof. The next to last translates what we are given about the sets to statements about numbers based on the definition. The last then uses that to establish what we need to say $A le C$
$endgroup$
– Ross Millikan
Jan 9 at 17:48
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
$begingroup$
I read your answer and your comment yesterday and they made no sense, I read them today and they are crystal clear. I guess I just needed to let it digest a little in my brain. Many thanks, yes, I guess it is all about that definition of one set being less than or equal to the other; I could define that operator in different ways.
$endgroup$
– orangesherbert
Jan 10 at 13:59
add a comment |
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