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Having a hard time wrapping my head around order relations
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I am having a hard time trying to understand what partial order and total order relations are. I need to answer a question in which I am given "two sequences S1 and S2 of n elements, possibly containing duplicates, on which a total order relation is defined".
Can someone give me a concrete example of what S1 and S2 might look like? Using that same example, please also explain what exactly makes this a total order relation.
order-theory
asked Aug 6 at 1:04
user928112
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I am having a hard time trying to understand what partial order and total order relations are. I need to answer a question in which I am given "two sequences S1 and S2 of n elements, possibly containing duplicates, on which a total order relation is defined".
Can someone give me a concrete example of what S1 and S2 might look like? Using that same example, please also explain what exactly makes this a total order relation.
order-theory
asked Aug 6 at 1:04
user928112
1716
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having a hard time trying to understand what partial order and total order relations are. I need to answer a question in which I am given "two sequences S1 and S2 of n elements, possibly containing duplicates, on which a total order relation is defined".
Can someone give me a concrete example of what S1 and S2 might look like? Using that same example, please also explain what exactly makes this a total order relation.
order-theory
asked Aug 6 at 1:04
user928112
1716
I am having a hard time trying to understand what partial order and total order relations are. I need to answer a question in which I am given "two sequences S1 and S2 of n elements, possibly containing duplicates, on which a total order relation is defined".
Can someone give me a concrete example of what S1 and S2 might look like? Using that same example, please also explain what exactly makes this a total order relation.
order-theory
asked Aug 6 at 1:04
user928112
1716
edited Aug 6 at 1:11
asked Aug 6 at 1:04
user928112
1716
asked Aug 6 at 1:04
user928112
1716
asked Aug 6 at 1:04
user928112
1716
1716
add a comment |Â
add a comment |Â
3 Answers
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The wording is somewhat ambiguous: is $n$ the length of the sequences? Is $n$ the number of elements on which the total order is defined? Is the total order defined on the two sequences or the $n$ elements? (The answer to this last question is surely the $n$ elements, since an order on two things is not interesting.) But the point is to understand orders, so...
A simple example of a total order is the usual $le$ relation on the integers. It wasn't specified in the question that the elements of the sequence had to be in order, but assuming they do, such a sequence might look like $(1,2,3,3,3,7)$. In a total order, it is clear how any two elements $x, y$ relate to each other: either $x < y$, $x = y$, or $x > y$.
In a partial order, we can't necessarily compare two elements. A natural example is the subset relation between sets. Some sets are comparable, e.g., $emptyset subset 1,2 subset 1,2,3,4,5$; others, however, cannot be compared: it is not true that $1,2 subset 1,3,5$, nor is it true that $1,3,5 subset 1,2$, nor are they equal. This is the partial. Nevertheless, this relation behaves in predictable ways: if $A subset B$ and $B subset C$, then we know that $A subset C$. This is the order.
answered Aug 6 at 1:21
Théophile
16.8k12438
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
 |Â
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up vote
2
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This means that all the elements of $S_1$ and $S_2$ (that can be seen as multisets) can be compared to one another.
For example, you could have sequences of integers, as integers are totally ordered (by the usual "smaller than" $≤$ or "greater than" $≥$ relations): e.g. $S_1 ≠(1, 2, 2, 4) quad S_2 ≠(2, 2, 6, 1)$
But the same can't be said of sequences of pairs of integers for the pointwise order: $$(a, b) ≤ (a', b') ⟺ a ≤ a' ∧ b ≤ b'$$
In this case, you only have a partial order: you can't compare $(0, 1)$ and $(1, 0)$ for example.
So, for example, $S_1 ≠big((0, 1), (1, 0)big) quad S_2 ≠big((0, 0), (1, 0), (0, 2)big)$ wouldn't satisfy your hypothesis.
answered Aug 6 at 1:27
youkad
964
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A totally ordered set is a set along with a binary relation $mathbb R$ which is transitive, antisymmetric and connex.
Transitive means $$ a mathbb R btext and b mathbb R c implies a mathbb R c$$
Antisymmetric means $$ a mathbb R btext and b mathbb R a implies a=b$$
Connex means For every pair $a$ and $b$, either $ a mathbb R b$ or $ b mathbb R a$
As an example $$A=5,12,34,52$$ with the regular $le$ is a totally ordered set.
Another example: $$A=a,b,c,d,e,f$$ with the dictionary order is a totally ordered set.
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The wording is somewhat ambiguous: is $n$ the length of the sequences? Is $n$ the number of elements on which the total order is defined? Is the total order defined on the two sequences or the $n$ elements? (The answer to this last question is surely the $n$ elements, since an order on two things is not interesting.) But the point is to understand orders, so...
A simple example of a total order is the usual $le$ relation on the integers. It wasn't specified in the question that the elements of the sequence had to be in order, but assuming they do, such a sequence might look like $(1,2,3,3,3,7)$. In a total order, it is clear how any two elements $x, y$ relate to each other: either $x < y$, $x = y$, or $x > y$.
In a partial order, we can't necessarily compare two elements. A natural example is the subset relation between sets. Some sets are comparable, e.g., $emptyset subset 1,2 subset 1,2,3,4,5$; others, however, cannot be compared: it is not true that $1,2 subset 1,3,5$, nor is it true that $1,3,5 subset 1,2$, nor are they equal. This is the partial. Nevertheless, this relation behaves in predictable ways: if $A subset B$ and $B subset C$, then we know that $A subset C$. This is the order.
answered Aug 6 at 1:21
Théophile
16.8k12438
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
 |Â
show 4 more comments
up vote
2
down vote
accepted
The wording is somewhat ambiguous: is $n$ the length of the sequences? Is $n$ the number of elements on which the total order is defined? Is the total order defined on the two sequences or the $n$ elements? (The answer to this last question is surely the $n$ elements, since an order on two things is not interesting.) But the point is to understand orders, so...
A simple example of a total order is the usual $le$ relation on the integers. It wasn't specified in the question that the elements of the sequence had to be in order, but assuming they do, such a sequence might look like $(1,2,3,3,3,7)$. In a total order, it is clear how any two elements $x, y$ relate to each other: either $x < y$, $x = y$, or $x > y$.
In a partial order, we can't necessarily compare two elements. A natural example is the subset relation between sets. Some sets are comparable, e.g., $emptyset subset 1,2 subset 1,2,3,4,5$; others, however, cannot be compared: it is not true that $1,2 subset 1,3,5$, nor is it true that $1,3,5 subset 1,2$, nor are they equal. This is the partial. Nevertheless, this relation behaves in predictable ways: if $A subset B$ and $B subset C$, then we know that $A subset C$. This is the order.
answered Aug 6 at 1:21
Théophile
16.8k12438
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
 |Â
show 4 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The wording is somewhat ambiguous: is $n$ the length of the sequences? Is $n$ the number of elements on which the total order is defined? Is the total order defined on the two sequences or the $n$ elements? (The answer to this last question is surely the $n$ elements, since an order on two things is not interesting.) But the point is to understand orders, so...
A simple example of a total order is the usual $le$ relation on the integers. It wasn't specified in the question that the elements of the sequence had to be in order, but assuming they do, such a sequence might look like $(1,2,3,3,3,7)$. In a total order, it is clear how any two elements $x, y$ relate to each other: either $x < y$, $x = y$, or $x > y$.
In a partial order, we can't necessarily compare two elements. A natural example is the subset relation between sets. Some sets are comparable, e.g., $emptyset subset 1,2 subset 1,2,3,4,5$; others, however, cannot be compared: it is not true that $1,2 subset 1,3,5$, nor is it true that $1,3,5 subset 1,2$, nor are they equal. This is the partial. Nevertheless, this relation behaves in predictable ways: if $A subset B$ and $B subset C$, then we know that $A subset C$. This is the order.
answered Aug 6 at 1:21
Théophile
16.8k12438
The wording is somewhat ambiguous: is $n$ the length of the sequences? Is $n$ the number of elements on which the total order is defined? Is the total order defined on the two sequences or the $n$ elements? (The answer to this last question is surely the $n$ elements, since an order on two things is not interesting.) But the point is to understand orders, so...
A simple example of a total order is the usual $le$ relation on the integers. It wasn't specified in the question that the elements of the sequence had to be in order, but assuming they do, such a sequence might look like $(1,2,3,3,3,7)$. In a total order, it is clear how any two elements $x, y$ relate to each other: either $x < y$, $x = y$, or $x > y$.
In a partial order, we can't necessarily compare two elements. A natural example is the subset relation between sets. Some sets are comparable, e.g., $emptyset subset 1,2 subset 1,2,3,4,5$; others, however, cannot be compared: it is not true that $1,2 subset 1,3,5$, nor is it true that $1,3,5 subset 1,2$, nor are they equal. This is the partial. Nevertheless, this relation behaves in predictable ways: if $A subset B$ and $B subset C$, then we know that $A subset C$. This is the order.
answered Aug 6 at 1:21
Théophile
16.8k12438
edited Aug 6 at 1:42
answered Aug 6 at 1:21
Théophile
16.8k12438
answered Aug 6 at 1:21
Théophile
16.8k12438
answered Aug 6 at 1:21
Théophile
16.8k12438
16.8k12438
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
 |Â
show 4 more comments
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
It might be worth noting that if the total order is defined on the $n$ elements of the sequences, one of the sequences might as well be something like $(1,7,3,3,2,5)$. I don't see anything in the quote to suggest that the sequence elements are in order with respect to the total order. Also, by the way, it's unclear from the OP's quote whether the sequences have $n$ elements or whether there is a set of $n$ elements from which the who-knows-what-length sequences' terms come from, though your answer doesn't mention $n$.
– Steve Kass
Aug 6 at 1:27
1
1
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
@SteveKass Indeed. I think the problem is extremely poorly worded; no wonder the OP is confused. I did mention $n$ at the beginning to express confusion over it, and because of that confusion I didn't feel it important to use it again.
– Théophile
Aug 6 at 1:31
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
Oops, noted. Ignore my last sentence in the comment.
– Steve Kass
Aug 6 at 1:34
1
1
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
@user928112 Yes, it would be perverse if the total order were defined on the two sequences alone, since there are only two. (But note that it's reasonable in general to define orders on many kinds of objects, like numbers, sets, sequences, etc.) Does the rest of the explanation make sense?
– Théophile
Aug 6 at 3:25
1
1
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
@user928112 In the case of numbers, yes, there is a natural way to compare any two of them, $le$, and we can order them accordingly. And yes, a total order effectively sorts the elements of a set. Now, not all sets contain numbers, which is why I gave the second example. How would you sort the set $1, 1,2, 1,3, 2,3, 1,2,3$, for example? A partial order, based on $subset$, is more appropriate here. (If you really wanted a total order, there are ways to do it, but they're not as natural as $subset$). Finally, what about complex numbers? You can't use $le$ anymore...
– Théophile
Aug 6 at 15:21
 |Â
show 4 more comments
up vote
2
down vote
This means that all the elements of $S_1$ and $S_2$ (that can be seen as multisets) can be compared to one another.
For example, you could have sequences of integers, as integers are totally ordered (by the usual "smaller than" $≤$ or "greater than" $≥$ relations): e.g. $S_1 ≠(1, 2, 2, 4) quad S_2 ≠(2, 2, 6, 1)$
But the same can't be said of sequences of pairs of integers for the pointwise order: $$(a, b) ≤ (a', b') ⟺ a ≤ a' ∧ b ≤ b'$$
In this case, you only have a partial order: you can't compare $(0, 1)$ and $(1, 0)$ for example.
So, for example, $S_1 ≠big((0, 1), (1, 0)big) quad S_2 ≠big((0, 0), (1, 0), (0, 2)big)$ wouldn't satisfy your hypothesis.
answered Aug 6 at 1:27
youkad
964
add a comment |Â
up vote
2
down vote
This means that all the elements of $S_1$ and $S_2$ (that can be seen as multisets) can be compared to one another.
For example, you could have sequences of integers, as integers are totally ordered (by the usual "smaller than" $≤$ or "greater than" $≥$ relations): e.g. $S_1 ≠(1, 2, 2, 4) quad S_2 ≠(2, 2, 6, 1)$
But the same can't be said of sequences of pairs of integers for the pointwise order: $$(a, b) ≤ (a', b') ⟺ a ≤ a' ∧ b ≤ b'$$
In this case, you only have a partial order: you can't compare $(0, 1)$ and $(1, 0)$ for example.
So, for example, $S_1 ≠big((0, 1), (1, 0)big) quad S_2 ≠big((0, 0), (1, 0), (0, 2)big)$ wouldn't satisfy your hypothesis.
answered Aug 6 at 1:27
youkad
964
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This means that all the elements of $S_1$ and $S_2$ (that can be seen as multisets) can be compared to one another.
For example, you could have sequences of integers, as integers are totally ordered (by the usual "smaller than" $≤$ or "greater than" $≥$ relations): e.g. $S_1 ≠(1, 2, 2, 4) quad S_2 ≠(2, 2, 6, 1)$
But the same can't be said of sequences of pairs of integers for the pointwise order: $$(a, b) ≤ (a', b') ⟺ a ≤ a' ∧ b ≤ b'$$
In this case, you only have a partial order: you can't compare $(0, 1)$ and $(1, 0)$ for example.
So, for example, $S_1 ≠big((0, 1), (1, 0)big) quad S_2 ≠big((0, 0), (1, 0), (0, 2)big)$ wouldn't satisfy your hypothesis.
answered Aug 6 at 1:27
youkad
964
This means that all the elements of $S_1$ and $S_2$ (that can be seen as multisets) can be compared to one another.
For example, you could have sequences of integers, as integers are totally ordered (by the usual "smaller than" $≤$ or "greater than" $≥$ relations): e.g. $S_1 ≠(1, 2, 2, 4) quad S_2 ≠(2, 2, 6, 1)$
But the same can't be said of sequences of pairs of integers for the pointwise order: $$(a, b) ≤ (a', b') ⟺ a ≤ a' ∧ b ≤ b'$$
In this case, you only have a partial order: you can't compare $(0, 1)$ and $(1, 0)$ for example.
So, for example, $S_1 ≠big((0, 1), (1, 0)big) quad S_2 ≠big((0, 0), (1, 0), (0, 2)big)$ wouldn't satisfy your hypothesis.
answered Aug 6 at 1:27
youkad
964
edited Aug 6 at 1:34
answered Aug 6 at 1:27
youkad
964
answered Aug 6 at 1:27
youkad
964
answered Aug 6 at 1:27
youkad
964
964
add a comment |Â
add a comment |Â
up vote
0
down vote
A totally ordered set is a set along with a binary relation $mathbb R$ which is transitive, antisymmetric and connex.
Transitive means $$ a mathbb R btext and b mathbb R c implies a mathbb R c$$
Antisymmetric means $$ a mathbb R btext and b mathbb R a implies a=b$$
Connex means For every pair $a$ and $b$, either $ a mathbb R b$ or $ b mathbb R a$
As an example $$A=5,12,34,52$$ with the regular $le$ is a totally ordered set.
Another example: $$A=a,b,c,d,e,f$$ with the dictionary order is a totally ordered set.
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
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A totally ordered set is a set along with a binary relation $mathbb R$ which is transitive, antisymmetric and connex.
Transitive means $$ a mathbb R btext and b mathbb R c implies a mathbb R c$$
Antisymmetric means $$ a mathbb R btext and b mathbb R a implies a=b$$
Connex means For every pair $a$ and $b$, either $ a mathbb R b$ or $ b mathbb R a$
As an example $$A=5,12,34,52$$ with the regular $le$ is a totally ordered set.
Another example: $$A=a,b,c,d,e,f$$ with the dictionary order is a totally ordered set.
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A totally ordered set is a set along with a binary relation $mathbb R$ which is transitive, antisymmetric and connex.
Transitive means $$ a mathbb R btext and b mathbb R c implies a mathbb R c$$
Antisymmetric means $$ a mathbb R btext and b mathbb R a implies a=b$$
Connex means For every pair $a$ and $b$, either $ a mathbb R b$ or $ b mathbb R a$
As an example $$A=5,12,34,52$$ with the regular $le$ is a totally ordered set.
Another example: $$A=a,b,c,d,e,f$$ with the dictionary order is a totally ordered set.
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
A totally ordered set is a set along with a binary relation $mathbb R$ which is transitive, antisymmetric and connex.
Transitive means $$ a mathbb R btext and b mathbb R c implies a mathbb R c$$
Antisymmetric means $$ a mathbb R btext and b mathbb R a implies a=b$$
Connex means For every pair $a$ and $b$, either $ a mathbb R b$ or $ b mathbb R a$
As an example $$A=5,12,34,52$$ with the regular $le$ is a totally ordered set.
Another example: $$A=a,b,c,d,e,f$$ with the dictionary order is a totally ordered set.
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
edited Aug 6 at 2:02
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 1:27
Mohammad Riazi-Kermani
27.8k41852
27.8k41852
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
add a comment |Â
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
1
1
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
@Théophile Of course, connex is correct. Thanks for the comment.
– Mohammad Riazi-Kermani
Aug 6 at 2:04
add a comment |Â
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