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Are “pair, triple, quadruple†considered to be sets?
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I have always interpreted "pair, triple, quadruple" as "sets containing two, three and four elements". I have never checked this assumption.
Consider the following examples:
The pair $(V, |cdot|)$ is a normed space. Another example is that of a graph with vertices and edges $(V, E)$.
The triple (triplet) $(S, (u_i)_i in S, (a_i)_i in S)$ is a
game.The quadruple $(V, F, +, times)$ is a vector space.
Are these things sets? For example, I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
It occurred to me that it might be strange to think of them as sets, because the set elements are vastly different from each other, e.g. the vector space example, or a digraph, where we insert an additional operation $o$ that specifies the orientation.
What sort of mathematical structures are these objects? Is there anyway to define operations on these objects? What are all the operations that can be defined on these objects?
elementary-set-theory definition
asked Jul 24 at 4:56
Roughly Stupid
405
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up vote
7
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I have always interpreted "pair, triple, quadruple" as "sets containing two, three and four elements". I have never checked this assumption.
Consider the following examples:
The pair $(V, |cdot|)$ is a normed space. Another example is that of a graph with vertices and edges $(V, E)$.
The triple (triplet) $(S, (u_i)_i in S, (a_i)_i in S)$ is a
game.The quadruple $(V, F, +, times)$ is a vector space.
Are these things sets? For example, I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
It occurred to me that it might be strange to think of them as sets, because the set elements are vastly different from each other, e.g. the vector space example, or a digraph, where we insert an additional operation $o$ that specifies the orientation.
What sort of mathematical structures are these objects? Is there anyway to define operations on these objects? What are all the operations that can be defined on these objects?
elementary-set-theory definition
asked Jul 24 at 4:56
Roughly Stupid
405
2
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I have always interpreted "pair, triple, quadruple" as "sets containing two, three and four elements". I have never checked this assumption.
Consider the following examples:
The pair $(V, |cdot|)$ is a normed space. Another example is that of a graph with vertices and edges $(V, E)$.
The triple (triplet) $(S, (u_i)_i in S, (a_i)_i in S)$ is a
game.The quadruple $(V, F, +, times)$ is a vector space.
Are these things sets? For example, I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
It occurred to me that it might be strange to think of them as sets, because the set elements are vastly different from each other, e.g. the vector space example, or a digraph, where we insert an additional operation $o$ that specifies the orientation.
What sort of mathematical structures are these objects? Is there anyway to define operations on these objects? What are all the operations that can be defined on these objects?
elementary-set-theory definition
asked Jul 24 at 4:56
Roughly Stupid
405
I have always interpreted "pair, triple, quadruple" as "sets containing two, three and four elements". I have never checked this assumption.
Consider the following examples:
The pair $(V, |cdot|)$ is a normed space. Another example is that of a graph with vertices and edges $(V, E)$.
The triple (triplet) $(S, (u_i)_i in S, (a_i)_i in S)$ is a
game.The quadruple $(V, F, +, times)$ is a vector space.
Are these things sets? For example, I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
It occurred to me that it might be strange to think of them as sets, because the set elements are vastly different from each other, e.g. the vector space example, or a digraph, where we insert an additional operation $o$ that specifies the orientation.
What sort of mathematical structures are these objects? Is there anyway to define operations on these objects? What are all the operations that can be defined on these objects?
elementary-set-theory definition
asked Jul 24 at 4:56
Roughly Stupid
405
edited Jul 24 at 4:59
asked Jul 24 at 4:56
Roughly Stupid
405
asked Jul 24 at 4:56
Roughly Stupid
405
asked Jul 24 at 4:56
Roughly Stupid
405
405
2
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34
add a comment |Â
2
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34
2
2
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34
add a comment |Â
3 Answers
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Are these things sets?
Yes and no…
A set is an unordered collection of elements. So when a definition says that "The ordered pair $(V,|cdot|)$ is a normed space, where…", the word "ordered" says that this is not a set. Moreover, the word "ordered" would often be omitted, because conventionally "pairs", "triples", …, "tuples" have the meaning of "ordered pairs", "ordered triples", …, "ordered tuples".
That being said, as already mentioned in another (really good!) answer, pairs and longer tuples can be modeled as sets, but that's not important here. People working with normed spaces defined as pairs $(V,|cdot|)$ hardly ever think of them as the set $V,$.
Note that in this example (and similarly in all the other examples that you mentioned), the two elements of the pair $(V,|cdot|)$ are very different things! $V$ itself is a set of elements of that space, while $|cdot|$ is a function $|cdot|colon VtomathbbR$.
Thanks to them being different things, we probably don't really need the order in "$(V,|cdot|)$", as it would be easy enough to interpret "$(|cdot|,V)$" as the same thing without any confusion. My guess (and I could be wrong) is that the tradition of setting up these definition as (ordered!) pairs or tuples is not a necessity but rather a matter of great convenience. For once, if we're defining something in math, it should be defined rigorously and unambiguously. Even if there are some random and not so important choices to be made, we should make them anyway, to ensure mutual understanding. Second, defining "$(|cdot|,V)$" would be terribly inconvenient, because the way we read it we would naturally want to define $|cdot|$ first — but we can't: we have to use $V$ in its description, but it hasn't been defined yet as it comes second in our pair. Isn't it more convenient if our notation reflects what we do? So let's put $V$ first!
I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
That's a very common abuse of language and notation that you'll have to get used to. In all these structures there's the underlying set, such as $V$, and additional items that describe a certain structure (the norm function for a normed space; the set of edges for a graph; the base field and the two operations for a vector space; etc.). It's very common to use the same name for the underlying set $V$ only when the context is clear. And since $V$ is a set, it has subsets, called "subspaces", "subgames", etc.
Here's an example. Once again, a normed space is a pair $(V,|cdot|)$, where $|cdot|colon VtomathbbR$ is a function (subject to certain properties, not important for this discussion). A "normed subspace" naturally should be a normed space in its own right. So the correct definition is that a normed subspace of a normed space $(V,|cdot|)$ is a normed space $(U,|cdot|')$, where $Usubseteq V$ and $|cdot|'colon UtomathbbR$ is defined by restricting the domain of $|cdot|$ to $U$. But since most of this is clear enough, it's quite standard to say "Let $U$ be a subspace of $V$" — which is abuse of language and notation, but it is an accepted shorthand to refer to the technically correct definition of being a "normed subspace".
answered Jul 24 at 6:15
zipirovich
10k11630
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
add a comment |Â
up vote
4
down vote
"tuple" is a general term and can mean any number of items, not a pair of items. For exactly two items we would instead say an ordered pair of items. The phrase $k$-tuple means an ordered arrangement of $k$ items, so $3$-tuple is an ordered triple, $4$-tuple is an ordered quadruple, etc.
Note that sets in themselves are unordered collections, so to represent items in an ordered collection, we need to construct sets in some special way that represents the ordering as well as the items in the collection.
There is no single best way to do this, but to start with you might read about the Kuratowski ordered pair, where $(a,b)$ is represented by the set $a,a,b$.
Once a way of representing an ordered pair as a set is decided, that can be iterated to create a $k$-tuple, i.e. a list of length $k$. For example, we might identify an ordered triple $(a,b,c)$ with the ordered pair $(a,(b,c))$. This can be continued recursively to whatever number of items (or length of list/tuple) is desired.
answered Jul 24 at 4:59
hardmath
28.1k94592
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
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0
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No.
Consider the following example:
A rational number is a number which can be represented by a pair of integers.
If a pair would be a set, there would be no element (1,1). So when using "pair" in this definition (I would say you read a lot of definitions like this for pairs/tuples), it cannot be a set as it can have duplicate entries.
Of course, something like a tuple (v, e) where e is an edge incident on v looks like you can represent it by a set, because the objects are clearly from different source sets and cannot be confused like when you have pairs of natural numbers.
answered Jul 24 at 12:29
allo
1012
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Are these things sets?
Yes and no…
A set is an unordered collection of elements. So when a definition says that "The ordered pair $(V,|cdot|)$ is a normed space, where…", the word "ordered" says that this is not a set. Moreover, the word "ordered" would often be omitted, because conventionally "pairs", "triples", …, "tuples" have the meaning of "ordered pairs", "ordered triples", …, "ordered tuples".
That being said, as already mentioned in another (really good!) answer, pairs and longer tuples can be modeled as sets, but that's not important here. People working with normed spaces defined as pairs $(V,|cdot|)$ hardly ever think of them as the set $V,$.
Note that in this example (and similarly in all the other examples that you mentioned), the two elements of the pair $(V,|cdot|)$ are very different things! $V$ itself is a set of elements of that space, while $|cdot|$ is a function $|cdot|colon VtomathbbR$.
Thanks to them being different things, we probably don't really need the order in "$(V,|cdot|)$", as it would be easy enough to interpret "$(|cdot|,V)$" as the same thing without any confusion. My guess (and I could be wrong) is that the tradition of setting up these definition as (ordered!) pairs or tuples is not a necessity but rather a matter of great convenience. For once, if we're defining something in math, it should be defined rigorously and unambiguously. Even if there are some random and not so important choices to be made, we should make them anyway, to ensure mutual understanding. Second, defining "$(|cdot|,V)$" would be terribly inconvenient, because the way we read it we would naturally want to define $|cdot|$ first — but we can't: we have to use $V$ in its description, but it hasn't been defined yet as it comes second in our pair. Isn't it more convenient if our notation reflects what we do? So let's put $V$ first!
I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
That's a very common abuse of language and notation that you'll have to get used to. In all these structures there's the underlying set, such as $V$, and additional items that describe a certain structure (the norm function for a normed space; the set of edges for a graph; the base field and the two operations for a vector space; etc.). It's very common to use the same name for the underlying set $V$ only when the context is clear. And since $V$ is a set, it has subsets, called "subspaces", "subgames", etc.
Here's an example. Once again, a normed space is a pair $(V,|cdot|)$, where $|cdot|colon VtomathbbR$ is a function (subject to certain properties, not important for this discussion). A "normed subspace" naturally should be a normed space in its own right. So the correct definition is that a normed subspace of a normed space $(V,|cdot|)$ is a normed space $(U,|cdot|')$, where $Usubseteq V$ and $|cdot|'colon UtomathbbR$ is defined by restricting the domain of $|cdot|$ to $U$. But since most of this is clear enough, it's quite standard to say "Let $U$ be a subspace of $V$" — which is abuse of language and notation, but it is an accepted shorthand to refer to the technically correct definition of being a "normed subspace".
answered Jul 24 at 6:15
zipirovich
10k11630
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
add a comment |Â
up vote
6
down vote
accepted
Are these things sets?
Yes and no…
A set is an unordered collection of elements. So when a definition says that "The ordered pair $(V,|cdot|)$ is a normed space, where…", the word "ordered" says that this is not a set. Moreover, the word "ordered" would often be omitted, because conventionally "pairs", "triples", …, "tuples" have the meaning of "ordered pairs", "ordered triples", …, "ordered tuples".
That being said, as already mentioned in another (really good!) answer, pairs and longer tuples can be modeled as sets, but that's not important here. People working with normed spaces defined as pairs $(V,|cdot|)$ hardly ever think of them as the set $V,$.
Note that in this example (and similarly in all the other examples that you mentioned), the two elements of the pair $(V,|cdot|)$ are very different things! $V$ itself is a set of elements of that space, while $|cdot|$ is a function $|cdot|colon VtomathbbR$.
Thanks to them being different things, we probably don't really need the order in "$(V,|cdot|)$", as it would be easy enough to interpret "$(|cdot|,V)$" as the same thing without any confusion. My guess (and I could be wrong) is that the tradition of setting up these definition as (ordered!) pairs or tuples is not a necessity but rather a matter of great convenience. For once, if we're defining something in math, it should be defined rigorously and unambiguously. Even if there are some random and not so important choices to be made, we should make them anyway, to ensure mutual understanding. Second, defining "$(|cdot|,V)$" would be terribly inconvenient, because the way we read it we would naturally want to define $|cdot|$ first — but we can't: we have to use $V$ in its description, but it hasn't been defined yet as it comes second in our pair. Isn't it more convenient if our notation reflects what we do? So let's put $V$ first!
I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
That's a very common abuse of language and notation that you'll have to get used to. In all these structures there's the underlying set, such as $V$, and additional items that describe a certain structure (the norm function for a normed space; the set of edges for a graph; the base field and the two operations for a vector space; etc.). It's very common to use the same name for the underlying set $V$ only when the context is clear. And since $V$ is a set, it has subsets, called "subspaces", "subgames", etc.
Here's an example. Once again, a normed space is a pair $(V,|cdot|)$, where $|cdot|colon VtomathbbR$ is a function (subject to certain properties, not important for this discussion). A "normed subspace" naturally should be a normed space in its own right. So the correct definition is that a normed subspace of a normed space $(V,|cdot|)$ is a normed space $(U,|cdot|')$, where $Usubseteq V$ and $|cdot|'colon UtomathbbR$ is defined by restricting the domain of $|cdot|$ to $U$. But since most of this is clear enough, it's quite standard to say "Let $U$ be a subspace of $V$" — which is abuse of language and notation, but it is an accepted shorthand to refer to the technically correct definition of being a "normed subspace".
answered Jul 24 at 6:15
zipirovich
10k11630
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Are these things sets?
Yes and no…
A set is an unordered collection of elements. So when a definition says that "The ordered pair $(V,|cdot|)$ is a normed space, where…", the word "ordered" says that this is not a set. Moreover, the word "ordered" would often be omitted, because conventionally "pairs", "triples", …, "tuples" have the meaning of "ordered pairs", "ordered triples", …, "ordered tuples".
That being said, as already mentioned in another (really good!) answer, pairs and longer tuples can be modeled as sets, but that's not important here. People working with normed spaces defined as pairs $(V,|cdot|)$ hardly ever think of them as the set $V,$.
Note that in this example (and similarly in all the other examples that you mentioned), the two elements of the pair $(V,|cdot|)$ are very different things! $V$ itself is a set of elements of that space, while $|cdot|$ is a function $|cdot|colon VtomathbbR$.
Thanks to them being different things, we probably don't really need the order in "$(V,|cdot|)$", as it would be easy enough to interpret "$(|cdot|,V)$" as the same thing without any confusion. My guess (and I could be wrong) is that the tradition of setting up these definition as (ordered!) pairs or tuples is not a necessity but rather a matter of great convenience. For once, if we're defining something in math, it should be defined rigorously and unambiguously. Even if there are some random and not so important choices to be made, we should make them anyway, to ensure mutual understanding. Second, defining "$(|cdot|,V)$" would be terribly inconvenient, because the way we read it we would naturally want to define $|cdot|$ first — but we can't: we have to use $V$ in its description, but it hasn't been defined yet as it comes second in our pair. Isn't it more convenient if our notation reflects what we do? So let's put $V$ first!
I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
That's a very common abuse of language and notation that you'll have to get used to. In all these structures there's the underlying set, such as $V$, and additional items that describe a certain structure (the norm function for a normed space; the set of edges for a graph; the base field and the two operations for a vector space; etc.). It's very common to use the same name for the underlying set $V$ only when the context is clear. And since $V$ is a set, it has subsets, called "subspaces", "subgames", etc.
Here's an example. Once again, a normed space is a pair $(V,|cdot|)$, where $|cdot|colon VtomathbbR$ is a function (subject to certain properties, not important for this discussion). A "normed subspace" naturally should be a normed space in its own right. So the correct definition is that a normed subspace of a normed space $(V,|cdot|)$ is a normed space $(U,|cdot|')$, where $Usubseteq V$ and $|cdot|'colon UtomathbbR$ is defined by restricting the domain of $|cdot|$ to $U$. But since most of this is clear enough, it's quite standard to say "Let $U$ be a subspace of $V$" — which is abuse of language and notation, but it is an accepted shorthand to refer to the technically correct definition of being a "normed subspace".
answered Jul 24 at 6:15
zipirovich
10k11630
Are these things sets?
Yes and no…
A set is an unordered collection of elements. So when a definition says that "The ordered pair $(V,|cdot|)$ is a normed space, where…", the word "ordered" says that this is not a set. Moreover, the word "ordered" would often be omitted, because conventionally "pairs", "triples", …, "tuples" have the meaning of "ordered pairs", "ordered triples", …, "ordered tuples".
That being said, as already mentioned in another (really good!) answer, pairs and longer tuples can be modeled as sets, but that's not important here. People working with normed spaces defined as pairs $(V,|cdot|)$ hardly ever think of them as the set $V,$.
Note that in this example (and similarly in all the other examples that you mentioned), the two elements of the pair $(V,|cdot|)$ are very different things! $V$ itself is a set of elements of that space, while $|cdot|$ is a function $|cdot|colon VtomathbbR$.
Thanks to them being different things, we probably don't really need the order in "$(V,|cdot|)$", as it would be easy enough to interpret "$(|cdot|,V)$" as the same thing without any confusion. My guess (and I could be wrong) is that the tradition of setting up these definition as (ordered!) pairs or tuples is not a necessity but rather a matter of great convenience. For once, if we're defining something in math, it should be defined rigorously and unambiguously. Even if there are some random and not so important choices to be made, we should make them anyway, to ensure mutual understanding. Second, defining "$(|cdot|,V)$" would be terribly inconvenient, because the way we read it we would naturally want to define $|cdot|$ first — but we can't: we have to use $V$ in its description, but it hasn't been defined yet as it comes second in our pair. Isn't it more convenient if our notation reflects what we do? So let's put $V$ first!
I have seen people defining "sub-graph", "sub-game", etc. which essentially implied to me that these things are sets.
That's a very common abuse of language and notation that you'll have to get used to. In all these structures there's the underlying set, such as $V$, and additional items that describe a certain structure (the norm function for a normed space; the set of edges for a graph; the base field and the two operations for a vector space; etc.). It's very common to use the same name for the underlying set $V$ only when the context is clear. And since $V$ is a set, it has subsets, called "subspaces", "subgames", etc.
Here's an example. Once again, a normed space is a pair $(V,|cdot|)$, where $|cdot|colon VtomathbbR$ is a function (subject to certain properties, not important for this discussion). A "normed subspace" naturally should be a normed space in its own right. So the correct definition is that a normed subspace of a normed space $(V,|cdot|)$ is a normed space $(U,|cdot|')$, where $Usubseteq V$ and $|cdot|'colon UtomathbbR$ is defined by restricting the domain of $|cdot|$ to $U$. But since most of this is clear enough, it's quite standard to say "Let $U$ be a subspace of $V$" — which is abuse of language and notation, but it is an accepted shorthand to refer to the technically correct definition of being a "normed subspace".
answered Jul 24 at 6:15
zipirovich
10k11630
edited Jul 24 at 18:23
answered Jul 24 at 6:15
zipirovich
10k11630
answered Jul 24 at 6:15
zipirovich
10k11630
answered Jul 24 at 6:15
zipirovich
10k11630
10k11630
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
add a comment |Â
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
1
1
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
Excellent answer; I especially appreciate the detail in clarifying the abuse of notation, something that used to bother me as well.
– Sambo
Jul 24 at 16:46
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
@Sambo: Thank you! I saw that this seemed to be a confusion for the OP that definitely needed to be addressed.
– zipirovich
Jul 24 at 18:25
add a comment |Â
up vote
4
down vote
"tuple" is a general term and can mean any number of items, not a pair of items. For exactly two items we would instead say an ordered pair of items. The phrase $k$-tuple means an ordered arrangement of $k$ items, so $3$-tuple is an ordered triple, $4$-tuple is an ordered quadruple, etc.
Note that sets in themselves are unordered collections, so to represent items in an ordered collection, we need to construct sets in some special way that represents the ordering as well as the items in the collection.
There is no single best way to do this, but to start with you might read about the Kuratowski ordered pair, where $(a,b)$ is represented by the set $a,a,b$.
Once a way of representing an ordered pair as a set is decided, that can be iterated to create a $k$-tuple, i.e. a list of length $k$. For example, we might identify an ordered triple $(a,b,c)$ with the ordered pair $(a,(b,c))$. This can be continued recursively to whatever number of items (or length of list/tuple) is desired.
answered Jul 24 at 4:59
hardmath
28.1k94592
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
add a comment |Â
up vote
4
down vote
"tuple" is a general term and can mean any number of items, not a pair of items. For exactly two items we would instead say an ordered pair of items. The phrase $k$-tuple means an ordered arrangement of $k$ items, so $3$-tuple is an ordered triple, $4$-tuple is an ordered quadruple, etc.
Note that sets in themselves are unordered collections, so to represent items in an ordered collection, we need to construct sets in some special way that represents the ordering as well as the items in the collection.
There is no single best way to do this, but to start with you might read about the Kuratowski ordered pair, where $(a,b)$ is represented by the set $a,a,b$.
Once a way of representing an ordered pair as a set is decided, that can be iterated to create a $k$-tuple, i.e. a list of length $k$. For example, we might identify an ordered triple $(a,b,c)$ with the ordered pair $(a,(b,c))$. This can be continued recursively to whatever number of items (or length of list/tuple) is desired.
answered Jul 24 at 4:59
hardmath
28.1k94592
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
add a comment |Â
up vote
4
down vote
up vote
4
down vote
"tuple" is a general term and can mean any number of items, not a pair of items. For exactly two items we would instead say an ordered pair of items. The phrase $k$-tuple means an ordered arrangement of $k$ items, so $3$-tuple is an ordered triple, $4$-tuple is an ordered quadruple, etc.
Note that sets in themselves are unordered collections, so to represent items in an ordered collection, we need to construct sets in some special way that represents the ordering as well as the items in the collection.
There is no single best way to do this, but to start with you might read about the Kuratowski ordered pair, where $(a,b)$ is represented by the set $a,a,b$.
Once a way of representing an ordered pair as a set is decided, that can be iterated to create a $k$-tuple, i.e. a list of length $k$. For example, we might identify an ordered triple $(a,b,c)$ with the ordered pair $(a,(b,c))$. This can be continued recursively to whatever number of items (or length of list/tuple) is desired.
answered Jul 24 at 4:59
hardmath
28.1k94592
"tuple" is a general term and can mean any number of items, not a pair of items. For exactly two items we would instead say an ordered pair of items. The phrase $k$-tuple means an ordered arrangement of $k$ items, so $3$-tuple is an ordered triple, $4$-tuple is an ordered quadruple, etc.
Note that sets in themselves are unordered collections, so to represent items in an ordered collection, we need to construct sets in some special way that represents the ordering as well as the items in the collection.
There is no single best way to do this, but to start with you might read about the Kuratowski ordered pair, where $(a,b)$ is represented by the set $a,a,b$.
Once a way of representing an ordered pair as a set is decided, that can be iterated to create a $k$-tuple, i.e. a list of length $k$. For example, we might identify an ordered triple $(a,b,c)$ with the ordered pair $(a,(b,c))$. This can be continued recursively to whatever number of items (or length of list/tuple) is desired.
answered Jul 24 at 4:59
hardmath
28.1k94592
edited Jul 24 at 5:23
answered Jul 24 at 4:59
hardmath
28.1k94592
answered Jul 24 at 4:59
hardmath
28.1k94592
answered Jul 24 at 4:59
hardmath
28.1k94592
28.1k94592
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
add a comment |Â
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
1
1
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
Sorry, I made a mistake in the title. By tuple I meant pair. Not a native speaker!
– Roughly Stupid
Jul 24 at 5:00
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
No, problem! Tuple does sound like "two"-ple. I've added a little information about how one might represent these finite ordered arrangements as sets.
– hardmath
Jul 24 at 5:04
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
Hi, can you elaborate on what you mean by ordered. For example, I understand that ordering is a way to distinguish the collection containing the same elements. For example, $(1,2)$ and $(2,1)$. Why does it make sense to order vector space by defining it as a quadruple? $(V, F, +, times)$ is the same as $(F, V, times, +)$ is the same as $(times, V, +, F)$? math.stackexchange.com/questions/1423955/…
– Roughly Stupid
Jul 24 at 5:08
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
By ordered I mean that the arrangement of the items makes a difference. $(1,2)$ is not the same as $(2,1)$, and if you define a vector space as a quadruple $(V,F,+,times)$, then it would be something entirely different if you reordered the items as $(F,V,times,+)$ or $(times,V,+,F)$. Ordered means that ordering makes a difference.
– hardmath
Jul 24 at 5:12
add a comment |Â
up vote
0
down vote
No.
Consider the following example:
A rational number is a number which can be represented by a pair of integers.
If a pair would be a set, there would be no element (1,1). So when using "pair" in this definition (I would say you read a lot of definitions like this for pairs/tuples), it cannot be a set as it can have duplicate entries.
Of course, something like a tuple (v, e) where e is an edge incident on v looks like you can represent it by a set, because the objects are clearly from different source sets and cannot be confused like when you have pairs of natural numbers.
answered Jul 24 at 12:29
allo
1012
add a comment |Â
up vote
0
down vote
No.
Consider the following example:
A rational number is a number which can be represented by a pair of integers.
If a pair would be a set, there would be no element (1,1). So when using "pair" in this definition (I would say you read a lot of definitions like this for pairs/tuples), it cannot be a set as it can have duplicate entries.
Of course, something like a tuple (v, e) where e is an edge incident on v looks like you can represent it by a set, because the objects are clearly from different source sets and cannot be confused like when you have pairs of natural numbers.
answered Jul 24 at 12:29
allo
1012
add a comment |Â
up vote
0
down vote
up vote
0
down vote
No.
Consider the following example:
A rational number is a number which can be represented by a pair of integers.
If a pair would be a set, there would be no element (1,1). So when using "pair" in this definition (I would say you read a lot of definitions like this for pairs/tuples), it cannot be a set as it can have duplicate entries.
Of course, something like a tuple (v, e) where e is an edge incident on v looks like you can represent it by a set, because the objects are clearly from different source sets and cannot be confused like when you have pairs of natural numbers.
answered Jul 24 at 12:29
allo
1012
No.
Consider the following example:
A rational number is a number which can be represented by a pair of integers.
If a pair would be a set, there would be no element (1,1). So when using "pair" in this definition (I would say you read a lot of definitions like this for pairs/tuples), it cannot be a set as it can have duplicate entries.
Of course, something like a tuple (v, e) where e is an edge incident on v looks like you can represent it by a set, because the objects are clearly from different source sets and cannot be confused like when you have pairs of natural numbers.
answered Jul 24 at 12:29
allo
1012
answered Jul 24 at 12:29
allo
1012
answered Jul 24 at 12:29
allo
1012
answered Jul 24 at 12:29
allo
1012
1012
add a comment |Â
add a comment |Â
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2
$(1,1)$ is a pair, but $1,1$ is a set containing one element. $(1,2)$ and $(2,1)$ are different pairs, but $1,2$ and $2,1$ are the same set.
– Rahul
Jul 24 at 6:17
All we really want from the definition of, say, a normed space is to be able to say "If you want to give me a normed space, you have to give me one thing $V$ and another thing $|cdot|$, and I need to be able to tell which is which." The formulation of an ordered pair $(V,|cdot|)$ is just a convenient abstraction in which to encode this idea.
– Rahul
Jul 24 at 6:34