Mixing
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Constructing lists from Ordered Pairs
$begingroup$
I've been searching online for a way of constructing lists from sets, but to no avail. However, I am aware of how to define ordered pairs and, more generally, ordered n-tuples from sets. My first preference being the Kuratowski definition:
$$ (a, b) := biglbracelbrace arbrace, lbrace a,brbrace bigrbrace $$
As I'm sure you are all aware, ordered pairs can then be used to define Cartesian products. Using Cartesian products, I believe I've find out how to construct/define lists using sets. It's essentially just repeated Cartesian products with sets containing only one constant member. E.g.
$$ [10, 3, emptyset] := lbrace 10 rbrace times lbrace 3 rbrace times lbrace emptyset rbrace $$
What I'm stuck on is how to represent the list as a single set for any given number of members. For three members I'm guessing
$$ [a,b,c] := Biglbrace biglbracebiglbrace lbrace a rbrace, lbrace a,b rbrace bigrbrace bigrbrace, biglbrace biglbracelbrace arbrace, lbrace a,brbrace bigrbrace, c bigrbrace Bigrbrace $$
based on the idea that $(a,b) times lbrace c rbrace = ((a,b),c)$. If you're having difficulty understanding the repeated nested braces, I recommend replacing the ordered pair $(a,b)$ with some unused, arbitrary symbol, e.g. $x$, figure out the ordered triplet as an ordered pair of $x$ and $c$, then use substitution to replace the arbitrary letter with the ordered pair.
Just to be clear, I'm using the following definition of list: an ordered collection of well-defined objects.
Two questions:
- Am I correct?
- How does one define a list in terms of sets for any arbitrary number of members of that list?
elementary-set-theory order-theory
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
$endgroup$
add a comment |
$begingroup$
I've been searching online for a way of constructing lists from sets, but to no avail. However, I am aware of how to define ordered pairs and, more generally, ordered n-tuples from sets. My first preference being the Kuratowski definition:
$$ (a, b) := biglbracelbrace arbrace, lbrace a,brbrace bigrbrace $$
As I'm sure you are all aware, ordered pairs can then be used to define Cartesian products. Using Cartesian products, I believe I've find out how to construct/define lists using sets. It's essentially just repeated Cartesian products with sets containing only one constant member. E.g.
$$ [10, 3, emptyset] := lbrace 10 rbrace times lbrace 3 rbrace times lbrace emptyset rbrace $$
What I'm stuck on is how to represent the list as a single set for any given number of members. For three members I'm guessing
$$ [a,b,c] := Biglbrace biglbracebiglbrace lbrace a rbrace, lbrace a,b rbrace bigrbrace bigrbrace, biglbrace biglbracelbrace arbrace, lbrace a,brbrace bigrbrace, c bigrbrace Bigrbrace $$
based on the idea that $(a,b) times lbrace c rbrace = ((a,b),c)$. If you're having difficulty understanding the repeated nested braces, I recommend replacing the ordered pair $(a,b)$ with some unused, arbitrary symbol, e.g. $x$, figure out the ordered triplet as an ordered pair of $x$ and $c$, then use substitution to replace the arbitrary letter with the ordered pair.
Just to be clear, I'm using the following definition of list: an ordered collection of well-defined objects.
Two questions:
- Am I correct?
- How does one define a list in terms of sets for any arbitrary number of members of that list?
elementary-set-theory order-theory
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
$endgroup$
add a comment |
$begingroup$
I've been searching online for a way of constructing lists from sets, but to no avail. However, I am aware of how to define ordered pairs and, more generally, ordered n-tuples from sets. My first preference being the Kuratowski definition:
$$ (a, b) := biglbracelbrace arbrace, lbrace a,brbrace bigrbrace $$
As I'm sure you are all aware, ordered pairs can then be used to define Cartesian products. Using Cartesian products, I believe I've find out how to construct/define lists using sets. It's essentially just repeated Cartesian products with sets containing only one constant member. E.g.
$$ [10, 3, emptyset] := lbrace 10 rbrace times lbrace 3 rbrace times lbrace emptyset rbrace $$
What I'm stuck on is how to represent the list as a single set for any given number of members. For three members I'm guessing
$$ [a,b,c] := Biglbrace biglbracebiglbrace lbrace a rbrace, lbrace a,b rbrace bigrbrace bigrbrace, biglbrace biglbracelbrace arbrace, lbrace a,brbrace bigrbrace, c bigrbrace Bigrbrace $$
based on the idea that $(a,b) times lbrace c rbrace = ((a,b),c)$. If you're having difficulty understanding the repeated nested braces, I recommend replacing the ordered pair $(a,b)$ with some unused, arbitrary symbol, e.g. $x$, figure out the ordered triplet as an ordered pair of $x$ and $c$, then use substitution to replace the arbitrary letter with the ordered pair.
Just to be clear, I'm using the following definition of list: an ordered collection of well-defined objects.
Two questions:
- Am I correct?
- How does one define a list in terms of sets for any arbitrary number of members of that list?
elementary-set-theory order-theory
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
$endgroup$
I've been searching online for a way of constructing lists from sets, but to no avail. However, I am aware of how to define ordered pairs and, more generally, ordered n-tuples from sets. My first preference being the Kuratowski definition:
$$ (a, b) := biglbracelbrace arbrace, lbrace a,brbrace bigrbrace $$
As I'm sure you are all aware, ordered pairs can then be used to define Cartesian products. Using Cartesian products, I believe I've find out how to construct/define lists using sets. It's essentially just repeated Cartesian products with sets containing only one constant member. E.g.
$$ [10, 3, emptyset] := lbrace 10 rbrace times lbrace 3 rbrace times lbrace emptyset rbrace $$
What I'm stuck on is how to represent the list as a single set for any given number of members. For three members I'm guessing
$$ [a,b,c] := Biglbrace biglbracebiglbrace lbrace a rbrace, lbrace a,b rbrace bigrbrace bigrbrace, biglbrace biglbracelbrace arbrace, lbrace a,brbrace bigrbrace, c bigrbrace Bigrbrace $$
based on the idea that $(a,b) times lbrace c rbrace = ((a,b),c)$. If you're having difficulty understanding the repeated nested braces, I recommend replacing the ordered pair $(a,b)$ with some unused, arbitrary symbol, e.g. $x$, figure out the ordered triplet as an ordered pair of $x$ and $c$, then use substitution to replace the arbitrary letter with the ordered pair.
Just to be clear, I'm using the following definition of list: an ordered collection of well-defined objects.
Two questions:
- Am I correct?
- How does one define a list in terms of sets for any arbitrary number of members of that list?
elementary-set-theory order-theory
elementary-set-theory order-theory
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
edited Jan 9 at 15:31
Andrés E. Caicedo
65.3k8158247
65.3k8158247
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
asked Jan 9 at 13:46
HarrisonOHarrisonO
464
464
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The most commonly used convention is to define the natural numbers first and then say that a list is a function whose domain is an initial (finite) segment of the naturals. Then,
$$ [a,b,c] = {(0,a), (1,b), (2,c)} $$
(This is convenient in set theory, e.g., because it lets you prove without Replacement that if $A$ is a set, then there is a set of all lists of elements from $A$).
If this is not to your liking you might also take a page from Lisp and represent the empty list as $varnothing$, and a non-empty list as the ordered pair of the first element and the representation of the rest of the list:
$$ [a,b,c] = (a,(b,(c,varnothing))) $$
(This works because $varnothing$ cannot be a Kuratowski pair).
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
$endgroup$
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The most commonly used convention is to define the natural numbers first and then say that a list is a function whose domain is an initial (finite) segment of the naturals. Then,
$$ [a,b,c] = {(0,a), (1,b), (2,c)} $$
(This is convenient in set theory, e.g., because it lets you prove without Replacement that if $A$ is a set, then there is a set of all lists of elements from $A$).
If this is not to your liking you might also take a page from Lisp and represent the empty list as $varnothing$, and a non-empty list as the ordered pair of the first element and the representation of the rest of the list:
$$ [a,b,c] = (a,(b,(c,varnothing))) $$
(This works because $varnothing$ cannot be a Kuratowski pair).
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
$endgroup$
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
add a comment |
$begingroup$
The most commonly used convention is to define the natural numbers first and then say that a list is a function whose domain is an initial (finite) segment of the naturals. Then,
$$ [a,b,c] = {(0,a), (1,b), (2,c)} $$
(This is convenient in set theory, e.g., because it lets you prove without Replacement that if $A$ is a set, then there is a set of all lists of elements from $A$).
If this is not to your liking you might also take a page from Lisp and represent the empty list as $varnothing$, and a non-empty list as the ordered pair of the first element and the representation of the rest of the list:
$$ [a,b,c] = (a,(b,(c,varnothing))) $$
(This works because $varnothing$ cannot be a Kuratowski pair).
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
$endgroup$
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
add a comment |
$begingroup$
The most commonly used convention is to define the natural numbers first and then say that a list is a function whose domain is an initial (finite) segment of the naturals. Then,
$$ [a,b,c] = {(0,a), (1,b), (2,c)} $$
(This is convenient in set theory, e.g., because it lets you prove without Replacement that if $A$ is a set, then there is a set of all lists of elements from $A$).
If this is not to your liking you might also take a page from Lisp and represent the empty list as $varnothing$, and a non-empty list as the ordered pair of the first element and the representation of the rest of the list:
$$ [a,b,c] = (a,(b,(c,varnothing))) $$
(This works because $varnothing$ cannot be a Kuratowski pair).
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
$endgroup$
The most commonly used convention is to define the natural numbers first and then say that a list is a function whose domain is an initial (finite) segment of the naturals. Then,
$$ [a,b,c] = {(0,a), (1,b), (2,c)} $$
(This is convenient in set theory, e.g., because it lets you prove without Replacement that if $A$ is a set, then there is a set of all lists of elements from $A$).
If this is not to your liking you might also take a page from Lisp and represent the empty list as $varnothing$, and a non-empty list as the ordered pair of the first element and the representation of the rest of the list:
$$ [a,b,c] = (a,(b,(c,varnothing))) $$
(This works because $varnothing$ cannot be a Kuratowski pair).
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
answered Jan 9 at 13:53
Henning MakholmHenning Makholm
240k17305541
240k17305541
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
add a comment |
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
1
1
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
$begingroup$
To add two small notes to this answer: 1. Another convention is to encode the length of the list as its first element, i.e. $[a, b, c] = (3, (a, (b, c)))$ -- with "special" case $ = emptyset$. 2. Ultimately it generally doesn't matter how you define this thing exactly, as there are bijections that are fiddly but not ultimately hard to write down directly.
$endgroup$
– Mees de Vries
Jan 9 at 13:56
add a comment |
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