Mixing
Properties of the Cartesian product of indexed sets
Clash Royale CLAN TAG#URR8PPP
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Given the indexed sets $(X_i)_iin I$, one usually defines their Cartesian product as:
$$prod_iin I X_i =_textdef;
left f: fin left(bigcup_iin I X_iright)^I wedge; forall i in I; f(i)in X_iright
labelcart1tag1$$
For $I=1,2$, it seems natural to set:
$$
X_1times X_2 =_textdef; prod_iin 1,2 X_i
$$
Then one expects, for $I=1ldots n$:
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n
labelcart2tag2$$
I am unable to derive refcart2. In fact, the generic element of
$
prodlimits_i=1^3 X_i
$
is:
$$
(1, f(1)), (2, f(2)), (3, f(3)), text where $f$ is a function in
left(bigcup_iin 1,2,3 X_iright)^1,2,3;;
$$
while the generic element of:
$$
left(prod_i=1^2 X_iright) times X_3
$$
is:
$$
(1, f(1)), (2, f(2)), text where $f$ is a function in
left( left(prod_i=1^2 X_iright) cup X_3right)^1,2;.
$$
So:
$$
prod_i=1^3 X_i neq left(prod_i=1^2 X_iright) times X_3
labelcart3tag3
$$
If that is correct, it extends to the Cartesian power:
$$
X^n+1 =
X^n times X
$$
If refcart3 is wrong, please, tell me how to prove the equality.
If refcart3 is correct, can we properly define refcart1 a Cartesian product, given that the standard/expected Cartesian product properties are missing?
UPDATE
The point of my question is:
since the infinite CP is a generalisation of the ordinary CP, for a finite index set:
$$prod_iin 1ldots n X_i = X_1timesldots X_n$$
In words: the finite versions of the infinite CP should work like the ordinary finite version. This might happen with a proper definition of $times$ for the generalised CP, and apparently not mine above.
In a similar question, @StefanH, wants a CP construction which observes the following properties:
i) it is associative;
ii) $A^0 times A^n = A^n$
iii) gives a natural choice for $A_0$.
Making the generalised $times$ associative is the opposite of what I am looking for, because the standard finite CP is not associative. As for ii) and iii), there is no $A^0$ in the standard CP and so no standard behaviour to generalise.
As noted: if the generalised CP (as defined in $refcart1$) does not share the essential properties enjoyed by the standard CP, then the name 'generalised CP' might be a misnomer.
elementary-set-theory
asked Jul 17 at 12:14
antonio
207210
add a comment |Â
up vote
2
down vote
favorite
Given the indexed sets $(X_i)_iin I$, one usually defines their Cartesian product as:
$$prod_iin I X_i =_textdef;
left f: fin left(bigcup_iin I X_iright)^I wedge; forall i in I; f(i)in X_iright
labelcart1tag1$$
For $I=1,2$, it seems natural to set:
$$
X_1times X_2 =_textdef; prod_iin 1,2 X_i
$$
Then one expects, for $I=1ldots n$:
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n
labelcart2tag2$$
I am unable to derive refcart2. In fact, the generic element of
$
prodlimits_i=1^3 X_i
$
is:
$$
(1, f(1)), (2, f(2)), (3, f(3)), text where $f$ is a function in
left(bigcup_iin 1,2,3 X_iright)^1,2,3;;
$$
while the generic element of:
$$
left(prod_i=1^2 X_iright) times X_3
$$
is:
$$
(1, f(1)), (2, f(2)), text where $f$ is a function in
left( left(prod_i=1^2 X_iright) cup X_3right)^1,2;.
$$
So:
$$
prod_i=1^3 X_i neq left(prod_i=1^2 X_iright) times X_3
labelcart3tag3
$$
If that is correct, it extends to the Cartesian power:
$$
X^n+1 =
X^n times X
$$
If refcart3 is wrong, please, tell me how to prove the equality.
If refcart3 is correct, can we properly define refcart1 a Cartesian product, given that the standard/expected Cartesian product properties are missing?
UPDATE
The point of my question is:
since the infinite CP is a generalisation of the ordinary CP, for a finite index set:
$$prod_iin 1ldots n X_i = X_1timesldots X_n$$
In words: the finite versions of the infinite CP should work like the ordinary finite version. This might happen with a proper definition of $times$ for the generalised CP, and apparently not mine above.
In a similar question, @StefanH, wants a CP construction which observes the following properties:
i) it is associative;
ii) $A^0 times A^n = A^n$
iii) gives a natural choice for $A_0$.
Making the generalised $times$ associative is the opposite of what I am looking for, because the standard finite CP is not associative. As for ii) and iii), there is no $A^0$ in the standard CP and so no standard behaviour to generalise.
As noted: if the generalised CP (as defined in $refcart1$) does not share the essential properties enjoyed by the standard CP, then the name 'generalised CP' might be a misnomer.
elementary-set-theory
asked Jul 17 at 12:14
antonio
207210
1
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the indexed sets $(X_i)_iin I$, one usually defines their Cartesian product as:
$$prod_iin I X_i =_textdef;
left f: fin left(bigcup_iin I X_iright)^I wedge; forall i in I; f(i)in X_iright
labelcart1tag1$$
For $I=1,2$, it seems natural to set:
$$
X_1times X_2 =_textdef; prod_iin 1,2 X_i
$$
Then one expects, for $I=1ldots n$:
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n
labelcart2tag2$$
I am unable to derive refcart2. In fact, the generic element of
$
prodlimits_i=1^3 X_i
$
is:
$$
(1, f(1)), (2, f(2)), (3, f(3)), text where $f$ is a function in
left(bigcup_iin 1,2,3 X_iright)^1,2,3;;
$$
while the generic element of:
$$
left(prod_i=1^2 X_iright) times X_3
$$
is:
$$
(1, f(1)), (2, f(2)), text where $f$ is a function in
left( left(prod_i=1^2 X_iright) cup X_3right)^1,2;.
$$
So:
$$
prod_i=1^3 X_i neq left(prod_i=1^2 X_iright) times X_3
labelcart3tag3
$$
If that is correct, it extends to the Cartesian power:
$$
X^n+1 =
X^n times X
$$
If refcart3 is wrong, please, tell me how to prove the equality.
If refcart3 is correct, can we properly define refcart1 a Cartesian product, given that the standard/expected Cartesian product properties are missing?
UPDATE
The point of my question is:
since the infinite CP is a generalisation of the ordinary CP, for a finite index set:
$$prod_iin 1ldots n X_i = X_1timesldots X_n$$
In words: the finite versions of the infinite CP should work like the ordinary finite version. This might happen with a proper definition of $times$ for the generalised CP, and apparently not mine above.
In a similar question, @StefanH, wants a CP construction which observes the following properties:
i) it is associative;
ii) $A^0 times A^n = A^n$
iii) gives a natural choice for $A_0$.
Making the generalised $times$ associative is the opposite of what I am looking for, because the standard finite CP is not associative. As for ii) and iii), there is no $A^0$ in the standard CP and so no standard behaviour to generalise.
As noted: if the generalised CP (as defined in $refcart1$) does not share the essential properties enjoyed by the standard CP, then the name 'generalised CP' might be a misnomer.
elementary-set-theory
asked Jul 17 at 12:14
antonio
207210
Given the indexed sets $(X_i)_iin I$, one usually defines their Cartesian product as:
$$prod_iin I X_i =_textdef;
left f: fin left(bigcup_iin I X_iright)^I wedge; forall i in I; f(i)in X_iright
labelcart1tag1$$
For $I=1,2$, it seems natural to set:
$$
X_1times X_2 =_textdef; prod_iin 1,2 X_i
$$
Then one expects, for $I=1ldots n$:
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n
labelcart2tag2$$
I am unable to derive refcart2. In fact, the generic element of
$
prodlimits_i=1^3 X_i
$
is:
$$
(1, f(1)), (2, f(2)), (3, f(3)), text where $f$ is a function in
left(bigcup_iin 1,2,3 X_iright)^1,2,3;;
$$
while the generic element of:
$$
left(prod_i=1^2 X_iright) times X_3
$$
is:
$$
(1, f(1)), (2, f(2)), text where $f$ is a function in
left( left(prod_i=1^2 X_iright) cup X_3right)^1,2;.
$$
So:
$$
prod_i=1^3 X_i neq left(prod_i=1^2 X_iright) times X_3
labelcart3tag3
$$
If that is correct, it extends to the Cartesian power:
$$
X^n+1 =
X^n times X
$$
If refcart3 is wrong, please, tell me how to prove the equality.
If refcart3 is correct, can we properly define refcart1 a Cartesian product, given that the standard/expected Cartesian product properties are missing?
UPDATE
The point of my question is:
since the infinite CP is a generalisation of the ordinary CP, for a finite index set:
$$prod_iin 1ldots n X_i = X_1timesldots X_n$$
In words: the finite versions of the infinite CP should work like the ordinary finite version. This might happen with a proper definition of $times$ for the generalised CP, and apparently not mine above.
In a similar question, @StefanH, wants a CP construction which observes the following properties:
i) it is associative;
ii) $A^0 times A^n = A^n$
iii) gives a natural choice for $A_0$.
Making the generalised $times$ associative is the opposite of what I am looking for, because the standard finite CP is not associative. As for ii) and iii), there is no $A^0$ in the standard CP and so no standard behaviour to generalise.
As noted: if the generalised CP (as defined in $refcart1$) does not share the essential properties enjoyed by the standard CP, then the name 'generalised CP' might be a misnomer.
elementary-set-theory
asked Jul 17 at 12:14
antonio
207210
edited Jul 17 at 16:23
asked Jul 17 at 12:14
antonio
207210
asked Jul 17 at 12:14
antonio
207210
asked Jul 17 at 12:14
antonio
207210
207210
1
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25
add a comment |Â
1
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25
1
1
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The equality
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n$$
is simply not true in general (the counter example you provided is valid).
What people usually do is to implicitly identify those two sets using the natural bijection $F : prod_i=1^n X_i longrightarrow left(prod_i=1^n-1 X_iright) times X_n $ defined as
$$ F(f) = left(1, f_big), (2, f(n))right$$
You have exactly the same problem if you define the cartesian product as set of ordered pairs, namely it's not associative. Furthermore the ordered pairs definition and functional definition are strictly different according to the standard notion of set equality.
However for the most part this is always ignored (by abuse of notation) since those sets being isomoprphic is good enough for most people using them.
answered Jul 17 at 12:55
gcc-6.0
58017
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
add a comment |Â
up vote
0
down vote
Note that your definition of $X_1times X_2$ to mean $prod_iin1,2 X_i$ is at odds with the usual definition in a standard development of set theory, where
$$ X_1times X_2 = langle x,yrangle mid xin X_1, yin X_2 $$
You may now try to fix that up by defining that $langle x,yrangle$ actually means the function that maps $1$ to $x$ and $2$ to $y$, such that the two definitions agree.
However, a function itself will usually have been defined as a set of ordered pairs already, so this gives us, for example
$$ langle 2,1rangle = langle 1,2rangle, langle 2,1rangle $$
which is ill-founded.
It's like straightening out a kink in a carpet -- you need to have one (ugly? limited?) concept of ordered pair before you can begin building a (nicer? more general?) theory of indexed products.
The best you can achieve in standard mathematics is to make each you "$=$" into a natural bijection between the sets which you'll need to keep track of.
There are some active research movements that seek to remedy this, or at least find a somewhat less awkward way it can play out. For example, under my imperfect understanding, the homotopy type theory program attempts a move away from strict equality into a weaker concept under which $X_1times X_2 = prod_iin1,2X_i$ is "just as good as any other equality you should ever want to care about" because every $=$ comes with a semi-invisible bijection attached.
answered Jul 17 at 12:39
Henning Makholm
226k16291520
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The equality
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n$$
is simply not true in general (the counter example you provided is valid).
What people usually do is to implicitly identify those two sets using the natural bijection $F : prod_i=1^n X_i longrightarrow left(prod_i=1^n-1 X_iright) times X_n $ defined as
$$ F(f) = left(1, f_big), (2, f(n))right$$
You have exactly the same problem if you define the cartesian product as set of ordered pairs, namely it's not associative. Furthermore the ordered pairs definition and functional definition are strictly different according to the standard notion of set equality.
However for the most part this is always ignored (by abuse of notation) since those sets being isomoprphic is good enough for most people using them.
answered Jul 17 at 12:55
gcc-6.0
58017
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
add a comment |Â
up vote
1
down vote
accepted
The equality
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n$$
is simply not true in general (the counter example you provided is valid).
What people usually do is to implicitly identify those two sets using the natural bijection $F : prod_i=1^n X_i longrightarrow left(prod_i=1^n-1 X_iright) times X_n $ defined as
$$ F(f) = left(1, f_big), (2, f(n))right$$
You have exactly the same problem if you define the cartesian product as set of ordered pairs, namely it's not associative. Furthermore the ordered pairs definition and functional definition are strictly different according to the standard notion of set equality.
However for the most part this is always ignored (by abuse of notation) since those sets being isomoprphic is good enough for most people using them.
answered Jul 17 at 12:55
gcc-6.0
58017
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The equality
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n$$
is simply not true in general (the counter example you provided is valid).
What people usually do is to implicitly identify those two sets using the natural bijection $F : prod_i=1^n X_i longrightarrow left(prod_i=1^n-1 X_iright) times X_n $ defined as
$$ F(f) = left(1, f_big), (2, f(n))right$$
You have exactly the same problem if you define the cartesian product as set of ordered pairs, namely it's not associative. Furthermore the ordered pairs definition and functional definition are strictly different according to the standard notion of set equality.
However for the most part this is always ignored (by abuse of notation) since those sets being isomoprphic is good enough for most people using them.
answered Jul 17 at 12:55
gcc-6.0
58017
The equality
$$prod_i=1^n X_i = left(prod_i=1^n-1 X_iright) times X_n$$
is simply not true in general (the counter example you provided is valid).
What people usually do is to implicitly identify those two sets using the natural bijection $F : prod_i=1^n X_i longrightarrow left(prod_i=1^n-1 X_iright) times X_n $ defined as
$$ F(f) = left(1, f_big), (2, f(n))right$$
You have exactly the same problem if you define the cartesian product as set of ordered pairs, namely it's not associative. Furthermore the ordered pairs definition and functional definition are strictly different according to the standard notion of set equality.
However for the most part this is always ignored (by abuse of notation) since those sets being isomoprphic is good enough for most people using them.
answered Jul 17 at 12:55
gcc-6.0
58017
answered Jul 17 at 12:55
gcc-6.0
58017
answered Jul 17 at 12:55
gcc-6.0
58017
answered Jul 17 at 12:55
gcc-6.0
58017
58017
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
add a comment |Â
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
Thanks, it helped. However could you give a definition for $f_big$? used in $$f_big$$ It could be something like $(f(1)ldots f(n-1))$, but I am not sure.
– antonio
Jul 17 at 15:09
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
@antonio Yes this is the restriction of the function (standard notation). It's the same function only defined on the first $n-1$ numbers. For more info en.wikipedia.org/wiki/Restriction_(mathematics) .
– gcc-6.0
Jul 17 at 15:40
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
To the proposer: I usually type $f|_S$ (with the "absolute-value" symbol not sub-scripted) to mean the restriction of $f $ to the sub-domain $S$, where $Ssubset dom(f)$. To a set-theorist a function $is$ its graph so $f|_S=(x,f(x)):xin S.$.. I suppose you could modify this to extend it to cases where $S$ is not necessarily a subset of $dom(f)$ but I don't think I've ever seen the need for it.
– DanielWainfleet
Jul 17 at 20:05
add a comment |Â
up vote
0
down vote
Note that your definition of $X_1times X_2$ to mean $prod_iin1,2 X_i$ is at odds with the usual definition in a standard development of set theory, where
$$ X_1times X_2 = langle x,yrangle mid xin X_1, yin X_2 $$
You may now try to fix that up by defining that $langle x,yrangle$ actually means the function that maps $1$ to $x$ and $2$ to $y$, such that the two definitions agree.
However, a function itself will usually have been defined as a set of ordered pairs already, so this gives us, for example
$$ langle 2,1rangle = langle 1,2rangle, langle 2,1rangle $$
which is ill-founded.
It's like straightening out a kink in a carpet -- you need to have one (ugly? limited?) concept of ordered pair before you can begin building a (nicer? more general?) theory of indexed products.
The best you can achieve in standard mathematics is to make each you "$=$" into a natural bijection between the sets which you'll need to keep track of.
There are some active research movements that seek to remedy this, or at least find a somewhat less awkward way it can play out. For example, under my imperfect understanding, the homotopy type theory program attempts a move away from strict equality into a weaker concept under which $X_1times X_2 = prod_iin1,2X_i$ is "just as good as any other equality you should ever want to care about" because every $=$ comes with a semi-invisible bijection attached.
answered Jul 17 at 12:39
Henning Makholm
226k16291520
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
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Note that your definition of $X_1times X_2$ to mean $prod_iin1,2 X_i$ is at odds with the usual definition in a standard development of set theory, where
$$ X_1times X_2 = langle x,yrangle mid xin X_1, yin X_2 $$
You may now try to fix that up by defining that $langle x,yrangle$ actually means the function that maps $1$ to $x$ and $2$ to $y$, such that the two definitions agree.
However, a function itself will usually have been defined as a set of ordered pairs already, so this gives us, for example
$$ langle 2,1rangle = langle 1,2rangle, langle 2,1rangle $$
which is ill-founded.
It's like straightening out a kink in a carpet -- you need to have one (ugly? limited?) concept of ordered pair before you can begin building a (nicer? more general?) theory of indexed products.
The best you can achieve in standard mathematics is to make each you "$=$" into a natural bijection between the sets which you'll need to keep track of.
There are some active research movements that seek to remedy this, or at least find a somewhat less awkward way it can play out. For example, under my imperfect understanding, the homotopy type theory program attempts a move away from strict equality into a weaker concept under which $X_1times X_2 = prod_iin1,2X_i$ is "just as good as any other equality you should ever want to care about" because every $=$ comes with a semi-invisible bijection attached.
answered Jul 17 at 12:39
Henning Makholm
226k16291520
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that your definition of $X_1times X_2$ to mean $prod_iin1,2 X_i$ is at odds with the usual definition in a standard development of set theory, where
$$ X_1times X_2 = langle x,yrangle mid xin X_1, yin X_2 $$
You may now try to fix that up by defining that $langle x,yrangle$ actually means the function that maps $1$ to $x$ and $2$ to $y$, such that the two definitions agree.
However, a function itself will usually have been defined as a set of ordered pairs already, so this gives us, for example
$$ langle 2,1rangle = langle 1,2rangle, langle 2,1rangle $$
which is ill-founded.
It's like straightening out a kink in a carpet -- you need to have one (ugly? limited?) concept of ordered pair before you can begin building a (nicer? more general?) theory of indexed products.
The best you can achieve in standard mathematics is to make each you "$=$" into a natural bijection between the sets which you'll need to keep track of.
There are some active research movements that seek to remedy this, or at least find a somewhat less awkward way it can play out. For example, under my imperfect understanding, the homotopy type theory program attempts a move away from strict equality into a weaker concept under which $X_1times X_2 = prod_iin1,2X_i$ is "just as good as any other equality you should ever want to care about" because every $=$ comes with a semi-invisible bijection attached.
answered Jul 17 at 12:39
Henning Makholm
226k16291520
Note that your definition of $X_1times X_2$ to mean $prod_iin1,2 X_i$ is at odds with the usual definition in a standard development of set theory, where
$$ X_1times X_2 = langle x,yrangle mid xin X_1, yin X_2 $$
You may now try to fix that up by defining that $langle x,yrangle$ actually means the function that maps $1$ to $x$ and $2$ to $y$, such that the two definitions agree.
However, a function itself will usually have been defined as a set of ordered pairs already, so this gives us, for example
$$ langle 2,1rangle = langle 1,2rangle, langle 2,1rangle $$
which is ill-founded.
It's like straightening out a kink in a carpet -- you need to have one (ugly? limited?) concept of ordered pair before you can begin building a (nicer? more general?) theory of indexed products.
The best you can achieve in standard mathematics is to make each you "$=$" into a natural bijection between the sets which you'll need to keep track of.
There are some active research movements that seek to remedy this, or at least find a somewhat less awkward way it can play out. For example, under my imperfect understanding, the homotopy type theory program attempts a move away from strict equality into a weaker concept under which $X_1times X_2 = prod_iin1,2X_i$ is "just as good as any other equality you should ever want to care about" because every $=$ comes with a semi-invisible bijection attached.
answered Jul 17 at 12:39
Henning Makholm
226k16291520
edited Jul 17 at 12:44
answered Jul 17 at 12:39
Henning Makholm
226k16291520
answered Jul 17 at 12:39
Henning Makholm
226k16291520
answered Jul 17 at 12:39
Henning Makholm
226k16291520
226k16291520
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
add a comment |Â
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
As regards the usual definition of $A times B$, in my recent journey in ST I got that no matter how one encodes things, the objects' properties count. For example, for OPs it just counts that $(a,b)=(c,d) leftrightarrow (a=c wedge b=d)$. To avoid abuse of notation or chicken-and-egg problems, one could define functions based on binary Cartesian products ($times$), ->then indexed families, ->then $prod$, ->then the a new binary CP: $times_prod$. However one should find a definition for $times_prod$ somehow matching the properties of $times$.
– antonio
Jul 17 at 15:02
add a comment |Â
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1
Possible duplicate of Seeking a new, more natural definition of the cartesian product of sets
– Luca Bressan
Jul 17 at 12:28
@LucaBressan: Please see the UPDATE section
– antonio
Jul 17 at 16:25