Axiom of choice and cartesian productThe Axiom of Choice and the Cartesian Product.Axiom of choice and the number of choice functionsAxiom of Choice and Cartesian ProductsConfusion regarding one formulation of the Axiom of Choice.Axiom of Choice (Naive Set Theory, Halmos)Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian productDoes the Cartesian product of an infinite family have all the elements we expect?Prob. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian productAxiom of Choice iff Every set has a choice function
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Axiom of choice and cartesian product
The Axiom of Choice and the Cartesian Product.Axiom of choice and the number of choice functionsAxiom of Choice and Cartesian ProductsConfusion regarding one formulation of the Axiom of Choice.Axiom of Choice (Naive Set Theory, Halmos)Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian productDoes the Cartesian product of an infinite family have all the elements we expect?Prob. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian productAxiom of Choice iff Every set has a choice function
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margin-bottom:0;
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$begingroup$
I learned that Axiom of Choice states that the cartesian product of a family of non-empty sets $X_i$ indexed by a non-empty set $I$ is non-empty.
I think I can accept this axiom.
But I don’t understand how it guarantees that cartesian product can have more than one elements! It’s because ‘non-empty’ sounds to me ‘having at least one element’ here. So I think AC just implies that there exists at least one element in the product.
How do we insure that there exist all the possible functions from $I$ to $cup X_i$?
Thanks in advance!
elementary-set-theory axiom-of-choice
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
I learned that Axiom of Choice states that the cartesian product of a family of non-empty sets $X_i$ indexed by a non-empty set $I$ is non-empty.
I think I can accept this axiom.
But I don’t understand how it guarantees that cartesian product can have more than one elements! It’s because ‘non-empty’ sounds to me ‘having at least one element’ here. So I think AC just implies that there exists at least one element in the product.
How do we insure that there exist all the possible functions from $I$ to $cup X_i$?
Thanks in advance!
elementary-set-theory axiom-of-choice
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
$endgroup$
$begingroup$
The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
$endgroup$
– Andrés E. Caicedo
11 hours ago
$begingroup$
@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
$endgroup$
– anadad
10 hours ago
1
$begingroup$
Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
$endgroup$
– anadad
9 hours ago
1
$begingroup$
Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
$endgroup$
– Andrés E. Caicedo
9 hours ago
|
show 2 more comments
$begingroup$
I learned that Axiom of Choice states that the cartesian product of a family of non-empty sets $X_i$ indexed by a non-empty set $I$ is non-empty.
I think I can accept this axiom.
But I don’t understand how it guarantees that cartesian product can have more than one elements! It’s because ‘non-empty’ sounds to me ‘having at least one element’ here. So I think AC just implies that there exists at least one element in the product.
How do we insure that there exist all the possible functions from $I$ to $cup X_i$?
Thanks in advance!
elementary-set-theory axiom-of-choice
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
$endgroup$
I learned that Axiom of Choice states that the cartesian product of a family of non-empty sets $X_i$ indexed by a non-empty set $I$ is non-empty.
I think I can accept this axiom.
But I don’t understand how it guarantees that cartesian product can have more than one elements! It’s because ‘non-empty’ sounds to me ‘having at least one element’ here. So I think AC just implies that there exists at least one element in the product.
How do we insure that there exist all the possible functions from $I$ to $cup X_i$?
Thanks in advance!
elementary-set-theory axiom-of-choice
elementary-set-theory axiom-of-choice
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
edited 11 hours ago
Andrés E. Caicedo
67.4k8 gold badges170 silver badges265 bronze badges
67.4k8 gold badges170 silver badges265 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
asked 11 hours ago
anadadanadad
476 bronze badges
476 bronze badges
$begingroup$
The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
$endgroup$
– Andrés E. Caicedo
11 hours ago
$begingroup$
@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
$endgroup$
– anadad
10 hours ago
1
$begingroup$
Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
$endgroup$
– anadad
9 hours ago
1
$begingroup$
Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
$endgroup$
– Andrés E. Caicedo
9 hours ago
|
show 2 more comments
$begingroup$
The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
$endgroup$
– Andrés E. Caicedo
11 hours ago
$begingroup$
@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
$endgroup$
– anadad
10 hours ago
1
$begingroup$
Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
$endgroup$
– anadad
9 hours ago
1
$begingroup$
Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
$endgroup$
– Andrés E. Caicedo
9 hours ago
$begingroup$
The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
$endgroup$
– Andrés E. Caicedo
11 hours ago
$begingroup$
The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
$endgroup$
– Andrés E. Caicedo
11 hours ago
$begingroup$
@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
$endgroup$
– anadad
10 hours ago
$begingroup$
@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
$endgroup$
– anadad
10 hours ago
1
1
$begingroup$
Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
$endgroup$
– anadad
9 hours ago
$begingroup$
Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
$endgroup$
– anadad
9 hours ago
1
1
$begingroup$
Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
$endgroup$
– Andrés E. Caicedo
9 hours ago
$begingroup$
Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
$endgroup$
– Andrés E. Caicedo
9 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The set of choice functions of domain $iin I$ satisfying $f(X_i)in X_i$ for all $iin I$ has cardinality $prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.
We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".
When you talk about functions from $I$ to $cup X_i$, you presumably mean the ones satisfying $f(i)in X_i$ for all $iin I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZFneg C$ implies the existence of $X_i$ for which zero of either exist.)
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
$endgroup$
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
|
show 3 more comments
$begingroup$
It sounds like you want to prove the following statement:
- Given a partial function $f:Itocup X_i$, there exists a (total) function $F:Itocup X_i$ that extends $f$.
Can you see how to define such an $F$, given the existence of some function $C:Itocup X_i$?
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
$endgroup$
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
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2 Answers
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2 Answers
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$begingroup$
The set of choice functions of domain $iin I$ satisfying $f(X_i)in X_i$ for all $iin I$ has cardinality $prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.
We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".
When you talk about functions from $I$ to $cup X_i$, you presumably mean the ones satisfying $f(i)in X_i$ for all $iin I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZFneg C$ implies the existence of $X_i$ for which zero of either exist.)
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
$endgroup$
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
|
show 3 more comments
$begingroup$
The set of choice functions of domain $iin I$ satisfying $f(X_i)in X_i$ for all $iin I$ has cardinality $prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.
We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".
When you talk about functions from $I$ to $cup X_i$, you presumably mean the ones satisfying $f(i)in X_i$ for all $iin I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZFneg C$ implies the existence of $X_i$ for which zero of either exist.)
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
$endgroup$
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
|
show 3 more comments
$begingroup$
The set of choice functions of domain $iin I$ satisfying $f(X_i)in X_i$ for all $iin I$ has cardinality $prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.
We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".
When you talk about functions from $I$ to $cup X_i$, you presumably mean the ones satisfying $f(i)in X_i$ for all $iin I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZFneg C$ implies the existence of $X_i$ for which zero of either exist.)
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
$endgroup$
The set of choice functions of domain $iin I$ satisfying $f(X_i)in X_i$ for all $iin I$ has cardinality $prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.
We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".
When you talk about functions from $I$ to $cup X_i$, you presumably mean the ones satisfying $f(i)in X_i$ for all $iin I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZFneg C$ implies the existence of $X_i$ for which zero of either exist.)
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
edited 11 hours ago
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
answered 11 hours ago
J.G.J.G.
49.3k2 gold badges43 silver badges65 bronze badges
49.3k2 gold badges43 silver badges65 bronze badges
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
|
show 3 more comments
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
Oh..! What I was curious about was if I can assume that there exists a choice function $C:P(cup X_i) to cup X_i$ such that $C(X_i)=x_i$. Because, in my textbook, it just says that every set has a choice function. (I’m not good at English so I don’t know how to interpret ‘a choice function’. So I’m confusing because I don’t know whether it means ‘at least one’ or ‘all possible choice functions from the collection of all non-empty subsets into the set’
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad A choice function on $S$ is a function $f$ of domain $Ssetminusemptyset$ satisfying $f(x)in x$ for all $xinoperatornamedomf$. Your textbook was either asserting the axiom of choice or explaining what it asserts; I'd need to see the exact words to know which. But yes, "a" means "at least one" in this case (or just about any "exists a" statement in mathematics; if uniqueness is intended, it's usually mentioned).
$endgroup$
– J.G.
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad When you write "there exists a choice function such that $C(X_i)=x_i$" (presumably, "for all $iin I$"), this really makes no sense, because you haven't told us what $x_i$ is. If you have access to some means of specifying an element $x_i$ for each $iin I$, then you actually have access to a choice function.
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
$begingroup$
@anadad There is a subtlety here, since set theory deals with consistency results, and with diverse models of the set-theoretic axioms. It may very well be that we are studying a model $M_1$ of set theory. In that model we know that there are choice functions for some family of sets. It may be that in some other model $M_2$ there is some specific choice function $f$ for that same family, and that that specific function is not a member of $M_1$. That happens. It doesn't mean that it is not true in $M_1$ that we do not have the set of all possible choice functions for that family. (Cont.)
$endgroup$
– Andrés E. Caicedo
10 hours ago
1
1
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
I thank you several more times.
$endgroup$
– Sebastiano
7 hours ago
|
show 3 more comments
$begingroup$
It sounds like you want to prove the following statement:
- Given a partial function $f:Itocup X_i$, there exists a (total) function $F:Itocup X_i$ that extends $f$.
Can you see how to define such an $F$, given the existence of some function $C:Itocup X_i$?
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
$endgroup$
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
add a comment
|
$begingroup$
It sounds like you want to prove the following statement:
- Given a partial function $f:Itocup X_i$, there exists a (total) function $F:Itocup X_i$ that extends $f$.
Can you see how to define such an $F$, given the existence of some function $C:Itocup X_i$?
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
$endgroup$
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
add a comment
|
$begingroup$
It sounds like you want to prove the following statement:
- Given a partial function $f:Itocup X_i$, there exists a (total) function $F:Itocup X_i$ that extends $f$.
Can you see how to define such an $F$, given the existence of some function $C:Itocup X_i$?
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
$endgroup$
It sounds like you want to prove the following statement:
- Given a partial function $f:Itocup X_i$, there exists a (total) function $F:Itocup X_i$ that extends $f$.
Can you see how to define such an $F$, given the existence of some function $C:Itocup X_i$?
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
edited 10 hours ago
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
answered 11 hours ago
Chris CulterChris Culter
23.2k4 gold badges40 silver badges93 bronze badges
23.2k4 gold badges40 silver badges93 bronze badges
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
add a comment
|
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
Sorry, I don’t know what partial function is! I want to prove that for each $x_i in X_i$, there exists a function $f:I to cup X_i$ defined by $f(i)=x_i$.
$endgroup$
– anadad
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad Do you mean 'for each set of $x_iin X_i$'? Right now your statement is a little light on the formalization. and if you attempt to properly formalize it you may discover that it trivializes...
$endgroup$
– Steven Stadnicki
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
@anadad It's just a function of a subset of $I$. See en.wikipedia.org/wiki/Partial_function
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Umm.. I don’t understand why $C: I to cup X_i$ is choice function. I learned that a choice function of A set S is a function from a collection of all non-empty subset of S into S itself. But $I$ doesn’t seem to the collection of all non-empty subset of $cup X_i$. Sorry but could you explain it more?
$endgroup$
– anadad
11 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
$begingroup$
Okay, $C$ isn't a choice function in that sense. Edited. The point is, you have one such function, as described in your question, and you want more of them.
$endgroup$
– Chris Culter
10 hours ago
add a comment
|
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The set of all possible functions is precisely the cartesian product, so what you are asking is tautologically true, even if the cartesian product is empty! I think what you mean to ask is how to ensure that the cartesian product is big (infinite, or even uncountably infinite) in typical cases, when all that the axiom ensures is that it is nonempty. As an example, suppose $X$ is one of the sets $X_i$. Can you show that, if nonempty, the cartesian product has size at least $|X|$?
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– Andrés E. Caicedo
11 hours ago
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@Andrés E. Caicedo I thought ‘the set of all possible functions is non-empty’ is needed because we don’t know such functions actually exist(if we know that they exist without anything else, we don’t need Axiom of Choice). My question is “Does ‘non-emptiness’ imply ‘the existence of all possible functions?’ I don’t know how to write my question clearly because my English isn’t good enough for that. I’m so sorry
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– anadad
10 hours ago
1
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Again, the cartesian product is by definition the set of all possible functions. What is significant is that, in any typical situation, although the axiom only states the existence of one such function, this automatically implies, without further appeal to the axiom, the existence of many.
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– Andrés E. Caicedo
10 hours ago
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Ahhh okay! Actually I was trying to prove that ,for $k in I$, $π_k : Π_i X_i to X_k$ is surjective. So I wanted to show, for each $a in X_k$, there exists a function $f_a : I to cup X_i$ such that $f_a(i)=a$ if i=k, $f_a(i) in X_i$ otherwise. And I was wondering if the fact that the cartesian product is non-empty guarantees the existence of f_a. And thank you for the help! Now I can write f_a exists in my note.
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– anadad
9 hours ago
1
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Yes, you can ensure that such a function $f_a$ exists: Once you have a function $f$ in the Cartesian product, you can use it to define a function $f_a$ as you want: Simply change the value that $f$ assigns to $i$ to the value that you want it to have ($a$). That is, you can take $f_a=(fsmallsetminus(i,f(i)))cup(i,a)$. This is precisely what Chris was suggesting (in a more general way) to do in his answer.
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– Andrés E. Caicedo
9 hours ago