Products and the axiom of choice The 2019 Stack Overflow Developer Survey Results Are InUniversal property of the direct product, proof verificationDoes the existence of products in the category of sets imply the Axiom of Choice?Axiom of Choice and Cartesian ProductsConfusion regarding one formulation of the Axiom of Choice.Axiom of Choice (Naive Set Theory, Halmos)Do we need Axiom of Choice to make infinite choices from a set?Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian productDefining relations without axiom of choiceProb. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian productSimple question: Which is the Wikipedia definition of axiom of choiceAxiom of Choice iff Every set has a choice function
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Products and the axiom of choice
The 2019 Stack Overflow Developer Survey Results Are InUniversal property of the direct product, proof verificationDoes the existence of products in the category of sets imply the Axiom of Choice?Axiom of Choice and Cartesian ProductsConfusion regarding one formulation of the Axiom of Choice.Axiom of Choice (Naive Set Theory, Halmos)Do we need Axiom of Choice to make infinite choices from a set?Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian productDefining relations without axiom of choiceProb. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian productSimple question: Which is the Wikipedia definition of axiom of choiceAxiom of Choice iff Every set has a choice function
$begingroup$
Here: Universal property of the direct product, proof verification
Matematleta noted in the comments, that the definition of the product uses the axiom of choice by default.
Why is that?
The definition i am looking at is simply:
For sets $X_1, X_2, dotso, X_n$ is $prod_i= 1^n X_i:=(x_1,dotso, x_n)$
Analogously for an arbitrary index set.
Also I always wondered, why you need the axiom of choice to state, that the product of not empty sets is not empty.
elementary-set-theory set-theory definition axiom-of-choice
asked Mar 23 at 2:41
CornmanCornman
3,40321229
$endgroup$
add a comment |
$begingroup$
Here: Universal property of the direct product, proof verification
Matematleta noted in the comments, that the definition of the product uses the axiom of choice by default.
Why is that?
The definition i am looking at is simply:
For sets $X_1, X_2, dotso, X_n$ is $prod_i= 1^n X_i:=(x_1,dotso, x_n)$
Analogously for an arbitrary index set.
Also I always wondered, why you need the axiom of choice to state, that the product of not empty sets is not empty.
elementary-set-theory set-theory definition axiom-of-choice
asked Mar 23 at 2:41
CornmanCornman
3,40321229
$endgroup$
1
$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46
add a comment |
$begingroup$
Here: Universal property of the direct product, proof verification
Matematleta noted in the comments, that the definition of the product uses the axiom of choice by default.
Why is that?
The definition i am looking at is simply:
For sets $X_1, X_2, dotso, X_n$ is $prod_i= 1^n X_i:=(x_1,dotso, x_n)$
Analogously for an arbitrary index set.
Also I always wondered, why you need the axiom of choice to state, that the product of not empty sets is not empty.
elementary-set-theory set-theory definition axiom-of-choice
asked Mar 23 at 2:41
CornmanCornman
3,40321229
$endgroup$
Here: Universal property of the direct product, proof verification
Matematleta noted in the comments, that the definition of the product uses the axiom of choice by default.
Why is that?
The definition i am looking at is simply:
For sets $X_1, X_2, dotso, X_n$ is $prod_i= 1^n X_i:=(x_1,dotso, x_n)$
Analogously for an arbitrary index set.
Also I always wondered, why you need the axiom of choice to state, that the product of not empty sets is not empty.
elementary-set-theory set-theory definition axiom-of-choice
elementary-set-theory set-theory definition axiom-of-choice
asked Mar 23 at 2:41
CornmanCornman
3,40321229
asked Mar 23 at 2:41
CornmanCornman
3,40321229
asked Mar 23 at 2:41
CornmanCornman
3,40321229
asked Mar 23 at 2:41
CornmanCornman
3,40321229
asked Mar 23 at 2:41
CornmanCornman
3,40321229
3,40321229
1
$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46
add a comment |
1
$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46
1
1
$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46
$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Because that's not a definition: what precisely is "$(x_1,...,x_n)$"? We might be tempted to hand-wave this away, and certainly nothing too surprising happens with finite tuples, but this is clearly unsatisfactory for arbitrary products - for example, what if I'm taking the product of a family of sets which isn't even linearly ordered in the first place?
- In general, never be satisfied when a definition you've written has all the "hard work" being done by the word "analogously" or something similar.
So we have to rigorously define what we mean by "large" ordered tuples.
Intuitively, an element of $prod_iin IX_i$ should be an "$I$-ary sequence" whose $i$th term comes from $X_i$. The simplest way to make this precise is to think of sequences as just being nicely-dressed functions - the sequence "First term $0$, second term $1$, third term $0$" is just the function with domain $1,2,3$ sending $1$ to $0$, $2$ to $1$, and $3$ to $0$.
Abstractly, then, we're talking about the following
An element of $prod_iin IX_i$ is a function $p:Irightarrowbigcup_iin I X_i$ with $p(i)in X_i$ for each $iin I$.
Now to talk about functions it suffices to talk about ordered pairs, which we can do by e.g. Kuratowski's definition. So we've converted a single "ordered $I$-tuple" into a set of ordered pairs.
That explains where the definition comes from. And with this in mind it's clear that the nonemptiness of the product of nonempty sets is equivalent to the axiom of choice: a choice function for a family of disjoint nonempty sets is literally an element of the product of that family of sets!
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
add a comment |
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$begingroup$
Because that's not a definition: what precisely is "$(x_1,...,x_n)$"? We might be tempted to hand-wave this away, and certainly nothing too surprising happens with finite tuples, but this is clearly unsatisfactory for arbitrary products - for example, what if I'm taking the product of a family of sets which isn't even linearly ordered in the first place?
- In general, never be satisfied when a definition you've written has all the "hard work" being done by the word "analogously" or something similar.
So we have to rigorously define what we mean by "large" ordered tuples.
Intuitively, an element of $prod_iin IX_i$ should be an "$I$-ary sequence" whose $i$th term comes from $X_i$. The simplest way to make this precise is to think of sequences as just being nicely-dressed functions - the sequence "First term $0$, second term $1$, third term $0$" is just the function with domain $1,2,3$ sending $1$ to $0$, $2$ to $1$, and $3$ to $0$.
Abstractly, then, we're talking about the following
An element of $prod_iin IX_i$ is a function $p:Irightarrowbigcup_iin I X_i$ with $p(i)in X_i$ for each $iin I$.
Now to talk about functions it suffices to talk about ordered pairs, which we can do by e.g. Kuratowski's definition. So we've converted a single "ordered $I$-tuple" into a set of ordered pairs.
That explains where the definition comes from. And with this in mind it's clear that the nonemptiness of the product of nonempty sets is equivalent to the axiom of choice: a choice function for a family of disjoint nonempty sets is literally an element of the product of that family of sets!
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
add a comment |
$begingroup$
Because that's not a definition: what precisely is "$(x_1,...,x_n)$"? We might be tempted to hand-wave this away, and certainly nothing too surprising happens with finite tuples, but this is clearly unsatisfactory for arbitrary products - for example, what if I'm taking the product of a family of sets which isn't even linearly ordered in the first place?
- In general, never be satisfied when a definition you've written has all the "hard work" being done by the word "analogously" or something similar.
So we have to rigorously define what we mean by "large" ordered tuples.
Intuitively, an element of $prod_iin IX_i$ should be an "$I$-ary sequence" whose $i$th term comes from $X_i$. The simplest way to make this precise is to think of sequences as just being nicely-dressed functions - the sequence "First term $0$, second term $1$, third term $0$" is just the function with domain $1,2,3$ sending $1$ to $0$, $2$ to $1$, and $3$ to $0$.
Abstractly, then, we're talking about the following
An element of $prod_iin IX_i$ is a function $p:Irightarrowbigcup_iin I X_i$ with $p(i)in X_i$ for each $iin I$.
Now to talk about functions it suffices to talk about ordered pairs, which we can do by e.g. Kuratowski's definition. So we've converted a single "ordered $I$-tuple" into a set of ordered pairs.
That explains where the definition comes from. And with this in mind it's clear that the nonemptiness of the product of nonempty sets is equivalent to the axiom of choice: a choice function for a family of disjoint nonempty sets is literally an element of the product of that family of sets!
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
add a comment |
$begingroup$
Because that's not a definition: what precisely is "$(x_1,...,x_n)$"? We might be tempted to hand-wave this away, and certainly nothing too surprising happens with finite tuples, but this is clearly unsatisfactory for arbitrary products - for example, what if I'm taking the product of a family of sets which isn't even linearly ordered in the first place?
- In general, never be satisfied when a definition you've written has all the "hard work" being done by the word "analogously" or something similar.
So we have to rigorously define what we mean by "large" ordered tuples.
Intuitively, an element of $prod_iin IX_i$ should be an "$I$-ary sequence" whose $i$th term comes from $X_i$. The simplest way to make this precise is to think of sequences as just being nicely-dressed functions - the sequence "First term $0$, second term $1$, third term $0$" is just the function with domain $1,2,3$ sending $1$ to $0$, $2$ to $1$, and $3$ to $0$.
Abstractly, then, we're talking about the following
An element of $prod_iin IX_i$ is a function $p:Irightarrowbigcup_iin I X_i$ with $p(i)in X_i$ for each $iin I$.
Now to talk about functions it suffices to talk about ordered pairs, which we can do by e.g. Kuratowski's definition. So we've converted a single "ordered $I$-tuple" into a set of ordered pairs.
That explains where the definition comes from. And with this in mind it's clear that the nonemptiness of the product of nonempty sets is equivalent to the axiom of choice: a choice function for a family of disjoint nonempty sets is literally an element of the product of that family of sets!
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
$endgroup$
Because that's not a definition: what precisely is "$(x_1,...,x_n)$"? We might be tempted to hand-wave this away, and certainly nothing too surprising happens with finite tuples, but this is clearly unsatisfactory for arbitrary products - for example, what if I'm taking the product of a family of sets which isn't even linearly ordered in the first place?
- In general, never be satisfied when a definition you've written has all the "hard work" being done by the word "analogously" or something similar.
So we have to rigorously define what we mean by "large" ordered tuples.
Intuitively, an element of $prod_iin IX_i$ should be an "$I$-ary sequence" whose $i$th term comes from $X_i$. The simplest way to make this precise is to think of sequences as just being nicely-dressed functions - the sequence "First term $0$, second term $1$, third term $0$" is just the function with domain $1,2,3$ sending $1$ to $0$, $2$ to $1$, and $3$ to $0$.
Abstractly, then, we're talking about the following
An element of $prod_iin IX_i$ is a function $p:Irightarrowbigcup_iin I X_i$ with $p(i)in X_i$ for each $iin I$.
Now to talk about functions it suffices to talk about ordered pairs, which we can do by e.g. Kuratowski's definition. So we've converted a single "ordered $I$-tuple" into a set of ordered pairs.
That explains where the definition comes from. And with this in mind it's clear that the nonemptiness of the product of nonempty sets is equivalent to the axiom of choice: a choice function for a family of disjoint nonempty sets is literally an element of the product of that family of sets!
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
answered Mar 23 at 2:52
Noah SchweberNoah Schweber
128k10152294
128k10152294
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
add a comment |
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
$begingroup$
Thanks for this answer. Very clear.
$endgroup$
– Cornman
Mar 23 at 2:55
add a comment |
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$begingroup$
Let $X$ be a set. Then look at $prod_AinmathcalP(X)setminusemptysetA$. This product is not empty is exactly saying "there exists a choice function"
$endgroup$
– Holo
Mar 23 at 2:46