Mixing
Correct formulation of axiom of choice
$begingroup$
In a paper, Asaf Karagila writes:
Definition 1 (The Axiom of Choice). If ${A_i
mid i ∈ I}$ is a set of non-empty sets, then there exists
a function $f$ with domain $I$ such that $f(i) ∈ A_i$
for all $i ∈ I$.
Does this formally make sense? Shouldn't it say
If $(A_i)_{iin I}$ be a family of non-empty sets …
?
Because if we have a set $M$ of non-empty sets there may be no unique family $(A_i)_{iin I}$ with ${A_imid iin I}= M$.
elementary-set-theory logic soft-question foundations
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
$endgroup$
add a comment |
$begingroup$
In a paper, Asaf Karagila writes:
Definition 1 (The Axiom of Choice). If ${A_i
mid i ∈ I}$ is a set of non-empty sets, then there exists
a function $f$ with domain $I$ such that $f(i) ∈ A_i$
for all $i ∈ I$.
Does this formally make sense? Shouldn't it say
If $(A_i)_{iin I}$ be a family of non-empty sets …
?
Because if we have a set $M$ of non-empty sets there may be no unique family $(A_i)_{iin I}$ with ${A_imid iin I}= M$.
elementary-set-theory logic soft-question foundations
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
$endgroup$
add a comment |
$begingroup$
In a paper, Asaf Karagila writes:
Definition 1 (The Axiom of Choice). If ${A_i
mid i ∈ I}$ is a set of non-empty sets, then there exists
a function $f$ with domain $I$ such that $f(i) ∈ A_i$
for all $i ∈ I$.
Does this formally make sense? Shouldn't it say
If $(A_i)_{iin I}$ be a family of non-empty sets …
?
Because if we have a set $M$ of non-empty sets there may be no unique family $(A_i)_{iin I}$ with ${A_imid iin I}= M$.
elementary-set-theory logic soft-question foundations
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
$endgroup$
In a paper, Asaf Karagila writes:
Definition 1 (The Axiom of Choice). If ${A_i
mid i ∈ I}$ is a set of non-empty sets, then there exists
a function $f$ with domain $I$ such that $f(i) ∈ A_i$
for all $i ∈ I$.
Does this formally make sense? Shouldn't it say
If $(A_i)_{iin I}$ be a family of non-empty sets …
?
Because if we have a set $M$ of non-empty sets there may be no unique family $(A_i)_{iin I}$ with ${A_imid iin I}= M$.
elementary-set-theory logic soft-question foundations
elementary-set-theory logic soft-question foundations
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
edited Jan 1 at 14:26
Asaf Karagila♦
302k32427757
302k32427757
asked Nov 6 '16 at 17:51
user384011
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It does make sense subject to a generous interpretation of the role of the indexing set $I$ and the indexing function $i mapsto A_i$. It would be much better style (in my opinion) either to write it as you suggested, stating explicitly that the indexing function $i mapsto A_i$ is part of the data or to write it without mentioning the indexing set at all: "if $U$ is a set of non-empty sets, then there is a function $f$ with domain $U$ such that $f(A) in A$ for every $A in U$". The two statements are equivalent (because any indexing function $i mapsto A_i$ of the set $U$ factors through the identity function on $U$, which you can regard as a sort of "minimalist" indexing function).
[Aside: I don't think there is any general agreement that "family" means indexed family. Indeed, some authors explicitly state that they use terms like "set", "family" and "collection" as synonyms. So it's safest to say "indexed set" or "indexed family".]
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
$endgroup$
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
add a comment |
$begingroup$
There are several common ways to express the axiom of choice in set theory:
If $M$ is a set of nonempty sets, there is a function $f$ such that $f(x) in x$ for each $x in M$.
- Variant: if $M$ is a set of pairwise disjoint nonempty sets, there is a set $C$ such that $|C cap x| = 1$ for all $x in M$.
If $g$ is a function and $I$ is a set such that $g(i)$ is nonempty for each $i in I$, there is a function $h$ such that $h(i) in g(i)$ for each $i in I$.
The latter could be viewed as a definition in terms of indexed families; we could look at $g(I)$ as an indexed family ${ G_i : i in I}$ where $G_i = g(i)$. The former is somewhat easier to state, and the variant is even easier because it does not require us to define a "function".
It is also at least somewhat common in set theory to look at each set as indexed by itself, when we want to treat a set as an indexed set. So we have $A = { A_a : a in A}$ where $A_a = a$ for $a in A$. (Say that five times fast...)
The definition linked in the question could be rephrased formally in terms of any of the definitions I mentioned. This is usually viewed as routine, particularly because all reasonable variants are equivalent to each other over ZF set theory, so in practice it does not make too much difference which formal sentence you take to represent the axiom of choice. (In fact, Kunen's classic textbook expressed the axiom of choice as "every set can be well ordered", which is well known to be equivalent over ZF.)
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
$endgroup$
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
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@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
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Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It does make sense subject to a generous interpretation of the role of the indexing set $I$ and the indexing function $i mapsto A_i$. It would be much better style (in my opinion) either to write it as you suggested, stating explicitly that the indexing function $i mapsto A_i$ is part of the data or to write it without mentioning the indexing set at all: "if $U$ is a set of non-empty sets, then there is a function $f$ with domain $U$ such that $f(A) in A$ for every $A in U$". The two statements are equivalent (because any indexing function $i mapsto A_i$ of the set $U$ factors through the identity function on $U$, which you can regard as a sort of "minimalist" indexing function).
[Aside: I don't think there is any general agreement that "family" means indexed family. Indeed, some authors explicitly state that they use terms like "set", "family" and "collection" as synonyms. So it's safest to say "indexed set" or "indexed family".]
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
$endgroup$
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
add a comment |
$begingroup$
It does make sense subject to a generous interpretation of the role of the indexing set $I$ and the indexing function $i mapsto A_i$. It would be much better style (in my opinion) either to write it as you suggested, stating explicitly that the indexing function $i mapsto A_i$ is part of the data or to write it without mentioning the indexing set at all: "if $U$ is a set of non-empty sets, then there is a function $f$ with domain $U$ such that $f(A) in A$ for every $A in U$". The two statements are equivalent (because any indexing function $i mapsto A_i$ of the set $U$ factors through the identity function on $U$, which you can regard as a sort of "minimalist" indexing function).
[Aside: I don't think there is any general agreement that "family" means indexed family. Indeed, some authors explicitly state that they use terms like "set", "family" and "collection" as synonyms. So it's safest to say "indexed set" or "indexed family".]
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
$endgroup$
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
add a comment |
$begingroup$
It does make sense subject to a generous interpretation of the role of the indexing set $I$ and the indexing function $i mapsto A_i$. It would be much better style (in my opinion) either to write it as you suggested, stating explicitly that the indexing function $i mapsto A_i$ is part of the data or to write it without mentioning the indexing set at all: "if $U$ is a set of non-empty sets, then there is a function $f$ with domain $U$ such that $f(A) in A$ for every $A in U$". The two statements are equivalent (because any indexing function $i mapsto A_i$ of the set $U$ factors through the identity function on $U$, which you can regard as a sort of "minimalist" indexing function).
[Aside: I don't think there is any general agreement that "family" means indexed family. Indeed, some authors explicitly state that they use terms like "set", "family" and "collection" as synonyms. So it's safest to say "indexed set" or "indexed family".]
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
$endgroup$
It does make sense subject to a generous interpretation of the role of the indexing set $I$ and the indexing function $i mapsto A_i$. It would be much better style (in my opinion) either to write it as you suggested, stating explicitly that the indexing function $i mapsto A_i$ is part of the data or to write it without mentioning the indexing set at all: "if $U$ is a set of non-empty sets, then there is a function $f$ with domain $U$ such that $f(A) in A$ for every $A in U$". The two statements are equivalent (because any indexing function $i mapsto A_i$ of the set $U$ factors through the identity function on $U$, which you can regard as a sort of "minimalist" indexing function).
[Aside: I don't think there is any general agreement that "family" means indexed family. Indeed, some authors explicitly state that they use terms like "set", "family" and "collection" as synonyms. So it's safest to say "indexed set" or "indexed family".]
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
edited Nov 6 '16 at 19:20
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
answered Nov 6 '16 at 19:13
Rob ArthanRob Arthan
29.1k42866
29.1k42866
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
add a comment |
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Could you explain the first sentence a bit? To me it does not make sense what Asaf wrote, since a set of nonempty sets doesn't uniquely determine an indexed family whose "range" is this set of nonempty sets.
$endgroup$
– user384011
Nov 6 '16 at 20:06
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
$begingroup$
Asaf should have begun by saying something like "let $I$ be some index set and let $A_i$ be a set/family/collection of sets indexed by $i in I$". If we are generous, we can take that as the only reasonable reading of what he actually wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 20:25
4
4
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
$begingroup$
Asaf also knows that every set can be indexed by itself. In case that your 18.4k points made you somehow miss a now-deleted answer. Or, you know, the fact that it is enough to consider choice functions for non-indexed families. Or well-orders. Or bases for vector spaces. Or, you know, Tychonoff's theorem. Or, maybe, every surjection splits. Or one of infinitely other equivalents of the axiom of choice. Don't presume to tell others what I should have written. Thank you very much.
$endgroup$
– Asaf Karagila♦
Nov 6 '16 at 20:40
1
1
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
$begingroup$
@AsafKaragila: I think you misunderstood him. I am a beginner and you are using advanced jargon. Rob just told me that if one would try to formulate the sentence so that beginners like me can understand it without confusion, then one should write it like this.
$endgroup$
– user384011
Nov 6 '16 at 20:58
5
5
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
$begingroup$
@AsafKaragila:you have misunderstood my comment: I meant that the definition you gave is implicit about the role of the index set $I$ and the indexing function $i mapsto A_i$. In that sense, you should (morally) have written more than you did to make the particular definition of AC that you chose in your paper clear. Nothing in my answer or my comment was intended to suggest that you should have written a completely different definition in your paper and I don't understand why you are (apparently) offended by what I wrote.
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:24
add a comment |
$begingroup$
There are several common ways to express the axiom of choice in set theory:
If $M$ is a set of nonempty sets, there is a function $f$ such that $f(x) in x$ for each $x in M$.
- Variant: if $M$ is a set of pairwise disjoint nonempty sets, there is a set $C$ such that $|C cap x| = 1$ for all $x in M$.
If $g$ is a function and $I$ is a set such that $g(i)$ is nonempty for each $i in I$, there is a function $h$ such that $h(i) in g(i)$ for each $i in I$.
The latter could be viewed as a definition in terms of indexed families; we could look at $g(I)$ as an indexed family ${ G_i : i in I}$ where $G_i = g(i)$. The former is somewhat easier to state, and the variant is even easier because it does not require us to define a "function".
It is also at least somewhat common in set theory to look at each set as indexed by itself, when we want to treat a set as an indexed set. So we have $A = { A_a : a in A}$ where $A_a = a$ for $a in A$. (Say that five times fast...)
The definition linked in the question could be rephrased formally in terms of any of the definitions I mentioned. This is usually viewed as routine, particularly because all reasonable variants are equivalent to each other over ZF set theory, so in practice it does not make too much difference which formal sentence you take to represent the axiom of choice. (In fact, Kunen's classic textbook expressed the axiom of choice as "every set can be well ordered", which is well known to be equivalent over ZF.)
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
$endgroup$
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
add a comment |
$begingroup$
There are several common ways to express the axiom of choice in set theory:
If $M$ is a set of nonempty sets, there is a function $f$ such that $f(x) in x$ for each $x in M$.
- Variant: if $M$ is a set of pairwise disjoint nonempty sets, there is a set $C$ such that $|C cap x| = 1$ for all $x in M$.
If $g$ is a function and $I$ is a set such that $g(i)$ is nonempty for each $i in I$, there is a function $h$ such that $h(i) in g(i)$ for each $i in I$.
The latter could be viewed as a definition in terms of indexed families; we could look at $g(I)$ as an indexed family ${ G_i : i in I}$ where $G_i = g(i)$. The former is somewhat easier to state, and the variant is even easier because it does not require us to define a "function".
It is also at least somewhat common in set theory to look at each set as indexed by itself, when we want to treat a set as an indexed set. So we have $A = { A_a : a in A}$ where $A_a = a$ for $a in A$. (Say that five times fast...)
The definition linked in the question could be rephrased formally in terms of any of the definitions I mentioned. This is usually viewed as routine, particularly because all reasonable variants are equivalent to each other over ZF set theory, so in practice it does not make too much difference which formal sentence you take to represent the axiom of choice. (In fact, Kunen's classic textbook expressed the axiom of choice as "every set can be well ordered", which is well known to be equivalent over ZF.)
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
$endgroup$
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
add a comment |
$begingroup$
There are several common ways to express the axiom of choice in set theory:
If $M$ is a set of nonempty sets, there is a function $f$ such that $f(x) in x$ for each $x in M$.
- Variant: if $M$ is a set of pairwise disjoint nonempty sets, there is a set $C$ such that $|C cap x| = 1$ for all $x in M$.
If $g$ is a function and $I$ is a set such that $g(i)$ is nonempty for each $i in I$, there is a function $h$ such that $h(i) in g(i)$ for each $i in I$.
The latter could be viewed as a definition in terms of indexed families; we could look at $g(I)$ as an indexed family ${ G_i : i in I}$ where $G_i = g(i)$. The former is somewhat easier to state, and the variant is even easier because it does not require us to define a "function".
It is also at least somewhat common in set theory to look at each set as indexed by itself, when we want to treat a set as an indexed set. So we have $A = { A_a : a in A}$ where $A_a = a$ for $a in A$. (Say that five times fast...)
The definition linked in the question could be rephrased formally in terms of any of the definitions I mentioned. This is usually viewed as routine, particularly because all reasonable variants are equivalent to each other over ZF set theory, so in practice it does not make too much difference which formal sentence you take to represent the axiom of choice. (In fact, Kunen's classic textbook expressed the axiom of choice as "every set can be well ordered", which is well known to be equivalent over ZF.)
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
$endgroup$
There are several common ways to express the axiom of choice in set theory:
If $M$ is a set of nonempty sets, there is a function $f$ such that $f(x) in x$ for each $x in M$.
- Variant: if $M$ is a set of pairwise disjoint nonempty sets, there is a set $C$ such that $|C cap x| = 1$ for all $x in M$.
If $g$ is a function and $I$ is a set such that $g(i)$ is nonempty for each $i in I$, there is a function $h$ such that $h(i) in g(i)$ for each $i in I$.
The latter could be viewed as a definition in terms of indexed families; we could look at $g(I)$ as an indexed family ${ G_i : i in I}$ where $G_i = g(i)$. The former is somewhat easier to state, and the variant is even easier because it does not require us to define a "function".
It is also at least somewhat common in set theory to look at each set as indexed by itself, when we want to treat a set as an indexed set. So we have $A = { A_a : a in A}$ where $A_a = a$ for $a in A$. (Say that five times fast...)
The definition linked in the question could be rephrased formally in terms of any of the definitions I mentioned. This is usually viewed as routine, particularly because all reasonable variants are equivalent to each other over ZF set theory, so in practice it does not make too much difference which formal sentence you take to represent the axiom of choice. (In fact, Kunen's classic textbook expressed the axiom of choice as "every set can be well ordered", which is well known to be equivalent over ZF.)
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
edited Nov 6 '16 at 21:46
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
answered Nov 6 '16 at 20:26
Carl MummertCarl Mummert
66.2k7131246
66.2k7131246
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
add a comment |
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
How does the second variant (the one with $C$) work out for $M = {{1}, {2}, {1, 2}}$?
$endgroup$
– Rob Arthan
Nov 6 '16 at 21:41
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
@Rob Arthan: thanks for reminding me about that omission
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:46
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
$begingroup$
Related: math.stackexchange.com/questions/97222/…
$endgroup$
– Carl Mummert
Nov 6 '16 at 21:47
add a comment |
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