Example of a set of real numbers that is Dedekind-finite but not finite
Without assuming $AC$, can we find an explicit example of a subset of $mathbb{R}$ such that it is not finite but it is Dedekind-finite?
set-theory infinity axiom-of-choice axioms
asked Jan 5 at 23:43
GödelGödel
1,413319
add a comment |
Without assuming $AC$, can we find an explicit example of a subset of $mathbb{R}$ such that it is not finite but it is Dedekind-finite?
set-theory infinity axiom-of-choice axioms
asked Jan 5 at 23:43
GödelGödel
1,413319
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
1
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31
add a comment |
Without assuming $AC$, can we find an explicit example of a subset of $mathbb{R}$ such that it is not finite but it is Dedekind-finite?
set-theory infinity axiom-of-choice axioms
asked Jan 5 at 23:43
GödelGödel
1,413319
Without assuming $AC$, can we find an explicit example of a subset of $mathbb{R}$ such that it is not finite but it is Dedekind-finite?
set-theory infinity axiom-of-choice axioms
set-theory infinity axiom-of-choice axioms
asked Jan 5 at 23:43
GödelGödel
1,413319
asked Jan 5 at 23:43
GödelGödel
1,413319
asked Jan 5 at 23:43
GödelGödel
1,413319
asked Jan 5 at 23:43
GödelGödel
1,413319
asked Jan 5 at 23:43
GödelGödel
1,413319
1,413319
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
1
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31
add a comment |
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
1
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
1
1
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31
add a comment |
2 Answers
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Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:
In the usual Cohen construction of a model of ZF+$neg$AC, we take a (countable transitie) "ground model" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. The resulting generic extension $M[mathcal{G}]$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $mathcal{G}$, adding only the set $G=ran(mathcal{G})$. In the resulting inner model $N$, that set $G$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.
A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
add a comment |
Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.
But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.
What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:
In the usual Cohen construction of a model of ZF+$neg$AC, we take a (countable transitie) "ground model" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. The resulting generic extension $M[mathcal{G}]$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $mathcal{G}$, adding only the set $G=ran(mathcal{G})$. In the resulting inner model $N$, that set $G$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.
A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
add a comment |
Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:
In the usual Cohen construction of a model of ZF+$neg$AC, we take a (countable transitie) "ground model" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. The resulting generic extension $M[mathcal{G}]$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $mathcal{G}$, adding only the set $G=ran(mathcal{G})$. In the resulting inner model $N$, that set $G$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.
A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
add a comment |
Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:
In the usual Cohen construction of a model of ZF+$neg$AC, we take a (countable transitie) "ground model" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. The resulting generic extension $M[mathcal{G}]$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $mathcal{G}$, adding only the set $G=ran(mathcal{G})$. In the resulting inner model $N$, that set $G$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.
A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:
In the usual Cohen construction of a model of ZF+$neg$AC, we take a (countable transitie) "ground model" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. The resulting generic extension $M[mathcal{G}]$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $mathcal{G}$, adding only the set $G=ran(mathcal{G})$. In the resulting inner model $N$, that set $G$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.
A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
answered Jan 6 at 0:00
Noah SchweberNoah Schweber
122k10149284
122k10149284
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
add a comment |
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
Thanks for your answer, It's so useful to me!
– Gödel
Jan 6 at 0:12
add a comment |
Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.
But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.
What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
add a comment |
Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.
But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.
What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
add a comment |
Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.
But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.
What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.
But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.
What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Jan 6 at 0:51
spaceisdarkgreenspaceisdarkgreen
32.5k21753
32.5k21753
add a comment |
add a comment |
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No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
– Andrés E. Caicedo
Jan 5 at 23:48
1
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
– Gödel
Jan 5 at 23:52
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
– Asaf Karagila♦
Jan 6 at 0:31