Can we state the existence of infinite set without infinity axiom? [duplicate]
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This question already has an answer here:
Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity
2 answers
How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?
2 answers
I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.
In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?
Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.
Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$
So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?
For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?
set-theory infinity axioms
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
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marked as duplicate by Asaf Karagila♦ set-theory
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Nov 28 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity
2 answers
How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?
2 answers
I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.
In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?
Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.
Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$
So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?
For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?
set-theory infinity axioms
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
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marked as duplicate by Asaf Karagila♦ set-theory
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Nov 28 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
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You might also want to read math.stackexchange.com/questions/1706661/…
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– Asaf Karagila♦
Nov 28 '18 at 13:44
add a comment |
$begingroup$
This question already has an answer here:
Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity
2 answers
How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?
2 answers
I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.
In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?
Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.
Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$
So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?
For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?
set-theory infinity axioms
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
$endgroup$
This question already has an answer here:
Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity
2 answers
How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?
2 answers
I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.
In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?
Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.
Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$
So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?
For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?
This question already has an answer here:
Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity
2 answers
How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?
2 answers
set-theory infinity axioms
set-theory infinity axioms
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
edited Nov 28 '18 at 13:14
Gödel
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
asked Nov 28 '18 at 13:09
GödelGödel
1,406319
1,406319
marked as duplicate by Asaf Karagila♦ set-theory
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Nov 28 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Asaf Karagila♦ set-theory
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Nov 28 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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You might also want to read math.stackexchange.com/questions/1706661/…
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– Asaf Karagila♦
Nov 28 '18 at 13:44
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You might also want to read math.stackexchange.com/questions/1706661/…
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– Asaf Karagila♦
Nov 28 '18 at 13:44
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– Asaf Karagila♦
Nov 28 '18 at 13:44
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– Asaf Karagila♦
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