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Cardinality less than the natural numbers is finite
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If we have a set S such that #(S)<#(N), where N is the set of the natural numbers, how would one prove that S must be finite? Is the axiom of choice relevant for such a proof? If so, what is a proof that doesn't require the axiom of choice?
elementary-set-theory set-theory axiom-of-choice
asked 59 mins ago
A. Smith
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If we have a set S such that #(S)<#(N), where N is the set of the natural numbers, how would one prove that S must be finite? Is the axiom of choice relevant for such a proof? If so, what is a proof that doesn't require the axiom of choice?
elementary-set-theory set-theory axiom-of-choice
asked 59 mins ago
A. Smith
503
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If we have a set S such that #(S)<#(N), where N is the set of the natural numbers, how would one prove that S must be finite? Is the axiom of choice relevant for such a proof? If so, what is a proof that doesn't require the axiom of choice?
elementary-set-theory set-theory axiom-of-choice
asked 59 mins ago
A. Smith
503
If we have a set S such that #(S)<#(N), where N is the set of the natural numbers, how would one prove that S must be finite? Is the axiom of choice relevant for such a proof? If so, what is a proof that doesn't require the axiom of choice?
elementary-set-theory set-theory axiom-of-choice
elementary-set-theory set-theory axiom-of-choice
asked 59 mins ago
A. Smith
503
asked 59 mins ago
A. Smith
503
asked 59 mins ago
A. Smith
503
asked 59 mins ago
A. Smith
503
asked 59 mins ago
A. Smith
503
503
add a comment |Â
add a comment |Â
3 Answers
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up vote
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By definition of $<$ for cardinalities, there exists an injection $fcolon Sto Bbb N$. For $ninBbb N$ let $h(n)=|,xin f(S)mid x<n,|$. If $h$ is bounded, $h(n)<M$ for all $n$, this shows that $f(s)<M$ for all $sin S$, hence $f$ can be viewed as map $to0,ldots, M$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $ninBbb N$ let $m=min,kinBbb Nmid h(j)ge n,$, observe that then $min f(S)$, and define $g(n)=f^-1(m)$. This gives us an injection (in fact, bijection) $Bbb Nto S$.
edited 15 mins ago
Alon Amit
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
add a comment |Â
up vote
1
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Assuming $$# (s)<#(Bbb N)$$ we do not need choice, we just need to take the smallest element after the previous one, if it is finite it will end, if it is not finite than it will create bijective from $S$ to $Bbb N$, and thus $# (S)=# (Bbb N)$, this is contradiction thus $# (s)<#(Bbb N)$ implies $S$ is finite.
But assuming $$# (s)notge#(Bbb N)$$is not enough to show $S$ is finite without choice! There exists infinite finite-dedekind set which is not comparable to $Bbb N$! But if you assume the axiom us choice it is provable that those sets do not exists.
answered 18 mins ago
Holo
4,6922729
add a comment |Â
up vote
0
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I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.
answered 37 mins ago
alper akyuz
1608
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
By definition of $<$ for cardinalities, there exists an injection $fcolon Sto Bbb N$. For $ninBbb N$ let $h(n)=|,xin f(S)mid x<n,|$. If $h$ is bounded, $h(n)<M$ for all $n$, this shows that $f(s)<M$ for all $sin S$, hence $f$ can be viewed as map $to0,ldots, M$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $ninBbb N$ let $m=min,kinBbb Nmid h(j)ge n,$, observe that then $min f(S)$, and define $g(n)=f^-1(m)$. This gives us an injection (in fact, bijection) $Bbb Nto S$.
edited 15 mins ago
Alon Amit
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
add a comment |Â
up vote
3
down vote
By definition of $<$ for cardinalities, there exists an injection $fcolon Sto Bbb N$. For $ninBbb N$ let $h(n)=|,xin f(S)mid x<n,|$. If $h$ is bounded, $h(n)<M$ for all $n$, this shows that $f(s)<M$ for all $sin S$, hence $f$ can be viewed as map $to0,ldots, M$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $ninBbb N$ let $m=min,kinBbb Nmid h(j)ge n,$, observe that then $min f(S)$, and define $g(n)=f^-1(m)$. This gives us an injection (in fact, bijection) $Bbb Nto S$.
edited 15 mins ago
Alon Amit
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
add a comment |Â
up vote
3
down vote
up vote
3
down vote
By definition of $<$ for cardinalities, there exists an injection $fcolon Sto Bbb N$. For $ninBbb N$ let $h(n)=|,xin f(S)mid x<n,|$. If $h$ is bounded, $h(n)<M$ for all $n$, this shows that $f(s)<M$ for all $sin S$, hence $f$ can be viewed as map $to0,ldots, M$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $ninBbb N$ let $m=min,kinBbb Nmid h(j)ge n,$, observe that then $min f(S)$, and define $g(n)=f^-1(m)$. This gives us an injection (in fact, bijection) $Bbb Nto S$.
edited 15 mins ago
Alon Amit
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
By definition of $<$ for cardinalities, there exists an injection $fcolon Sto Bbb N$. For $ninBbb N$ let $h(n)=|,xin f(S)mid x<n,|$. If $h$ is bounded, $h(n)<M$ for all $n$, this shows that $f(s)<M$ for all $sin S$, hence $f$ can be viewed as map $to0,ldots, M$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $ninBbb N$ let $m=min,kinBbb Nmid h(j)ge n,$, observe that then $min f(S)$, and define $g(n)=f^-1(m)$. This gives us an injection (in fact, bijection) $Bbb Nto S$.
edited 15 mins ago
Alon Amit
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
edited 15 mins ago
Alon Amit
10.5k3766
edited 15 mins ago
Alon Amit
10.5k3766
edited 15 mins ago
Alon Amit
10.5k3766
10.5k3766
answered 43 mins ago
Hagen von Eitzen
267k21259482
answered 43 mins ago
Hagen von Eitzen
267k21259482
answered 43 mins ago
Hagen von Eitzen
267k21259482
267k21259482
add a comment |Â
add a comment |Â
up vote
1
down vote
Assuming $$# (s)<#(Bbb N)$$ we do not need choice, we just need to take the smallest element after the previous one, if it is finite it will end, if it is not finite than it will create bijective from $S$ to $Bbb N$, and thus $# (S)=# (Bbb N)$, this is contradiction thus $# (s)<#(Bbb N)$ implies $S$ is finite.
But assuming $$# (s)notge#(Bbb N)$$is not enough to show $S$ is finite without choice! There exists infinite finite-dedekind set which is not comparable to $Bbb N$! But if you assume the axiom us choice it is provable that those sets do not exists.
answered 18 mins ago
Holo
4,6922729
add a comment |Â
up vote
1
down vote
Assuming $$# (s)<#(Bbb N)$$ we do not need choice, we just need to take the smallest element after the previous one, if it is finite it will end, if it is not finite than it will create bijective from $S$ to $Bbb N$, and thus $# (S)=# (Bbb N)$, this is contradiction thus $# (s)<#(Bbb N)$ implies $S$ is finite.
But assuming $$# (s)notge#(Bbb N)$$is not enough to show $S$ is finite without choice! There exists infinite finite-dedekind set which is not comparable to $Bbb N$! But if you assume the axiom us choice it is provable that those sets do not exists.
answered 18 mins ago
Holo
4,6922729
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assuming $$# (s)<#(Bbb N)$$ we do not need choice, we just need to take the smallest element after the previous one, if it is finite it will end, if it is not finite than it will create bijective from $S$ to $Bbb N$, and thus $# (S)=# (Bbb N)$, this is contradiction thus $# (s)<#(Bbb N)$ implies $S$ is finite.
But assuming $$# (s)notge#(Bbb N)$$is not enough to show $S$ is finite without choice! There exists infinite finite-dedekind set which is not comparable to $Bbb N$! But if you assume the axiom us choice it is provable that those sets do not exists.
answered 18 mins ago
Holo
4,6922729
Assuming $$# (s)<#(Bbb N)$$ we do not need choice, we just need to take the smallest element after the previous one, if it is finite it will end, if it is not finite than it will create bijective from $S$ to $Bbb N$, and thus $# (S)=# (Bbb N)$, this is contradiction thus $# (s)<#(Bbb N)$ implies $S$ is finite.
But assuming $$# (s)notge#(Bbb N)$$is not enough to show $S$ is finite without choice! There exists infinite finite-dedekind set which is not comparable to $Bbb N$! But if you assume the axiom us choice it is provable that those sets do not exists.
answered 18 mins ago
Holo
4,6922729
edited 12 mins ago
answered 18 mins ago
Holo
4,6922729
answered 18 mins ago
Holo
4,6922729
answered 18 mins ago
Holo
4,6922729
4,6922729
add a comment |Â
add a comment |Â
up vote
0
down vote
I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.
answered 37 mins ago
alper akyuz
1608
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
add a comment |Â
up vote
0
down vote
I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.
answered 37 mins ago
alper akyuz
1608
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.
answered 37 mins ago
alper akyuz
1608
I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.
answered 37 mins ago
alper akyuz
1608
answered 37 mins ago
alper akyuz
1608
answered 37 mins ago
alper akyuz
1608
answered 37 mins ago
alper akyuz
1608
1608
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
add a comment |Â
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
1
1
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
That is exactly the proof that requires the axiom of choice.
– Asaf Karagila♦
27 mins ago
add a comment |Â
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