For two ordinals, $|A|nleq |B|$, can one conclude $|A|>|B|$?
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For two ordinals, $|A|nleq |B|$, can one conclude $|A|>|B|$?
Notice the $<$ here represent the cardinal order.
I was thinking use the linear order relationship of $A$ and $B$ in ordinal, but I'm not sure if it's possible without the discussion of AC.
elementary-set-theory
asked Nov 29 at 3:31
user9976437
608
add a comment |
up vote
1
down vote
favorite
For two ordinals, $|A|nleq |B|$, can one conclude $|A|>|B|$?
Notice the $<$ here represent the cardinal order.
I was thinking use the linear order relationship of $A$ and $B$ in ordinal, but I'm not sure if it's possible without the discussion of AC.
elementary-set-theory
asked Nov 29 at 3:31
user9976437
608
3
Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
1
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For two ordinals, $|A|nleq |B|$, can one conclude $|A|>|B|$?
Notice the $<$ here represent the cardinal order.
I was thinking use the linear order relationship of $A$ and $B$ in ordinal, but I'm not sure if it's possible without the discussion of AC.
elementary-set-theory
asked Nov 29 at 3:31
user9976437
608
For two ordinals, $|A|nleq |B|$, can one conclude $|A|>|B|$?
Notice the $<$ here represent the cardinal order.
I was thinking use the linear order relationship of $A$ and $B$ in ordinal, but I'm not sure if it's possible without the discussion of AC.
elementary-set-theory
elementary-set-theory
asked Nov 29 at 3:31
user9976437
608
asked Nov 29 at 3:31
user9976437
608
asked Nov 29 at 3:31
user9976437
608
asked Nov 29 at 3:31
user9976437
608
asked Nov 29 at 3:31
user9976437
608
608
3
Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
1
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13
add a comment |
3
Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
1
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13
3
3
Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
1
1
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Let me rephrase your question first: Fix ordinals $alpha$ and $beta$. Assume there does not exist an injection from $alpha$ to $beta$. Can one conclude that there exists an injection from $beta$ to $alpha$?
The answer is yes. And as the comments have pointed out the axiom of choice is not required here.
An argument can take advantage of the well-order structure of $alpha$ and $beta$ to construct the desired injection from $beta$ to $alpha$.
Via recursion define $f:betatoalpha:ximapstobegin{cases}
min{(alphasetminus{f(delta):delta<xi})}&text{if }(alphasetminus{f(delta):delta<xi})nevarnothing\
z&text{else}
end{cases}$
where $z$ is some dummy set which isn't an ordinal. One can show that $f$ never takes the value $z$ because of the assumption that there does not exist an injection from $alpha$ to $beta$. Because $znotinmathrm{range}(f)$, it follows that $f$ is an order-preserving injection as desired.
answered Nov 30 at 8:07
Alberto Takase
1,738414
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let me rephrase your question first: Fix ordinals $alpha$ and $beta$. Assume there does not exist an injection from $alpha$ to $beta$. Can one conclude that there exists an injection from $beta$ to $alpha$?
The answer is yes. And as the comments have pointed out the axiom of choice is not required here.
An argument can take advantage of the well-order structure of $alpha$ and $beta$ to construct the desired injection from $beta$ to $alpha$.
Via recursion define $f:betatoalpha:ximapstobegin{cases}
min{(alphasetminus{f(delta):delta<xi})}&text{if }(alphasetminus{f(delta):delta<xi})nevarnothing\
z&text{else}
end{cases}$
where $z$ is some dummy set which isn't an ordinal. One can show that $f$ never takes the value $z$ because of the assumption that there does not exist an injection from $alpha$ to $beta$. Because $znotinmathrm{range}(f)$, it follows that $f$ is an order-preserving injection as desired.
answered Nov 30 at 8:07
Alberto Takase
1,738414
add a comment |
up vote
0
down vote
accepted
Let me rephrase your question first: Fix ordinals $alpha$ and $beta$. Assume there does not exist an injection from $alpha$ to $beta$. Can one conclude that there exists an injection from $beta$ to $alpha$?
The answer is yes. And as the comments have pointed out the axiom of choice is not required here.
An argument can take advantage of the well-order structure of $alpha$ and $beta$ to construct the desired injection from $beta$ to $alpha$.
Via recursion define $f:betatoalpha:ximapstobegin{cases}
min{(alphasetminus{f(delta):delta<xi})}&text{if }(alphasetminus{f(delta):delta<xi})nevarnothing\
z&text{else}
end{cases}$
where $z$ is some dummy set which isn't an ordinal. One can show that $f$ never takes the value $z$ because of the assumption that there does not exist an injection from $alpha$ to $beta$. Because $znotinmathrm{range}(f)$, it follows that $f$ is an order-preserving injection as desired.
answered Nov 30 at 8:07
Alberto Takase
1,738414
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let me rephrase your question first: Fix ordinals $alpha$ and $beta$. Assume there does not exist an injection from $alpha$ to $beta$. Can one conclude that there exists an injection from $beta$ to $alpha$?
The answer is yes. And as the comments have pointed out the axiom of choice is not required here.
An argument can take advantage of the well-order structure of $alpha$ and $beta$ to construct the desired injection from $beta$ to $alpha$.
Via recursion define $f:betatoalpha:ximapstobegin{cases}
min{(alphasetminus{f(delta):delta<xi})}&text{if }(alphasetminus{f(delta):delta<xi})nevarnothing\
z&text{else}
end{cases}$
where $z$ is some dummy set which isn't an ordinal. One can show that $f$ never takes the value $z$ because of the assumption that there does not exist an injection from $alpha$ to $beta$. Because $znotinmathrm{range}(f)$, it follows that $f$ is an order-preserving injection as desired.
answered Nov 30 at 8:07
Alberto Takase
1,738414
Let me rephrase your question first: Fix ordinals $alpha$ and $beta$. Assume there does not exist an injection from $alpha$ to $beta$. Can one conclude that there exists an injection from $beta$ to $alpha$?
The answer is yes. And as the comments have pointed out the axiom of choice is not required here.
An argument can take advantage of the well-order structure of $alpha$ and $beta$ to construct the desired injection from $beta$ to $alpha$.
Via recursion define $f:betatoalpha:ximapstobegin{cases}
min{(alphasetminus{f(delta):delta<xi})}&text{if }(alphasetminus{f(delta):delta<xi})nevarnothing\
z&text{else}
end{cases}$
where $z$ is some dummy set which isn't an ordinal. One can show that $f$ never takes the value $z$ because of the assumption that there does not exist an injection from $alpha$ to $beta$. Because $znotinmathrm{range}(f)$, it follows that $f$ is an order-preserving injection as desired.
answered Nov 30 at 8:07
Alberto Takase
1,738414
answered Nov 30 at 8:07
Alberto Takase
1,738414
answered Nov 30 at 8:07
Alberto Takase
1,738414
answered Nov 30 at 8:07
Alberto Takase
1,738414
1,738414
add a comment |
add a comment |
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Yes, and this does not require any form of choice.
– Andrés E. Caicedo
Nov 29 at 3:50
1
Do you know that, for any two well-ordered sets, there is an isomorphism between one of them and an initial segment of the other?
– bof
Nov 29 at 6:13