bijections and order types
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Suppose $kappa$ is an infinite cardinal and $alpha$ is an ordinal of cardinality $kappa$. Is it possible to find a bijection $f : kappa to alpha$ such that for all $x subseteqkappa$, $mathrm{ot}(x) leq mathrm{ot}(f[x])$? (Here, $mathrm{ot}(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
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Suppose $kappa$ is an infinite cardinal and $alpha$ is an ordinal of cardinality $kappa$. Is it possible to find a bijection $f : kappa to alpha$ such that for all $x subseteqkappa$, $mathrm{ot}(x) leq mathrm{ot}(f[x])$? (Here, $mathrm{ot}(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose $kappa$ is an infinite cardinal and $alpha$ is an ordinal of cardinality $kappa$. Is it possible to find a bijection $f : kappa to alpha$ such that for all $x subseteqkappa$, $mathrm{ot}(x) leq mathrm{ot}(f[x])$? (Here, $mathrm{ot}(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
Suppose $kappa$ is an infinite cardinal and $alpha$ is an ordinal of cardinality $kappa$. Is it possible to find a bijection $f : kappa to alpha$ such that for all $x subseteqkappa$, $mathrm{ot}(x) leq mathrm{ot}(f[x])$? (Here, $mathrm{ot}(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
set-theory lo.logic
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
asked Nov 8 at 1:24
Monroe Eskew
7,51512057
7,51512057
add a comment |
add a comment |
1 Answer
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9
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When $kappa = aleph_0$, any bijection works.
When $kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $fcolon kappato kappa+omega$. Indeed, let $X = f^{-1}((kappa+omega)setminus kappa)$. Then $X$ is not cofinal in $kappa$, so we can pick some $alphain kappa$ greater than every element of $X$, and $f(alpha)<kappa$. Then $text{ot}(Xcup {alpha})>omega$, but $text{ot}(f(Xcup {alpha})) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
When $kappa = aleph_0$, any bijection works.
When $kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $fcolon kappato kappa+omega$. Indeed, let $X = f^{-1}((kappa+omega)setminus kappa)$. Then $X$ is not cofinal in $kappa$, so we can pick some $alphain kappa$ greater than every element of $X$, and $f(alpha)<kappa$. Then $text{ot}(Xcup {alpha})>omega$, but $text{ot}(f(Xcup {alpha})) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
add a comment |
up vote
9
down vote
accepted
When $kappa = aleph_0$, any bijection works.
When $kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $fcolon kappato kappa+omega$. Indeed, let $X = f^{-1}((kappa+omega)setminus kappa)$. Then $X$ is not cofinal in $kappa$, so we can pick some $alphain kappa$ greater than every element of $X$, and $f(alpha)<kappa$. Then $text{ot}(Xcup {alpha})>omega$, but $text{ot}(f(Xcup {alpha})) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
When $kappa = aleph_0$, any bijection works.
When $kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $fcolon kappato kappa+omega$. Indeed, let $X = f^{-1}((kappa+omega)setminus kappa)$. Then $X$ is not cofinal in $kappa$, so we can pick some $alphain kappa$ greater than every element of $X$, and $f(alpha)<kappa$. Then $text{ot}(Xcup {alpha})>omega$, but $text{ot}(f(Xcup {alpha})) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
When $kappa = aleph_0$, any bijection works.
When $kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $fcolon kappato kappa+omega$. Indeed, let $X = f^{-1}((kappa+omega)setminus kappa)$. Then $X$ is not cofinal in $kappa$, so we can pick some $alphain kappa$ greater than every element of $X$, and $f(alpha)<kappa$. Then $text{ot}(Xcup {alpha})>omega$, but $text{ot}(f(Xcup {alpha})) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
answered Nov 8 at 2:11
Alex Kruckman
1,46411012
1,46411012
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
add a comment |
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
6
6
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
I think you can proceed similarly when $kappa$ has countable cofinality, using $kappa+omega_1$ instead of $kappa+omega$. The preimage $X$ of the final $omega_1$ must have order-type $leqomega_1$ by the condition in the problem, but it can't be $<omega_1$ by cardinality. So $X$ has order-type exactly $omega_1$, and then $X$ can't be cofinal in $kappa$. Then pick a larger element $alpha<kappa$ and notice that $Xcup{alpha}$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
Nov 8 at 2:58
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
Nov 8 at 3:00
add a comment |
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