Mixing
bijections and order types
Clash Royale CLAN TAG#URR8PPP
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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 $x subseteqkappa$, $mathrmot(x) leq mathrmot(f[x])$? (Here, $mathrmot(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked 2 hours ago
Monroe Eskew
7,45012056
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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 $x subseteqkappa$, $mathrmot(x) leq mathrmot(f[x])$? (Here, $mathrmot(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked 2 hours ago
Monroe Eskew
7,45012056
add a comment |Â
up vote
1
down vote
favorite
up vote
1
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 $x subseteqkappa$, $mathrmot(x) leq mathrmot(f[x])$? (Here, $mathrmot(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
asked 2 hours ago
Monroe Eskew
7,45012056
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 $x subseteqkappa$, $mathrmot(x) leq mathrmot(f[x])$? (Here, $mathrmot(y)$ is the order type of a set of ordinals $y$.)
set-theory lo.logic
set-theory lo.logic
asked 2 hours ago
Monroe Eskew
7,45012056
asked 2 hours ago
Monroe Eskew
7,45012056
asked 2 hours ago
Monroe Eskew
7,45012056
asked 2 hours ago
Monroe Eskew
7,45012056
asked 2 hours ago
Monroe Eskew
7,45012056
7,45012056
add a comment |Â
add a comment |Â
1 Answer
1
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4
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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 $textot(Xcup alpha)>omega$, but $textot(f(Xcup alpha)) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered 1 hour ago
Alex Kruckman
1,41411012
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
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 $textot(Xcup alpha)>omega$, but $textot(f(Xcup alpha)) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered 1 hour ago
Alex Kruckman
1,41411012
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
add a comment |Â
up vote
4
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 $textot(Xcup alpha)>omega$, but $textot(f(Xcup alpha)) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered 1 hour ago
Alex Kruckman
1,41411012
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
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 $textot(Xcup alpha)>omega$, but $textot(f(Xcup alpha)) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered 1 hour ago
Alex Kruckman
1,41411012
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 $textot(Xcup alpha)>omega$, but $textot(f(Xcup alpha)) = omega$.
I'm not sure about the case when $kappa$ is uncountable with cofinality $omega$.
answered 1 hour ago
Alex Kruckman
1,41411012
answered 1 hour ago
Alex Kruckman
1,41411012
answered 1 hour ago
Alex Kruckman
1,41411012
answered 1 hour ago
Alex Kruckman
1,41411012
1,41411012
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
add a comment |Â
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
2
2
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
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 $Xcupalpha$ has order-type $omega_1+1$ while its image has order-type only $omega_1$, just as in your proof.
– Andreas Blass
40 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
@AndreasBlass Ah great, thanks for supplying the argument.
– Alex Kruckman
38 mins ago
add a comment |Â
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