Function on the set of limit countable ordinalsA question about $dotS^Q$-semiproperness and revised countable support iterated forcing of length a limit ordinalCofinality of countable ordinals in ZF, and in toposesOn a characterization of the recursively inaccessible ordinalsMemorable ordinalsProblem with definability in the constructible hierarchySplitting of ordinals of oscillation ranks of a Baire $1$ functionDo these ordinals exist?Connection between countable ordinals and Turing degreesFormalizations of The Matchstick Diagram Representation of OrdinalsIs the smallest $L_alpha$ with undefinable ordinals always countable?
Function on the set of limit countable ordinals
A question about $dotS^Q$-semiproperness and revised countable support iterated forcing of length a limit ordinalCofinality of countable ordinals in ZF, and in toposesOn a characterization of the recursively inaccessible ordinalsMemorable ordinalsProblem with definability in the constructible hierarchySplitting of ordinals of oscillation ranks of a Baire $1$ functionDo these ordinals exist?Connection between countable ordinals and Turing degreesFormalizations of The Matchstick Diagram Representation of OrdinalsIs the smallest $L_alpha$ with undefinable ordinals always countable?
$begingroup$
Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:
- $forall lambdainLambda:~f(lambda)<lambda$
$forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?
set-theory ordinal-numbers
asked Apr 24 at 6:13
ar.grigar.grig
1
$endgroup$
add a comment |
$begingroup$
Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:
- $forall lambdainLambda:~f(lambda)<lambda$
$forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?
set-theory ordinal-numbers
asked Apr 24 at 6:13
ar.grigar.grig
1
$endgroup$
5
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25
add a comment |
$begingroup$
Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:
- $forall lambdainLambda:~f(lambda)<lambda$
$forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?
set-theory ordinal-numbers
asked Apr 24 at 6:13
ar.grigar.grig
1
$endgroup$
Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:
- $forall lambdainLambda:~f(lambda)<lambda$
$forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?
set-theory ordinal-numbers
set-theory ordinal-numbers
asked Apr 24 at 6:13
ar.grigar.grig
1
asked Apr 24 at 6:13
ar.grigar.grig
1
asked Apr 24 at 6:13
ar.grigar.grig
1
asked Apr 24 at 6:13
ar.grigar.grig
1
asked Apr 24 at 6:13
ar.grigar.grig
1
1
5
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25
add a comment |
5
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25
5
5
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
$endgroup$
add a comment |
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1 Answer
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$begingroup$
No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
$endgroup$
add a comment |
$begingroup$
No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
$endgroup$
add a comment |
$begingroup$
No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
$endgroup$
No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
answered Apr 24 at 6:54
Monroe EskewMonroe Eskew
8,03922565
8,03922565
add a comment |
add a comment |
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5
$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25