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Science Forum Index » Logic Forum » A large-cardinal axiom
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| Rupert |
 Posted: Sun Jan 11, 2004 11:31 am |
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Make the convention that filters over a cardinal kappa include all
sets of the form (alpha: lambda<=alpha<kappa) for all lambda. If X is
a subset of kappa, define H_0(X)=(alpha: X intersect alpha is
stationary in alpha), H_alpha+1(X)=(beta: there is a normal filter on
beta containing X intersect beta closed under H_alpha),
H_beta(X)=intersection for all alpha<beta of H_alpha(X) for limit
ordinals beta. Call kappa alpha-super-Mahlo if there is a normal
filter over kappa containing the set of inaccessibles less than kappa
closed under H_alpha. Call kappa enormous if kappa is
kappa-super-Mahlo.
I would be interested in knowing how these large-cardinal properties
relate to the other large-cardinal properties. I don't see how to
prove any of the "standard" large cardinals even dominate a
2-super-Mahlo cardinal. |
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| Robert E. Beaudoin |
| Posted: Tue Jan 13, 2004 7:59 pm |
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Rupert wrote:
Quote: Make the convention that filters over a cardinal kappa include all
sets of the form (alpha: lambda<=alpha<kappa) for all lambda. If X is
a subset of kappa, define H_0(X)=(alpha: X intersect alpha is
stationary in alpha), H_alpha+1(X)=(beta: there is a normal filter on
beta containing X intersect beta closed under H_alpha),
H_beta(X)=intersection for all alpha<beta of H_alpha(X) for limit
ordinals beta. Call kappa alpha-super-Mahlo if there is a normal
filter over kappa containing the set of inaccessibles less than kappa
closed under H_alpha. Call kappa enormous if kappa is
kappa-super-Mahlo.
I would be interested in knowing how these large-cardinal properties
relate to the other large-cardinal properties. I don't see how to
prove any of the "standard" large cardinals even dominate a
2-super-Mahlo cardinal.
I assume you want H_alpha(X) to contain only ordinals beta less
than sup(X). Otherwise H_alpha(beta) = beta + 1 for any
beta < kappa, making it impossible for any normal filter on beta
to be closed under H_alpha. Also I assume you want all the filters
you mention to be proper filters. Then there are no enormous
cardinals. For suppose kappa is the least enormous cardinal, and
let I be the set of inaccessibles less than kappa. Then one can
show, by induction on alpha < kappa, that alpha is disjoint from
H_alpha(I), whence H_kappa(I) is empty. The induction is easy
when alpha is not a successor. On the other hand, were beta to be
a member of both alpha + 1 and H_alpha+1(I), then by the definition
of H_alpha+1 beta would be enormous, contradicting minimality of
kappa.
If you back off slightly from your definition of enormous you get
something attainable: If kappa is a (kappa+2)-strong cardinal then
for all alpha < kappa kappa is alpha-super-Mahlo. Fix an elementary
j:V-->M with critical point kappa such that R(kappa+2) is a subset of
M. (That's the definition of (kappa+2)-strongness.) Then let U be
the usual normal measure on kappa: a subset x of kappa is in U
if and only if kappa is an element of j(x). By strongness U belongs
to M, the M-powerset of kappa is the same as the V-powerset of kappa,
and for every alpha < kappa the restriction of j(H_alpha) to the
powerset of kappa is exactly H_alpha. It then follows by induction on
alpha < kappa that U is closed under H_alpha. Limit stages are easy
as U is kappa-complete. For the successor stage alpha + 1, suppose
x is in U; then kappa is an element of j(x). Let y = H_alpha+1(x).
Now the intersection of j(x) with kappa is x, U lies in M, and
j(H_alpha) restricted to P(kappa) is H_alpha; it follows that kappa
belongs to j(y), i.e. y is in U. So kappa is alpha-super-Mahlo
for all alpha < kappa, via the same normal filter U.
Robert E. Beaudoin |
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