Re: Subsets of cardinals in a well-ordering
- From: Robert Sheskey <rsheskey@xxxxxxxxxx>
- Date: Fri, 25 May 2007 15:50:03 -0700
In article <1179945153.082475.161790@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, justinpa84@xxxxxxxxx says...
(from Kunen - exercise 19) Let kappa be an infinite cardinal, let W
well-order kappa. Prove there's a subset X of kappa with cardinality
kappa on which W agrees with the usual ordering of ordinals.
One way to rephrase this is that any kappa length sequence of distinct
ordinals less than kappa has an increasing subsequence of length
kappa.
I can prove it for regular cardinals easily enough, but I'm pretty
much stumped on proving it for the more general case (although I've
reduced it to several statements trying some other strategies, but I
think these are deadends).
I'll suggest an argument for the special case kappa = aleph_omega.
I think it's adaptable to the general case, though I haven't looked
at the details.
Assume the result is true for regular cardinals. Let
kappa = aleph_omega and suppose f:kappa -> kappa is a permutation.
We want to show that there is a subset B of kappa such that
|B| = kappa and f|B is order-preserving.
Let X_0 = aleph_0, X_{n+1} = aleph_{n+1} \ aleph_n,
and A(n,k) = {eta : eta in X_n and f(eta) in X_k }
Note that |A(n,k)| <= aleph_k and X_n = Union_k A(n,k).
Since aleph_n = |X_n| is regular, we must have at least one k such
that A(n,k) has power aleph_n. Let g(n) be that k, that is, g is
a function omega -> omega such that |A(n,g(n))| = aleph_n.
g is unbounded, so there is an infinite subset J of omega on which
g is order-preserving. Note that if m<n, both m,n in J, then for all
eta in A(m,g(m)), xi in A(n,g(n)), we have eta < xi and f(eta) < f(xi).
Now apply the result for regular cardinals to get a subset B_n of
A(n,g(n)) such that |B_n| = aleph_n, and f|B_n is order-preserving.
Finally, piece together the B_n's to get B.
Robert Sheskey
.
- References:
- Subsets of cardinals in a well-ordering
- From: Justin Palumbo
- Subsets of cardinals in a well-ordering
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