Subsets of cardinals in a well-ordering

(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).

.

Relevant Pages

  • Re: New Physics Theory
    ... Are there any actual physicists here? ... people claiming to have proven that relativity doesn't exist, ... | frame exist; the other frame has an elevated gamma, ... While the special relativity formula ...
    (sci.physics)
  • Re: Androcles asks for Derivation of LT
    ... >>Sam and Joe are at rest relative to frame B. ... >>Sally and Jane are at rest relative to frame G. ... > So our equations so far work for Galilean relativity, ... Not real physicists, mind you. ...
    (sci.physics)
  • Re: Dilworths Theorem
    ... A covering can become a partition by ... that the cardinalities of antichains may not have a supremum, ... also works if kappa is a singular cardinal. ... Thus some cardinality bound less than all the singletons, ...
    (sci.math)
  • Re: Finitely Valid Formula
    ... Let K be a set of cardinality kappa with neither 0 nor 1. ... A lattice is an ordered set for which every two elements have an infinum ...
    (sci.logic)
  • Re: Continuum Hypothesis Query
    ... > I thought every cardinal was an equivalence class of ordinals. ... > For example, w and w+1 are in the same equivalence class, denoted Aleph_0. ... problem that the cardinality of your Aleph_0 is Aleph_1, ... Now if kappa is a cardinal, then the cardinality of kappa is ...
    (sci.math)