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(Definition)
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To get some sense of what this means, observe that for any
,
, so the set of
is stationary (in ). More strongly, suppose
. Then any subset of
is bounded in so
on a stationary set. Since
, it follows that
. Hence
, the most common form (often written as just ), implies CH.
C. Akemann and N. Weaver used to construct a -algebra serving as a counterexample to Naimark's problem.
- 1
- Akemann, C., and N. Weaver, Consistency of a counterexample to Naimark's problem. Preprint available on the arXiv at http://arxiv.org/abs/math.OA/0312135.
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" " is owned by Henry. [ full author list (2) | owner history (2) ]
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(view preamble)
Cross-references: counterexample, CH, implies, bounded, subset, stationary, sequence, combinatorial principle, stationary set
There are 5 references to this entry.
This is version 5 of , born on 2002-07-31, modified 2004-04-10.
Object id is 3245, canonical name is Diamond.
Accessed 4281 times total.
Classification:
| AMS MSC: | 03E65 (Mathematical logic and foundations :: Set theory :: Other hypotheses and axioms) |
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Pending Errata and Addenda
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