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Jensen's covering theorem states that if $0^\sharp$ doesn't exist, then every uncountable set of ordinals can be covered by a constructible set of the same cardinality.

Now consider the following (somewhat) dual statements:

  1. Every uncountable set of ordinals covers a constructible uncountable set of ordinals.
  2. Every uncountable set of ordinals covers a constructible set of ordinals of the same cardinality.

I have two questions:

  1. Does any of the above statements follow from the non-existence of $0^\sharp$?
  2. If the answer is "no", are they still known to be consistent (in order to avoid trivialities, we may assume that V=L doesn't hold)?
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2 Answers 2

up vote 5 down vote accepted

The answer to the first question is no: You can add a club subset of a stationary subset of $\omega_1$ by forcing. The closure of any uncountable subset of the generic club is a club contained in the stationary set, so if we begin in $L$ with a stationary-costationary subset of $\omega_1$, and add a club through it, the extension contradicts both statements.

As for your second question, add a Cohen real to $L$. Suppose $A\subseteq{\rm ORD}$ and $|A|$ has uncountable cofinality in the extension. Since Cohen's forcing is countable, there is a condition $p$ that is in the generic and such that $A_p=\{\alpha\mid p$ forces $\alpha\in\dot A\}$ has the same size as $A$. Note that this is a constructible subset of $A$ in the ground model.

In fact, the same holds for uncountable sets of cofinality $\omega$: Suppose now that $A$ has countable cofinality, let $\alpha$ be its supremum, let $\gamma$ be least such that $A\cap\gamma$ is uncountable. For each $\beta$ between $\gamma$ and $\alpha$ there is a $p_\beta$ in the generic that decides a subset of $A\cap\beta$ that is in $L$ and has size $|A\cap\beta|$. There must be a $p$ that appears as $p_\beta$ unboundedly often, and $A_p$ is as wanted.

In summary: The extension of $L$ obtained by adding a Cohen real is a model of your two statements.

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Thanks, Andres. :) –  Haim Jul 31 '11 at 14:57
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Perhaps Magidor's covering lemma may be mentioned: if $0^\sharp$ does not exist and $A$ is a set of ordinals which is closed under the primitive recursive set functions, then $A$ is the union of countably many constructible sets. M. Magidor: Representing sets of ordinals..., Transactions of the AMS, 317(1990), 91-126.

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