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)?
> 
> 
> 



Dual covering theorem
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.


