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Solovay's Theorem on stationary sets states that any stationary subset of a regular uncountable cardinal $\kappa$ is the disjoint union of $\kappa$ stationary subsets.

In Jech's "Set Theory", it is proved that any stationary subset of $E_{\lambda}^{\kappa}=\{\alpha<\kappa:\operatorname{cf}\alpha=\lambda\}$, for any regular $\lambda<\kappa$, can be decomposed into $\kappa$ disjoint stationary sets, and the same is proved for any stationary subset of $\{\alpha<\kappa:\operatorname{cf}\alpha<\alpha\}$.

Also, the author shows that if $S$ is a stationary subset of $\kappa$ consisting only of regular cardinals, then $\{\alpha\in S: S\cap \alpha$ is not a stationary subset of $\alpha\}$ is a stattionary set.

When the author begins the proof of Solovay's Theorem he states "Let $A$ be a stationary subset of $\kappa$, then we may assume that the set $W$ of all $\alpha\in A$ such that $\alpha$ is a regular cardinal and $A\cap \alpha$ is not stationary in $\alpha$, is stationary", and he says this is because of the facts I have written above, but I don't see why we can make such assumption, any help will be appreciated.

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He’s not actually saying that $W$ itself has that property. However, $W$ has a stationary subset with that property, and a decomposition of a subset $S$ of $W$ into $\kappa$ pairwise disjoint stationary sets is easily extended to such a decomposition of $W$: just throw $W\setminus S$ into one of the stationary pieces.

Let $V=\{\alpha\in W:\operatorname{cf}\alpha=\alpha\}$. $V$ is a set of regular cardinals, so if $V$ is stationary, then $$\{\alpha\in V:V\cap\alpha\text{ is not stationary in }\alpha\}$$ is stationary, and we may substitute it for $W$, as explained above.

Suppose, then, that $V$ is not stationary. $W$ is stationary, so $W\setminus V$ must be stationary as well. But $W\setminus V=\{\alpha\in W:\operatorname{cf}\alpha<\alpha\}$, so it’s a stationary subset of $\{\alpha<\kappa:\operatorname{cf}\alpha<\alpha\}$, and you already know that it can be decomposed into $\kappa$ pairwise disjoint stationary sets.

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a few moments ago I thought of the same argument; namely, if $V$ is not stationary, is $W-V$ stationary?, but what if there is some closed unbounded set $C$ such that $C\cap V\neq \emptyset$ but $C\cap (W-V)=\emptyset$, this does not contradict the fact that $W$ is stationary –  Camilo Arosemena Jan 4 '13 at 20:08
    
@Camilo: There’s no contradiction: if $C\cap V\ne\varnothing$, then $C\cap W\ne\varnothing$, since $V\subseteq W$. –  Brian M. Scott Jan 4 '13 at 20:14
    
@Camilo: If $V$ is non-stationary, then $\kappa\setminus V$ contains some cub $C$. $W$ is stationary, so $W\setminus V\supseteq W\cap C$ is stationary: the intersection of a stationary set with a cub is always stationary. –  Brian M. Scott Jan 4 '13 at 20:16
    
@Camilo: I just proved that it can’t happen. –  Brian M. Scott Jan 4 '13 at 20:18
    
you're right!, thank you very much for your answer :) –  Camilo Arosemena Jan 4 '13 at 20:20

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