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If $X$ is normal, it is well known that for any open-covering $(U_i)$ of $X$, there exist closed subspaces $F_i$ and $G_i$ and an open subspaces $O_i$ such that $$F_i\subset O_i\subset G_i\subset U_i\text{ with }X=\bigcup F_i \quad (*)$$ Let $d\geq 1$. A covering $(U_i)$ of $X$ is called $d$-covering if any $x\in X$ is contained at least in $d$ sets among $U_i$.

My question is: For any open-covering $(U_i)$, one can find (or build) a $d$-covering $(F_i)$ such that (*) still true?

Thanks for your answers, any comments or/and references are welcome

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  • $\begingroup$ Say $d=2$ while the cover consist of a single element, $X$, that is $(U_i)=\{X\}$. In this case we could take $F=O=G=X$. Unless you count $F$ more than once, the covering $\{F\}$ would not be a $d$-covering, if I understand your definition. $\endgroup$
    – Mirko
    Commented Oct 27, 2014 at 3:15
  • $\begingroup$ I have to add that $d\geq n$ where $n$ denotes the number of initial open sets $U_i$ $\endgroup$
    – MyIsmail
    Commented Oct 27, 2014 at 12:12

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