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I'm having troubles solving this problem, any help will be appreciate :)

Let $X$ be a compact metric space and for all open cover $U$ of $X$ we denote $N(U)=min\{|V| : V $subcover of $ U\}$.

Let $\alpha$ and $\beta$ be two finite partition of $X$ and $U$ a finite open cover of $X$ such that $\alpha=\beta \vee U=\{A\cap B: A\in \beta,  B\in U\}$ prove that $\forall B\in\beta, \quad |\{A\in \alpha: A\subseteq B\}|\leq N(U)$

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For $\alpha$ to be a partition it's necessary that $U$ be also a partition. (Otherwise if $U_i\cap U_j\neq\varnothing$ then for some $A\in\beta$ we have $U_i\cap U_j\cap A\neq\varnothing$ and therefore $(U_i\cap A)\cap (U_j\cap A)\neq\varnothing$). Then $N(U)=|U|$. Fixing $B\in\beta$, we also have that $|\{A\in\alpha : A\subseteq B\}|=|\{j : U_j\cap B\neq \varnothing\}|$, which is trivially less than $|U|$.

Personal comment: I really find this exercise strange. We do not use the hypothesis that $X$ is even a topological space. Maybe I am missing something.

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