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Claim: If $(X,\tau)$ is a topological space, $\mathcal B$ a base for $\tau$ and $\mathcal U$ an open cover of $X$ then there is a subcover $\mathcal V \subset \mathcal U$ whose cardinality is not larger than that of $\mathcal B$.

I appreciate some hints, since since is homework. The idea is to remove redundand open sets in $\mathcal U$. My first idea was to remove all $U \in \mathcal U$ which are subeset of the union of other elements of $\mathcal U$, but I failed to show that the remaining subset of $\mathcal U$ still covers $X$.

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  • $\begingroup$ Partition the base $\mathcal{B}$ into those sets that are contained in some $U\in\mathcal{U}$ and those that aren't. $\endgroup$ Nov 19, 2013 at 11:31

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Hint: Let $\mathcal{B}^\prime = \{ W \in \mathcal{B} : ( \exists U \in \mathcal{U} ) ( W \subseteq U ) \}$. Now associate to each $W \in \mathcal{B}^\prime$ one element of $\mathcal{U}$ in an appropriate fashion.

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    $\begingroup$ (I guess this is essentially D. Fischer's comment above. I'm just A. Fischer.$\;\;$;-)$\;\;$) $\endgroup$
    – user642796
    Nov 19, 2013 at 11:33
  • $\begingroup$ What will be $\mathcal V$ then ? $\endgroup$
    – user42761
    Nov 19, 2013 at 11:35
  • $\begingroup$ @André: $\mathcal{V}$ will be the set of all of the elements of $\mathcal{U}$ associated with some $W \in \mathcal{B}^\prime$. $\endgroup$
    – user642796
    Nov 19, 2013 at 11:36
  • $\begingroup$ Using AC we can make a function $f: \mathcal B' \to \mathcal U$ s.t. $W \subset f(W)$, but we need an injection $\mathcal V \to \mathcal B'$ where $\mathcal V$ should be $f(\mathcal B')$ as I understand it right. $\endgroup$
    – user42761
    Nov 19, 2013 at 11:50
  • $\begingroup$ @André: You have the right function $f$. There is no need to have an injection $\mathcal{V} \to \mathcal{B}^\prime$. If you're worried about the cardinalities being properly related, recall that since $f$ maps $\mathcal{B}^\prime$ onto $\mathcal{V}$ this tells us something about the cardinalities of these two sets. (Use all the Choice you want.) (Technically, you choose for each $U \in \mathcal{V}$ some $W_U \in \mathcal{B}^\prime$ with $f( W_U ) = U$, and let $\mathcal{B}^{\prime\prime} = \{ W_U : U \in \mathcal{V} \}$, and then $f \restriction \mathcal{B}^{\prime\prime}$ is a bijection.) $\endgroup$
    – user642796
    Nov 19, 2013 at 11:58

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