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Math Help - open cover

  1. #1
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    open cover

    Suppose that a metric space  X has a countable dense cover  Q . Show that any open cover of  X has a subcover which is at most countable.

    So consider an arbitrary cover of  X , \{U_{\alpha}: \alpha \in A \} . Let  \mathcal{B} be the set of open balls  B_{r}(q) where  r \in \mathbb{Q} and  q \in Q . Let  E \subset \mathcal{B} be the set of open balls such that  U_{\alpha} is contained in one of the  \{U_{\alpha} \} . From here can we say the these union of open balls is  X because of the denseness of  Q ? From here how do we show that the subcover is either finite or countable?
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  2. #2
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    I'm not very good in English, but I don't think Q is a cover...

    I think you mean X is separable, and Q is countable and dense in X... I don't know if there is a english name for this kind of set...

    First of all, you should note \mathcal{B} is countable:
    |\mathcal{B}| = |Q \times \mathbb{Q}| = \aleph_0 \aleph_0 = \aleph_0

    Secondly E is not a subcover of \{U_{\alpha}: \alpha \in A \}.

    Thirdly, the idea is good, and E covers X:
    Let x \in X. Then x \in U_\alpha for some \alpha. U_\alpha is neighbourhood hence there is the ball B_{r}(x) \subset U_\alpha. Q is dense in X so there is q \in Q such that d(x, q) < r/3. There exists ball B_\epsilon(q) \in \mathcal{B} where r/3< \epsilon < r/2 . Thus x \in B_\epsilon(q), and B_\epsilon(q) \subset B_r(x).

    Of course B_\epsilon(q) \in E. Hence E (and \mathcal{B} too) is cover of X.
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