Cover of a metric space.

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Let $E$ be a separable and complete metric space. Let $epsilon > 0.$ I want to find a cover of $E$ consisting of balls $B(q_n, epsilon), n in mathbbN, q_n in E,$ such that every $e in E$ is contained only in finitely many balls $B(q_n,epsilon).$ Is this possible?







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    Let $E$ be a separable and complete metric space. Let $epsilon > 0.$ I want to find a cover of $E$ consisting of balls $B(q_n, epsilon), n in mathbbN, q_n in E,$ such that every $e in E$ is contained only in finitely many balls $B(q_n,epsilon).$ Is this possible?







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      Let $E$ be a separable and complete metric space. Let $epsilon > 0.$ I want to find a cover of $E$ consisting of balls $B(q_n, epsilon), n in mathbbN, q_n in E,$ such that every $e in E$ is contained only in finitely many balls $B(q_n,epsilon).$ Is this possible?







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      Let $E$ be a separable and complete metric space. Let $epsilon > 0.$ I want to find a cover of $E$ consisting of balls $B(q_n, epsilon), n in mathbbN, q_n in E,$ such that every $e in E$ is contained only in finitely many balls $B(q_n,epsilon).$ Is this possible?









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      asked Aug 7 at 15:03









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          I suppose that there is a more elementary proof than this one, but here it goes: every metric space is paracompact (this was proved by A. H. Stone) and every paracompact space is metacompact.






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            up vote
            2
            down vote



            accepted










            I suppose that there is a more elementary proof than this one, but here it goes: every metric space is paracompact (this was proved by A. H. Stone) and every paracompact space is metacompact.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              I suppose that there is a more elementary proof than this one, but here it goes: every metric space is paracompact (this was proved by A. H. Stone) and every paracompact space is metacompact.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                I suppose that there is a more elementary proof than this one, but here it goes: every metric space is paracompact (this was proved by A. H. Stone) and every paracompact space is metacompact.






                share|cite|improve this answer













                I suppose that there is a more elementary proof than this one, but here it goes: every metric space is paracompact (this was proved by A. H. Stone) and every paracompact space is metacompact.







                share|cite|improve this answer













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                answered Aug 7 at 15:11









                José Carlos Santos

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