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I want to prove that every metric space which is Lindelöf has a countable basis. First I tried to show that a countable cover, which exists by the Lindelöf property, is a countable basis, but for the second property of basis, which is about the intersection of two basis elements, I have no idea.

Thanks for your help.

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  • $\begingroup$ Please check your posts for typos. $\endgroup$ – Mariano Suárez-Álvarez Jun 10 '16 at 17:54
  • $\begingroup$ @MarianoSuárez-Alvarez: While you are right in principle, let us be more forgiving of users the native language of whom is not English (which seems to be the case here, just check the OP's other posts, in particular his verbs). $\endgroup$ – Alex M. Jun 10 '16 at 17:56
  • $\begingroup$ I don't know why you think I was not being forgiving, @AlexM. $\endgroup$ – Mariano Suárez-Álvarez Jun 10 '16 at 18:02
  • $\begingroup$ @MarianoSuárez-Alvarez: What I meant to say is that, probably, the OP believes that the current spelling really is correct, so from his point of view there is nothing to correct. $\endgroup$ – Alex M. Jun 10 '16 at 18:04
  • $\begingroup$ Well, my comment was a hint that it is not. $\endgroup$ – Mariano Suárez-Álvarez Jun 10 '16 at 18:05
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Since $X$ is a Lindelöf space, for every $n$ there exist countably many $x_{n,j}$ with $$\bigcup_jB(x_{n,j},1/n)=X.$$So $X$ is separable; in fact $\{x_{n,j}:n,j\in\Bbb N\}$ is dense, and hence $$\{B((x_{n,j},1/k):n,j,k\in\Bbb N\}$$is a basis.

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