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Let metric topology be the topology generated by metric balls of a metrizable space $X$

Is there a subbase $S$ that generates the metric topology?

I am asking because in most textbooks, it seems that people only discuss the base generated by the metric balls, and no attention is paid to subbase and no definition is given.

Thanks

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    $\begingroup$ Well, any base is a subbase, so... $\endgroup$ – Alex Provost Jul 22 '16 at 19:01
  • $\begingroup$ Are you looking for a subbase which isn't a base? $\endgroup$ – Alex Provost Jul 22 '16 at 19:14
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    $\begingroup$ There's no point in mentioning a (sub)base unless it is actually useful. See my answer here. $\endgroup$ – Eric Wofsey Jul 22 '16 at 20:22
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A possible sub-base would be to take balls of only rational radius. Any ball of irrational radius can be realized as the infinite union of balls of rational radius.

E.g. $ \ \ B_\pi(x) = B_3(x) \cup B_{3.1}(x) \cup B_{3.14}(x) \cup \cdots$

I hope I've interpreted the definitions correctly. I'm taking "sub-base" to mean a subset $S \subset B$ that serves as a basis for the metric topology, where $B$ is the basis for the metric topology consisting of all open balls.

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    $\begingroup$ This is even a base for the metric topology. $\endgroup$ – Alex Provost Jul 22 '16 at 19:09
  • $\begingroup$ Indeed @AlexProvost. I thought that's what OP was requesting. $\endgroup$ – Kaj Hansen Jul 22 '16 at 19:13

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