Metric spaces/Talk

I can't understand this:

A subset X of M is open if for every point x of X there is a strictly positive number r such that B (x ; r) is contained in X. With this definition, every metric space is automatically a topological space.

What does B (x ; r) saying?

Is it saying that the set of all open balls of the metric space is a topology for the metric space? -- Simon J Kissane

No. I've rewritten the explanation. Do you understand it now?
Zundark, 2001-08-11

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While the new definition is certainly correct, maybe we should point out that the union can (and usually is) a union of infinitely many sets balls.

Also, maybe the example Rⁿ with euclidean metric could be given. --AxelBoldt

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Would it be possible to move this page to metric space (no s)?

Yes, it needs to be changed. I would have done it already, except that it really requires changing all the links to it (links to redirected pages being somewhat undesirable for a number of reasons). --Zundark, 2001 Sep 24