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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
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