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Every metric space is Hausdorff

  1. Nov 22, 2012 #1
    The usual proof of this theorem seems to assume that the topology of the metric space is the one generated by the metric. But if I use another topology, for example the trivial, the space need not be Hausdorff but the metric stays the same. Am I missing something or is the statement of the theorem just sloppy?
     
  2. jcsd
  3. Nov 22, 2012 #2
    If you say that you have a metric space [itex](X,d)[/itex], then it is always assumed that we take the topology generated by the metric. If we happen to take another topology, then we always state this explicitely.

    It's the same as saying that you work with [itex]\mathbb{R}^n[/itex]. The topology is always assumed to be Euclidean unless otherwise stated.
     
  4. Nov 22, 2012 #3
    Thanks. :smile:
     
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