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Empty Set Metric Space

  1. Nov 3, 2011 #1
    Can we define a metric space [itex](\emptyset, d)[/itex]? The metric is the part that confuses me, since it seems like all of the required properties of d are satisfied since they are "not not satisfied", but I'm not sure.

    Thank you!
     
  2. jcsd
  3. Nov 5, 2011 #2
    Yes, I don't see the problem with that. However, I think you would probably want to include "non-empty" in the axioms for a metric space, it's just that you wouldn't usually bother because you don't gain anything of interest by looking at an empty metric space.
     
  4. Nov 5, 2011 #3

    disregardthat

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    It's a perfectly fine metic space and it's a good thing to keep that convention. So we don't always have to make awkward exceptions to theorems, such as "every non-empty subspace of a metric space is a metric space."
     
  5. Nov 5, 2011 #4
    OK, thanks to you both.
     
  6. Nov 6, 2011 #5

    Deveno

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    how are they not satisfied? every condition for a metric holds for every element of Ø, no matter how we define the metric (although if you must have a definition, use d(x,x) = 0, d(x,y) = 1, for all x,y not in Ø).
     
  7. Nov 6, 2011 #6
    ... this all comes to vacuous truths again, as discussed elsewhere.
     
  8. Nov 6, 2011 #7
    What is a metric?

    It's a function from X cross X to ℝ, plus some conditions.

    If X is the empty set, it's a function from the empty set to ℝ.

    What's a function from a set S to a set T?

    Formally, it's a subset of S cross T, satisfying some condition.

    So, the empty set can be viewed as a function from the empty set to any other set. The empty function. So, that's your metric and it vacuously satisfies all the conditions.
     
  9. Nov 6, 2011 #8
    I wrote "not not satisfied", which is a slightly stupider way of saying "vacuously true" as Jamma and homeomorphic have clarified.
     
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