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A metric on finite sets

  1. Nov 10, 2008 #1

    recently I stumbled across the question whether for finite sets [itex]A,B[/itex] the function

    [tex]d(A,B):=|A\cup B| - |A\cap B|[/itex]

    is a http://en.wikipedia.org/wiki/Metric_distance" [Broken]? Trivially, [itex]d(A,A)=0[/itex] and of course [itex]d[/itex] is symmetric, but how about the triangle inequality? Does it hold?

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 10, 2008 #2


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    yes it is. For the triangle inequality see the formula [itex](A \Delta B) \Delta (B \Delta C) = A \Delta C[/itex] here - http://en.wikipedia.org/wiki/Symmetric_difference" [Broken].

    Also, for infinite sets you can replace the size |A| of a set with its measure (eg, length, volume, etc) to get a pseudometric, as mentioned in the wikipedia link.
    Last edited by a moderator: May 3, 2017
  4. Nov 11, 2008 #3
    I never really came across symmetric difference as an explicit operator in set theory.

    Thanks for the information,
  5. Nov 11, 2008 #4


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    It's certainly not a metric in the usual sense, since metrics are real-valued not set-valued.
  6. Nov 11, 2008 #5
    you didn't notice the bars around the sets, denoting cardinality. for finite sets this will be an integer number, hence it is a metric in a usual sense.
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