Is the d(A,B) Function a Valid Metric for Finite Sets?

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birulami
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Hi,

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

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.
 
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I never really came across symmetric difference as an explicit operator in set theory.

Thanks for the information,
Harald.
 
It's certainly not a metric in the usual sense, since metrics are real-valued not set-valued.
 
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.