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

[birulami]

Hi,

recently I stumbled across the question whether for finite sets A,B the function

d(A,B):=|A\cup B| - |A\cap B|[/itex]<br /> <br /> is a http://en.wikipedia.org/wiki/Metric_distance&quot; ? Trivially, d(A,A)=0 and of course d is symmetric, but how about the triangle inequality? Does it hold?<br /> <br /> Harald.

 

[gel]

yes it is. For the triangle inequality see the formula (A \Delta B) \Delta (B \Delta C) = A \Delta C 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.

 

[birulami]

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

Thanks for the information,
Harald.

 

[CRGreathouse]

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

 

[matticus]

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.