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Difference between revisions of "Talk:Multiset"

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(No and yes)
(That seems to be the prevailing convention)
 
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:There seems no need for the carrier to be a finite set, but the Parikh vector is normally only defined in this case.  And yes, "simple set" means set here.  [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 22:30, 12 January 2016 (CET)
 
:There seems no need for the carrier to be a finite set, but the Parikh vector is normally only defined in this case.  And yes, "simple set" means set here.  [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 22:30, 12 January 2016 (CET)
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::When the carrier is infinite, it seems natural to ask, why the multiplicity must be finite. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 11:17, 13 January 2016 (CET)
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:::That seems to be the prevailing convention.  Richard Stanley and Gian-Carlo Rota, to name but two, use this definition.  [[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 18:51, 13 January 2016 (CET)

Latest revision as of 17:51, 13 January 2016

I guess, the carrier $C(X)$ is required to be a finite set. Really?

Also I guess that "simple set" (in this context) means "a set" (as understood in most other contexts); but is this clear enough? Boris Tsirelson (talk) 21:29, 12 January 2016 (CET)

There seems no need for the carrier to be a finite set, but the Parikh vector is normally only defined in this case. And yes, "simple set" means set here. Richard Pinch (talk) 22:30, 12 January 2016 (CET)
When the carrier is infinite, it seems natural to ask, why the multiplicity must be finite. Boris Tsirelson (talk) 11:17, 13 January 2016 (CET)
That seems to be the prevailing convention. Richard Stanley and Gian-Carlo Rota, to name but two, use this definition. Richard Pinch (talk) 18:51, 13 January 2016 (CET)
How to Cite This Entry:
Multiset. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Multiset&oldid=37510