Difference between revisions of "Talk:Multiset"
From Encyclopedia of Mathematics
(No and yes) |
(When the carrier is infinite, why the multiplicity must be finite?) |
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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) |
Revision as of 10:17, 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)
How to Cite This Entry:
Multiset. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Multiset&oldid=37510
Multiset. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Multiset&oldid=37510