# My cycle definitionof Schwartz set is incorrect.

*To*: debian-vote@lists.debian.org
*Subject*: My cycle definitionof Schwartz set is incorrect.
*From*: Michael Ossipoff <email9648742@gmail.com>
*Date*: Tue, 14 May 2013 16:17:10 -0400
*Message-id*: <CAOKDY5DfyXo0-Pfj6qmgR51H=_TyYNk2x4+ff0a73cUPO7Yu4g@mail.gmail.com>

Sorry to bother you again, but I want to correct an error that I made,
in a definition that I posted:
I wanted to express the beatpath definition of the Schwartz set in a
simpler and more compelling or appealing way, and the cycle definition
(that I've posted here) seemed such a simplification.
But the cycle definition doesn't define the Schwartz set. A candidate
that doesn't have a defeat that isn't in a cycle isn't necessarily in
the Scwhartz set (as defined by the unbeaten set definition and the
beatpath definition].
Of the two definitions (unbeaten set and beatpath), the beatpath
definition desn't have much compellingness. For compellingness, I much
prefer the unbeaten set definition.
Let me state both definitions here:
Unbeaten set definition of the Schwartz set::
1. An unbeaten sets is a set of alternatives none of which are beaten
by anything outside the set.
2. An innermost unbeaten set is an unbeaten set that doesn't contain a
smaller unbeaten set.
3.The Schwartz set is the set of alternatives that are in innermost
unbeaten sets.
[end of unbeaten set definition of Schwartz set]
---------------------------------------
Beatpath definition of Schwartz set:
There is a beatpath from X to Y if X beats Y, or if X beats A and
there is a beatpath from A to Y.
If there is a beatpath from Y to X, but not from X to Y, then X is not
in the Schwartz set.
Otherwise X is in the Schwartz set.
[end of beatpath definition of the Schwartz set]
Michael Ossipoff

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