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My cycle definitionof Schwartz set is incorrect.

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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