Hi, Raul Miller: > > > Is it accurate to say that if there is no y such that x>>y (i.e., x > > > defeats nothing), then x is NOT in the set? > > Yes. > However, it's not true for the general case. Imagine you have two > options and they're tied. Then, both options would be in the schwartz > set, and yet no option beats any other option. > You're right. For simplification-of-rules-to-keep-track-of-in-my-head reasons, I had "x==y is equivalent to x>>y AND y>>x" in the back of my head, which, while useful, is not the generally-understood definition. -- Matthias Urlichs | noris network AG | http://smurf.noris.de/
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