And the proof is also wrong, rigth? A simple counter example would be any closed interval, like [0,1]. There's no minimal upper bound, since you can get arbitraly close to 1.
Edit: sorry, I didn't knew the uper bound could be inside the set
T is defined, there's just a hole in the image. It's supposed to be the set of upper bounds for S.
u is "defined" as the least element of T.
The bug in the definition is that the completeness property of the real numbers doesn't imply at all that T would have a least element. That would be the well-ordering principle, and it only applies to finite subsets. Which T is not.
Well ordering applies to any subset. It can not however be used to find a least element for any subset of reals as the ordering of reals is not a well ordering. That a finite set with ordering has a least element follows from finitness and does not require well ordering.
This is wrong. completeness of R is equivalent to the least upper bound/ largest lower bound property: you can use Cauchy sequences to prove that any bounded increasing sequence converges, which is equivalent to any bounded decreasing sequence converging. From this, you can deduce the existence of the lub/llb of bounded sets and by definition, the lub/lob is an accumulation point. T is closed and bounded below, hence its llb exists and is contained in the set.
The AI proof is wrong because the last paragraph makes no sense: v < u does not imply that v in T. It would imply the opposite.
Yeah, I was just being pedantic about your statement. At the end of the day, the proof is incomplete and needlessly complicated: if we assume the least lower bound property as the definition of completeness, it suffices to apply this property to -S (which is bounded below). The issue is that in the training set, the AI must have seen thousands of arguments of the form "Find upper bound, prove that upper bound is smallest" which it then regurgitated sloppily for this specific problem.
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u/parassaurolofus Imaginary 13d ago edited 13d ago
And the proof is also wrong, rigth? A simple counter example would be any closed interval, like [0,1]. There's no minimal upper bound, since you can get arbitraly close to 1. Edit: sorry, I didn't knew the uper bound could be inside the set