supremum proof in predicate calculus

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October 10th 2008, 06:47 AM
archidi

supremum proof in predicate calculus

My lecturer must have gone made, he asked us to do a proof of the following problem in predicate calculus:
Let S be a non empty subset of real Nos bounded from above.
Let A={as:sεS,a>0}.Then prove Sup(A)=aSup(S)

In the proof we must show explicitly any quantifier,propositional logic,real Nos axioms,theorems,definitions involved
I don't know if any of you guys understands whats going on here
October 10th 2008, 08:34 AM
Plato

Here is what I know as a proof. But I know nothing of the way your instructor expects it done. Perhaps you can fill in the details.

Suppose that $\sigma = \sup (S)$ by definition $\left( {\forall s \in S} \right)\left[ {s \leqslant \sigma } \right]$ and $\left( {\forall s \in S} \right)\left[ {as \leqslant a\sigma } \right]$ .
But that means $\left( {\forall x \in A} \right)\left( {\exists t \in S} \right)\left[ {x = at \leqslant a\sigma } \right]$ or $a\sigma$ is an upper bound for $A$ .
Let $\alpha = \sup (A)$ so that $\alpha \le a\sigma$ .
Here we work for a contradiction: assume that $\alpha < a\sigma$ .
Then $\frac{\alpha }{a} < \sigma \Rightarrow \left( {\exists u \in S} \right)\left[ {\frac{\alpha }{a} < u \leqslant \sigma \Rightarrow \alpha < au \leqslant a\sigma } \right]$ .
But $au \in A \wedge \alpha < au$ ! CONTRADICTION.
October 10th 2008, 06:46 PM
archidi

Thanks for the clever proof particularly the contradiction part.

What details,you got to be joking,can you fill the details.

The man is totally mad how can i do the proof by explicitly mentioning all the laws of logic involved theorems axioms of supremum e.t.c

proofs in analysis are so complicated,as they are .
Its an impossible task to do them the way he want us to

I think i am gonna drop the course