Results 1 to 3 of 3

Math Help - supremum proof in predicate calculus

  1. #1
    Banned
    Joined
    Oct 2008
    Posts
    71

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2008
    Posts
    71
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. predicate calculus
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: December 13th 2009, 11:01 AM
  2. Predicate calculus - propositions
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 2nd 2009, 08:50 PM
  3. Predicate Calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 21st 2009, 05:19 PM
  4. Predicate Calculus Question
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: May 20th 2009, 05:39 PM
  5. Predicate Calculus help
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: March 6th 2009, 10:19 PM

Search Tags


/mathhelpforum @mathhelpforum