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Math Help - Sets question.

  1. #1
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    Sets question.

    Let a1, d2, a3, ... an, ... be a sequence of real numbers. The sequence converges to a real number a if:

    [see attachment]

    Work out the negation of the sentence, ie a definition of what it means to say that the sequence does not converge to a.

    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Sets question.-question1.jpg  
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  2. #2
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    Hi
    The negation for a " \forall\ \exists\ \forall\ PROPOSITION " formula will be " \exists\ \forall\ \exists\ \text{no}(PROPOSITION) "

    In your case:

    \exists\epsilon >0\ \forall N\in\mathbb{N}\ \exists n>N\ \text{s.t.}\ |a_n-a|\geq\epsilon
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    Hi
    The negation for a " \forall\ \exists\ \forall\ PROPOSITION " formula will be " \exists\ \forall\ \exists\ \text{no}(PROPOSITION) "

    In your case:

    \exists\epsilon >0\ \forall N\in\mathbb{N}\ \exists n>N\ \text{s.t.}\ |a_n-a|\geq\epsilon
    clic-clac can you prove what you have just written down?? i.e

    ~ \forall\epsilon[\epsilon>0\rightarrow\exists k(k\in N\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))] is equivalent to:

    ......... \exists\epsilon[\epsilon>0\wedge\forall k(k\in N\rightarrow\exists n(n\geq k\wedge |a_n-a|\geq\epsilon))].................................................. ................................
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  4. #4
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    Hi
    Well same thing. Since \neg( A\Rightarrow B)\equiv A\wedge \neg B and \neg (A\wedge B)\equiv \neg A\vee\neg B\equiv A\Rightarrow\neg B, we have:

    \neg(\forall\epsilon[\epsilon>0\rightarrow\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))])
    \equiv\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\exists n(n\geq k\wedge |a_n-a|>\epsilon))]
    Last edited by clic-clac; February 16th 2009 at 12:25 AM. Reason: cor
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  5. #5
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    Quote Originally Posted by clic-clac View Post
    Hi
    Well same thing. Since \neg( A\Rightarrow B)\equiv A\wedge \neg B and \neg (A\wedge B)\equiv \neg A\vee\neg B\equiv A\Rightarrow\neg B, we have:

    \neg(\forall\epsilon[\epsilon>0\rightarrow\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))])
    \equiv\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\exists n(n\geq k\wedge |a_n-a|>\epsilon))]
    Good proof ,but from the 2nd step on wards i tried to find suitable laws of logic to justify each step and i could not.

    O.k for the first step since there is the theorem in predicate calculus to account for that,i.e

    ..............~ \forall u P\Longleftrightarrow\exists u\neg P.......................................

    But from there i could not find any laws to justify each step ,since you kept the quantification symbols at each step and you did work inside the formula ,without getting rid of the quantification symbols.

    There are laws in predicate calculus for removing and introducing the quantification symbols and that in combination with the laws of propositional calculus can give you a proof
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  6. #6
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    The only things I used are what I said:
    and
    You don't have to get rid of the quantification symbols, just replace them by the other one.

    For instance, from second step to third:

    \exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]


    \equiv\exists\epsilon[\epsilon>0\wedge\forall k\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]


    But
    \neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)) has the form \neg(A\wedge B) so it's the same than k\in \mathbb{N}\Rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))

    we replace it in the whole formula:

    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
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  7. #7
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    Quote Originally Posted by clic-clac View Post
    You don't have to get rid of the quantification symbols, just replace them by the other one.

    But there is no law of logic that allow you to do that ,at least as far as i know.

    If however you happen to know of any such law ,mention it so that we could see how that law is derived from the axioms of the predicate calculus
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  8. #8
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    Quote Originally Posted by archidi View Post
    But there is no law of logic that allow you to do that ,at least as far as i know.
    archidi;267897, I have tried to stay out of this thread but now I must make a comment.
    archidi;267897 do you know any real logic beyond the level of SCHAUM’S OUTLINE?
    I can suggest some much better texts for you to study.
    Symbolic Logic by Irving Copi is still the most teachable textbook.
    It will give you a deeper understanding of quantifier theory.

    The other classics are two texts by Quine: Methods of logic and Set Theory and its Logic.
    Set Theory and Logic
    by Stoll is also a classic.

    You seem to fancy yourself as a logician, but your postings in this tread belie that notion.
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  9. #9
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    Quote Originally Posted by Plato View Post
    archidi;267897, I have tried to stay out of this thread but now I must make a comment.
    archidi;267897 do you know any real logic beyond the level of SCHAUMíS OUTLINE?
    I can suggest some much better texts for you to study.
    Symbolic Logic by Irving Copi is still the most teachable textbook.
    It will give you a deeper understanding of quantifier theory.

    The other classics are two texts by Quine: Methods of logic and Set Theory and its Logic.
    Set Theory and Logic by Stoll is also a classic.

    You seem to fancy yourself as a logician, but your postings in this tread belie that notion.

    Thank you plato for your kind words and advise,but definitely i never said that i fancy my self as a logician.

    I simply ask questions when i see something that i think it is wrong and then produce my arguments to substantiate the case .


    My dear plato i suppose you remember that when i first joined the forum you where the one i was asking questions.


    But i also happen to have a great teacher in logic whose knowledge in logic is internationally recognized.


    Under his patient advise and unrepeatable teaching i learned a lot about logic within a very short time.

    The books now in my library are:

    1) Introduction To Symbolic Logic and its Applications,by Rudolf Camp

    2) First Course in Mathematical Logic ,by Patrick Suppes and Shirley Hill

    3) Popular Lectures on Mathematical Logic, by Hao Wang

    4)Theory of Formal Systems, by RAYMOND M. SMULLYAN


    5)Logic as Algebra, by PAUL HALMOS and STEVEN GIVANT

    6) Set Theory and Logic ,by ROBERT R. STOLL

    7)Logic Computing Machines and Automation,by Alice Mary Hilton

    8)Introduction to Elementary Mathematical Logic ,by Abram Aronovich Stolyar

    9)Introduction To Logic fifth edition ,by IRVING M. COPI

    10)Symbolic Logic fourth edition ,by IRVING M. COPI

    11)Foundations without Foundationalism A CASE FOR SECOND ORDER LOGIC ,by Stewart Shapiro

    12) Introduction To Logic AND TO THE METHODOLOGY OF DEDUCTIVE SCIENCES ,by Alfred Tarski

    13)Introduction To Logic , by PATRICK SUPPES

    14) What is Mathematical Logic ,by J N Crossley,C.J.Ash,C.J.Brickhill,J.C.Stillwell,N.H.W illiams

    15)FIRST ORDER LOGIC ,by Raymond M.Smullyan

    16) FIRST ORDER MATHEMATICAL LOGIC by Angelo Margaris

    17)BOOLEAN ALGEBRA and SWITHING CIRCUITS IN SCHAUM'S OUTLINE SERIES.by ELLIOT MENDELSON

    18)LOGIC IN SCHAUM'S OUTLINE SERIES by John Nolt and Dennis Rohatyn

    plus other books in other languages.

    The question in the thread is still not answered
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  10. #10
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    Quote Originally Posted by archidi View Post
    But i also happen to have a great teacher in logic whose knowledge in logic is internationally recognized.
    The question in the thread is still not answered
    Well I doubt that first statement. Is your idea of international is limited?
    What has you instructor helped you with in quantification theory?
    From what you have posted, the answer to that question is: VERY LITTLE.
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  11. #11
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    Quote Originally Posted by Plato View Post
    Well I doubt that first statement. Is your idea of international is limited?
    What has you instructor helped you with in quantification theory?
    From what you have posted, the answer to that question is: VERY LITTLE.
    My dear Plato you are simply curious to see what the right proof to that problem is .

    Why ,don't you know the answer, or you agree with clic clac's proof??
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  12. #12
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    Mhh.. Even before studying logic, one of the first rules in "advanced" mathematics you may see is: if P is a proposition, \neg(\forall xP)\equiv\exists x\neg P and \neg(\exists xP)\equiv \forall x\neg P, and that makes sense (I think most people will agree with that).
    Moreover, as you wrote yourself:
    O.k for the first step since there is the theorem in predicate calculus to account for that,i.e

    ..............~.......................................
    So I don't see the problem, you "replaced" \forall by \exists ...

    My dear Plato you are simply curious to see what the right proof to that problem is
    The right proof?

    As you seem to know about predicate calculus and have a great teacher, Plato's reaction is quite understandable!
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  13. #13
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    Quote Originally Posted by clic-clac View Post
    The only things I used are what I said:
    You don't have to get rid of the quantification symbols, just replace them by the other one.

    For instance, from second step to third:

    \exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]

    \equiv\exists\epsilon[\epsilon>0\wedge\forall k\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]

    But \neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)) has the form \neg(A\wedge B) so it's the same than k\in \mathbb{N}\Rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))

    we replace it in the whole formula:

    \equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]
    Quote Originally Posted by archidi View Post
    But there is no law of logic that allow you to do that ,at least as far as i know.

    If however you happen to know of any such law ,mention it so that we could see how that law is derived from the axioms of the predicate calculus

    Clic-clac that was my reply to your arguments.

    All i said is that as far as i know there are not any laws of logic that allow you to replace any part of a formula by another equivalent formula


    Do you know any such laws?Just state them and lets discuss them


    Does that mean that i fancy my self as a logician?? As plato said ??

    Really i dont see the point of writing useless posts confusing the viewers at the same time.

    I do apologize to the viewers for writing posts completely irrelevant to the main issue of the problem
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  14. #14
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    All i said is that as far as i know there are not any laws of logic that allow you to replace any part of a formula by another equivalent formula
    Since your first post did not mention you were asking about replacing a formula by an equivalent one, I couldn't see what part of the proof you did not agree with.

    Of course we have the right to do that! The proof isn't so simple and I can't write a whole course. That can be done by induction, first proving it for atomic formulas (that doesn't demand a lot of efforts) and then checking the different cases.
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  15. #15
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    Quote Originally Posted by clic-clac View Post
    Since your first post did not mention you were asking about replacing a formula by an equivalent one, I couldn't see what part of the proof you did not agree with.

    Of course we have the right to do that! The proof isn't so simple and I can't write a whole course. That can be done by induction, first proving it for atomic formulas (that doesn't demand a lot of efforts) and then checking the different cases.

    The only rules for substitution are the following ones:

    1) \vdash r=s\rightarrow t_{r}=t_{s},where r,s,  t_{r},t_{s} are terms,r occurs in  t_{r} and  t_{s} is the result of replacing one or more specified occurences ( but not necessarily all occurrences) of r in  t_{r} by occurrences of s.

    ......for e.g if r=s =====> x+r= x+s..............................................

    Where now  t_{r} is :x+r,and  t_{s} is x+s............................................... ................................................


    2) \vdash r=s\rightarrow (P_{r}\Longleftrightarrow P_{s}),where  P_{r},P_{s} are formulas,r and s are terms,r occurs in  P_{r} and  P_{s} is the result of replacing one or more specified occurrences ( but not necessarily all occurrences) of r in  p_{r} by occurrences of s,provided that no variable that occurs in r or s is a bound variable of  P_{r}.

    And a simple example is;

    ...........r=s ,then x<r <====>x<s......................................... ................................................


    Where now  P_{r} is x<r ,and  P_{s} is x<s.....................................

    Both of the above theorems are proved by induction.


    As you can see none of them can be used in your proof
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