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Math Help - Characterisation of a point of accumulation - Serge Lang - Undergraduate Analysis

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    Super Member Bernhard's Avatar
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    Characterisation of a point of accumulation - Serge Lang - Undergraduate Analysis

    I am an amateur math hobbyist working through Serger Lang's book "Undergraduate Analysis":

    On page 39 Lang writes the following:

    "Let { x_{n}} (n = 1,2, ... ) be a sequence and x a number. We shall say that x is a point of accumulation of the sequence if given \epsilon there exists infinitely many integers n such that:

    | x_{n} x| < \epsilon
    ..... "

    Lang then goes on to characterise a point of accumulation in a second way, namely as follows:

    "In the definition of point of accumulation we could have said that given \epsilon and given N there exists some n \geq N such that | x_{n} x| < \epsilon "

    I have tried to prove that these two conceptualisations are the same but cannot form a convincing and exact proof,

    I would really appreciate help in this matter.

    Bernhard
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  2. #2
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    Quote Originally Posted by Bernhard View Post
    I am an amateur math hobbyist working through Serger Lang's book "Undergraduate Analysis":

    On page 39 Lang writes the following:

    "Let { x_{n}} (n = 1,2, ... ) be a sequence and x a number. We shall say that x is a point of accumulation of the sequence if given \epsilon there exists infinitely many integers n such that:

    | x_{n} x| < \epsilon
    ..... "

    Lang then goes on to characterise a point of accumulation in a second way, namely as follows:

    "In the definition of point of accumulation we could have said that given \epsilon and given N there exists some n \geq N such that | x_{n} x| < \epsilon "

    I have tried to prove that these two conceptualisations are the same but cannot form a convincing and exact proof,

    I would really appreciate help in this matter.

    Bernhard

    Assume first form, and let be N\in\mathbb{N} . Since there are infinite indexes that fulfill the first form there must be some n>N that fulfills it, otherwise only those indexes s.t. n\leq N would fulfill it , and there'd be a finite number of them...

    Assume the second form...and now you try this by yourself.

    Tonio
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