Question about sequences

Discussion in 'Undergraduate Math' started by khosaa, Feb 3, 2011.

  1. khosaa

    khosaa Guest


    I am studying from Malik's book "Mathematical Analysis", and I came
    across a definition that seems strange to me. I want to check and see
    if I am not missing something.
    In his section on "non-convergent sequences" Malik defines a "sequence
    unbounded on the left" as follows:

    "If the sequence {S_n} is unbounded on the left then we say that -
    infinity is a limit point of the sequence, and to each positive number
    G, however large, there corresponds a positive integer m such that S_n
    < -G for every n >=m"

    I think this does not sound right since it for any given negative
    number, it requires *all* the sequence members past some index to lie
    below said negative number. Hence, by Malik's definition, the sequence
    S_n = n * (-1)^n, would not be unbounded on the left. I would think
    that a better definition would have been that for every positive
    number G, there exists a natural number m such that S_m < -G.

    It also seems somewhat strange to talk about infinity (or negative
    infinity) as a limit point of a sequence. I could see how -infinity
    could be the lim inf of a sequence, but Malik's defn of limit point
    of a sequence is that for every epsilon, an infinite number of terms
    in the sequence are required to be in the open epsilon nbd of the
    limit point. I would like to think he is making this definition for a
    real-valued limit point and one would adopt a different definition for
    a limit point in the extended reals (although he does not seem to do
    this at any point in the text).

    Thank you for any insights,
    khosaa, Feb 3, 2011
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  2. I'd call his definition strongly unbounded below.
    Your definition is the usual negation of bounded below.
    In elementary calculus, limit to infinity is distinguished differently
    that finite limit. There are of course similarities that the extended
    reals unify. A nhood base for oo in the extended reals is
    { (r,oo) | r in R } where (r,oo) = { x | r < x }. This is
    the interval topology for the extended reals. Of course the definition
    of limit to oo in the extended reals and the separate definition with
    beginning calculus are the same.

    A disadvantage of the extended reals is, among many, addition
    isn't a group.
    William Elliot, Feb 3, 2011
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  3. khosaa

    Paul Sperry Guest

    He goes on to say "i.e. the sequence has an infinity of terms below -G."
    You would be correct. He is trying to say what it means to say -oo is a
    limit point of a sequence (since " -oo " really isn't _anything_).

    Subsequently, he says that if the sequence is bounded above but not
    below and has no limit point other that -oo (i.e. no limit point at all
    in the usual sense) then, by definition, the sequence "converges to
    -oo". That is equivalent to "for every positive number G, there exists
    a natural number m such that S_m < -G".
    He is defining the phrase " -oo is a limit point of a sequence" as a
    whole. He is not giving a property of the (non-existent) thing -oo nor
    is he saying that -oo is a limit point in the sense that he has
    previously defined limit point. In the shaded box at the end of the
    section he emphasizes that oo is not a real number.

    "Limit point" is used two different ways: It is a property of a real
    number relative to a set of real numbers and it is used to describe a
    property of a sequence of real numbers. The two are only intuitively
    Paul Sperry, Feb 3, 2011
  4. khosaa

    khosaa Guest

    Thank you for this interpretation: it ties in well with Malik's
    standard method for showing a "value" is a limit point of a sequence
    (which is that very ndb of the point has an infinite number of
    sequence terms).

    khosaa, Feb 4, 2011
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