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A question about analysis

  1. Oct 20, 2012 #1
    In this link:

    http://math.berkeley.edu/~scanez/courses/math104/fall11/homework/hw7-solns.pdf

    In number 3, the text says "By the denseness of Q in R, there exists a sequence [itex](r_n)[/itex] of rationals converging to x."

    I have several questions about this:

    1) Why/how does denseness imply that there exists a sequence? Becuase Q is dense in R, we know that any two real numbers have a rational number between them. But how does that tell us that there is a "sequence" of rational numbers?

    2) Do we have to assume that x is an irrational number?

    3) Does this statement mean that for every irrational number, there is a sequence of rational number converging to it?

    Thanks in advance...
     
  2. jcsd
  3. Oct 20, 2012 #2

    jbunniii

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    Well, that's not all that denseness tells you. Between any two even integers is an odd integer, but that doesn't mean that the odd integers are dense in the even integers. What is the definition of denseness?

    The first paragraph proves that f(x) = x for any rational x. The second paragraph uses that fact, but you don't have to assume that x is irrational in the second paragraph. The argument also works for rational x, because given a rational number, you can certainly find a sequence of rationals converging to that number.

    I'm not sure what "this statement" refers to, but it's true that for every irrational number, there is a sequence of rational numbers converging to it.

    If [itex]x[/itex] is irrational, start by picking any rational, say [itex]r_1[/itex]. Since [itex]x \neq r_1[/itex], the distance between these numbers is positive: [itex]|x - r_1| > 0[/itex]. Choose a rational [itex]r_2[/itex] such that [itex]|x - r_2| < (1/2)|x - r_1|[/itex], i.e. [itex]r_2[/itex] is less than half the distance from [itex]x[/itex] as [itex]r_1[/itex].

    Then just keep repeating this process. At the n'th step, you will choose a rational [itex]r_n[/itex] such that [itex]|x - r_n| < (1/2)^n |x - r_1|[/itex]. Clearly the sequence [itex]r_n[/itex] converges to [itex]x[/itex].

    Another, more concrete (but less rigorous) way to see this is to consider the decimal expansion of [itex]x[/itex]. Any real number has a decimal expansion. You can form a sequence of rationals converging to the real number by simply taking more and more digits, e.g.

    [itex]x_1 = 3[/itex]
    [itex]x_2 = 3.1[/itex]
    [itex]x_3 = 3.14[/itex]
    [itex]x_4 = 3.141[/itex]
    [itex]x_5 = 3.1415[/itex]
    [itex]x_6 = 3.14159[/itex]
    and so on
     
  4. Oct 20, 2012 #3
    This is the definition that my textbook gives us "If a,b in R and a < b, then there is a rational r in Q such that a < r < b".

    But I think I understand why sequences exist, from the answer you gave to my 3rd question. Thanks a lot. :)
     
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