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Real Analysis - Prove f is discontinuous

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Let f(x) = 1 for rational numbers x and f(x) = 0 for irrational numbers. Show that f is discontinuous at every x in R.

    2. Relevant equations
    Definition of continuity.


    3. The attempt at a solution
    I want to find a sequence (x_n) that converges to x_0 but that x_n is rational for even n, and irrational for odd n. This will show that (f(x_n)) cannot converge, since it will alternate between 0 and 1, and thus f is discontinuous. My problem is that I can't think of a sequence that does this!

    Thanks for any help. :smile:
     
  2. jcsd
  3. Apr 8, 2008 #2

    NateTG

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    Step 1: Find a rational sequence that converges to [itex]x[/itex].
    Step 2: Find an alternating rational/irrational sequence that converges to [itex]0[/itex].
    (Hint: I suggest something of the form [itex]0,y_1,0,y_3...[/itex].)
    Step 3: Add the two together.
     
  4. Apr 8, 2008 #3
    How about (x_n) = x, and (y_n) = 0, 1/pi, 0, 1/pi^2, ... ? This should work, right? Thanks.
     
  5. Apr 8, 2008 #4

    lurflurf

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    forget seqences
    let U be an open set containing x
    show f(U)={0,1}

    If you must procede on the original path...
    consider a standard rational sequence like
    x_n=(n^-1)(floor(x*n)+(1/2)(1+(-1)^n))
    altered like
    x_n=(n^-1)(floor(x*n)+(1/sqrt(2))(1+(-1)^n))
     
  6. Apr 8, 2008 #5

    lurflurf

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    If you know pi is irrational
    maybe it is 4...
     
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