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Continuity question

  1. May 17, 2009 #1

    C.E

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    1. Let X be R be a finite set and define f : R [tex]\rightarrow[/tex] R by f(x) = 1 if x [tex]\in[/tex] X and f(x) = 0 otherwise. At which points c in R is f continuous? Give proofs.

    3. I don't know how to start this, do you think it is ok to assume that X represents an interval of R? If not how can you possibly deduce the points continuity?
     
  2. jcsd
  3. May 17, 2009 #2

    quasar987

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    X is finite. Meaning it contains a finite number of points. So X is certainly not an interval.

    Try some concrete examples with increasing degree of complexity. Say X={0}. What then? (i.e., where is f continuous?) Now what if X={-1,1}, etc. If you've solved the problem in these two particular cases, then surely you can guess the answer to the general case and back your intuition with a proof.
     
  4. May 17, 2009 #3

    C.E

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    Is this right?

    The function is discontinous for all x in X and continuous elsewhere.
    To prove discontinuity at x in X let x_1, x_2, ... x_n be the points in X then if we assume X_2 is the member of X closest to x_1. Then taking episilon =0.5 and only considering delta less than 0.5|x_1 - x_2| we prove discontinuity.

    To prove continuity elsewhere we use a similar argument letting X_1 be the closest member of X to the point x not in X then setting delta= 0.5|X_1-x| completes the proof.

    Any comments?
     
  5. May 17, 2009 #4

    quasar987

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    You seem to have set x=x_1 in the first paragraph but never said so explicitly, which is confusing. Also, when you say "Then taking epsilon =0.5 and only considering delta less than 0.5|x_1 - x_2| we prove discontinuity.", I suspect that you have the right idea, but your sentence expresses it poorly. How about instead: "Then, taking epsilon=0.5, notice that for delta=0.5|x_1 - x_2|, we have 0<|x_1-y|<delta implies |f(x_1)-f(y)|=|1-0|=1>epsilon, thus proving that f is discontinuous at x_1."

    In the second paragraph, I suggest adding "then for any epsilon>0, take delta= 0.5|X_1-x|, thus completing the proof.", but it can't hurt to write things more explicitely either.
     
  6. May 18, 2009 #5

    HallsofIvy

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    I think simpler is: since X is finite, there exist [itex]\epsilon> 0[/itex] such that the distance between any two points in X is greater than [itex]\epsilon[/itex].
     
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