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Check for basic concepts

  1. Dec 10, 2013 #1
    1) B = (0,1) U {2}, the derived set of B which contains all limit points in B is [0,1], right?

    2) B = (0,1) intersect Q,the closure of B is [0,1] too right?

    3) a subset of R whose only limit point is 1... Will {1+ 1/n: n is natural number} work?

    4)
    an interval on which the function f(x) = 1/(1-x^2} is uniformly continuous

    R \ {1} work?
     
  2. jcsd
  3. Dec 10, 2013 #2
    Number 2- the closure of B is [0,1] because it's the smallest closed set that contains B and it's limit points which are 0 and 1.

    Number 4- There's a little error. Solving for x: [tex]1-x^{2}=0\Longleftrightarrow1=x^{2}\Longleftrightarrow x=\pm1[/tex] . So in your case you'll have to choose [tex]\mathbb{R}{-1,1}[/tex]
     
  4. Dec 10, 2013 #3

    Will [100, 101] work for 4?
     
  5. Dec 10, 2013 #4

    jbunniii

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    Yes, that works fine. Indeed, any compact (closed and bounded) interval on which the function is continuous will work. So you just need to avoid the two points where the denominator is zero.
     
  6. Dec 10, 2013 #5

    jbunniii

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    In fact, 0 and 1 are not the only limit points of ##B##. Every point in ##[0,1]## is a limit point of ##B##.
     
  7. Dec 10, 2013 #6
    Yes i know, what I meant to say is that 0 and 1 or the two missing limit points so the closed interval [0,1] is the smallest closed interval that contains all of the limit points of (0,1).
     
  8. Dec 10, 2013 #7

    jbunniii

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    But ##0## and ##1## are not the only limit points missing from ##B##. All of the irrationals in ##(0,1)## are also limit points of ##B## which are not in ##B##. The right answer is still ##[0,1]##, but it's important to recognize that ##[0,1]## is providing a lot more than two missing points.
     
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