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Math Help - open/closed intervals

  1. #1
    Member Jskid's Avatar
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    open/closed intervals

    Find A^c with respect to U=R in the following case. A=(1, \infty) \cup (- \infty, -2]

    The answer is ( - \infty , a] \cup [b, \infty) How do you know if it's an open interval or a closed interval?
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  2. #2
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    Re: open/closed intervals

    Since open sets do not contain their boundary, their compliment must contain all points that are in the boundary of the original set (where open) and not contain the boundary when it's there (when closed).

    Think of the interval as a set of points. Note that -2 is in A and 1 is not in A. This means A^c must contain -2, but can't contain 1.

    Your answer is not correct. The answer is (-2, 1]. -2 can't be in A^c because it's in A. 1 must be in A^c because it isn't in A. Is that clear? The compliment is just the set of points that aren't in its base.
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  3. #3
    Member Jskid's Avatar
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    Re: open/closed intervals

    Yes that is clear, just one point is not. Why must it be an open interval for + \infty, - \infty?
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    Re: open/closed intervals

    Quote Originally Posted by Jskid View Post
    Yes that is clear, just one point is not. Why must it be an open interval for + \infty, - \infty?
    You were told that you were only dealing with real numbers, and + \infty and -\infty are not real numbers. Therefore they are neither in A, nor its compliment.
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  5. #5
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    Re: open/closed intervals

    Quote Originally Posted by Jskid View Post
    Why must it be an open interval for + \infty, - \infty?
    What does "to be an open interval for" mean? I also assume you mean (-\infty,+ \infty). Finally, this question is not related to the problem in the OP, is it?
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    Re: open/closed intervals

    Quote Originally Posted by emakarov View Post
    What does "to be an open interval for" mean? I also assume you mean (-\infty,+ \infty). Finally, this question is not related to the problem in the OP, is it?
    They mean that infinity always appears as an open interval, I presume. Unless you're using the extended real line (which adds one extra point for each of -\infty and +\infty), those values are not real numbers and as such they cannot be in the compliment.

    Another way of saying that: the real line has no greatest or least element. It cannot be both closed and unbounded at the same time in the same direction, so "infinity" will never appear as the closed endpoint of an interval on the real numbers.
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