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Math Help - Advanced Calculus - Epsilon-Neighborhoods

  1. #1
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    Advanced Calculus - Epsilon-Neighborhoods

    This is from a real analysis textbook so I'm not 100% sure if it should go in here or in analysis.

    The problem:
    Show that if a,b belong in R, and a=/=b, then there exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals to an empty set.

    This is what I have so far:
    Assume that there does not exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals an empty set. Let, 0< epsilon < absolute value of (1/2(b-a)). Therefore, since the intersection of U and V does not equal an empty set, there must exist a y with which y-a < epsilon and y-b < epsilon.

    Then I calculated the bounds of the union of all epsilon neighborhoods within the range of 0< epsilon < absolute value of (1/2(b-a)) giving U the bounds 1/2(3a - b) and 1/2(a+b); V the bounds 1/2(3b-a) and 1/2(a+b).

    Am I on the right track? I know I have to somehow prove that if y-a < epsilon and y-b < epsilon then it cannot be within both of those bounds, but I don't how to do that...
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  2. #2
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    Just take \varepsilon=\frac{|b-a|}{4}.
    That single value works for both.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Just take \varepsilon=\frac{|b-a|}{4}.
    That single value works for both.
    Just out of curiosity, maybe I am reading the problem wrong but why does \delta=\frac{d(a,b)}{2} work? I am only asking because usually one picks the minimum radius for the open balls to be disjoint.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Just out of curiosity, maybe I am reading the problem wrong but why does \delta=\frac{d(a,b)}{2} work? I am only asking because usually one picks the minimum radius for the open balls to be disjoint.
    Yes it works. But there are times I want their closures to be disjoint.
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  5. #5
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    So taking this new approach where, \varepsilon=\frac{1}{2}|b-a|. I find that the two sets are:

    U:=\{x \epsilon R : |x-a| < \frac{1}{4}|b-a|\}
    V:=\{y \epsilon R: |y-b| < \frac{1}{4}|b-a|\}

    To prove they are disjoint, I would have to find that no element of U can be found in V.

    So I let x_1 \epsilon U.

    Let us assume that x_1 \epsilon V, also.
    Therefore, I have to prove that |x-b| > \frac{1}{4} |b-a| to find the 'contradiction'.

    Or am I thinking about it too complicated again?
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  6. #6
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    Suppose that z \in U \cap V then
    \left| {a - b} \right| \leqslant \left| {a - z} \right| + \left| {z - b} \right| <\frac{{\left| {a - b} \right|}}{4} + \frac{{\left| {a - b} \right|}}{4} = \frac{{\left| {a - b} \right|}}{2}

    Do you see the contradiction?
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  7. #7
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    Yeah, I was just about to post that I got it. Triangle Inequality, right?

    Thanks!
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