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Real Analysis Convergence Question

  1. Oct 12, 2013 #1
    show that if a and b are distinct real numbers, then there exists a number ε > 0 such that the ε -neighorboods Vε (a) and Vε (b) are disjoint.

    How to solve this question?

    Thank you
  2. jcsd
  3. Oct 13, 2013 #2


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    That's a pretty simple question. What have you thought about it so far?
  4. Oct 13, 2013 #3

    would it be the following?

    let |b-a|/2 = ε
    Assume x ∈ Vε (a) and Vε (b).
    |b-a| = |(b-Xn)+(Xn-a)| ≦ |(b-Xn)| + |(Xn - a)| < |b-a|/2 + |b-2|/2 = |b - a|
    A contradiction
  5. Oct 14, 2013 #4


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    Your notation isn't great - You use an Xn when you previously defined x, and you have a |b-2| when you probably mean |b-a|, but other than that it looks good to me.
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