1. Oct 12, 2013 #1

   Askhwhelp

   

   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

    

   Askhwhelp, Oct 12, 2013
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3. Oct 13, 2013 #2

   vanhees71

   

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   That's a pretty simple question. What have you thought about it so far?

    

   vanhees71, Oct 13, 2013
4. Oct 13, 2013 #3

   Askhwhelp

   

   
   
   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

    

   Askhwhelp, Oct 13, 2013
5. Oct 14, 2013 #4

   Office_Shredder

   

   Staff Emeritus

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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.

    

   Office_Shredder, Oct 14, 2013

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