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Real Analysis Convergence Question

  • Thread starter Askhwhelp
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
vanhees71
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That's a pretty simple question. What have you thought about it so far?
 
  • #3
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That's a pretty simple question. What have you thought about it so far?


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
 
  • #4
Office_Shredder
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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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