ε-neighborhoods of distinct real numbers are disjoint

[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

 

[vanhees71]

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

 

[Askhwhelp]

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

 

[Office_Shredder]

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.