(Tricky) Absolute Value Inequalities

[vertciel]

Hello everyone,

I'm posting here since I'm only having trouble with an intermediate step in proving that

\sqrt{x} \text{ is uniformly continuous on } [0, \infty].

By definition, |x - x_0| < ε^2 \Longleftrightarrow -ε^2 < x - x_0 < ε^2 \Longleftrightarrow -ε^2 + x_0 < x < ε^2 + x_0

1. How does this imply the inequality in red?

\text{ Since } ε > 0 \text{ then } x_0 - ε^2 < x_0

However, I do not know more about x₀ vs x.

2. Also, how does the above imply the case involving the orange; what "else" is there?

Thank you very much!

 

[Mark44]

The inequality |x - x₀| < ε² doesn't specify whether x is to the right of x₀ or to the left of it. That's the reason for the two inequalities.

 

[vertciel]

Thank you for your response, Mark44.

Could you please explain the red box?

 

[SammyS]

Thank you for your response, Mark44.

Could you please explain the red box?

It looks like that's exactly what he did !