The absolute value version is equivalent to having both $a < b + \varepsilon$ and $a > b - \varepsilon$, both for all positive epsilons. To see this, use the definition of the absolute value.
Also, $a < b + \varepsilon$ for all $\varepsilon > 0$ and $a \le b + \varepsilon$ for all $\varepsilon > 0$ are equivalent. Clearly the strict version implies the non-strict one. For the other way, since the claims hold for all positive epsilons, putting $\varepsilon/2$ in the non-strict inequality implies the strict one with epsilon. Arbitrary epsilon, holds for all of them, as usual.
To get a handle on these, I recommend studying any course or book in rigorous analysis with the epsilon-delta method. These are very standard arguments.
tommi
·
2023-11-02T08:57:23Z (about 1 year ago)