$\left(\forall \varepsilon >0: |a-b| < \varepsilon\right) \iff a=b$ vs. $\left(\forall \varepsilon > 0: a \le b + \varepsilon \right) \iff a \le b$
How does $\left(\forall \varepsilon >0: |a-b| < \varepsilon\right) \iff a=b$ relate to $\left(\forall \varepsilon > 0: a \le b + \varepsilon \right) \iff a \le b$? Does one equivalence imply the other? Are they equivalent?
I feel they're related because they're both equivalences, they both involve $\varepsilon > 0$, and they both involve inequalities.
1 answer
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.
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