I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:
If $a \le b + \varepsilon$ for all $\varepsilon > 0$, then $a \le b$.
Essentially, I am wondering why the theorem assumes $a \le b + \varepsilon$ rather than $a < b + \varepsilon$. Of course, the former follows from the latter, so nothing is lost. But I don't really see what the $=$ case is providing here.
When Apostol uses the theorem in proving Theorem 1.15, he has the situation $a+b < c + 2 \varepsilon$. I'm wondering if this is generally the case when Theorem 1.1 is used in analysis, i.e. if one typically has $a < b +\varepsilon$ rather than $ a \le b + \varepsilon$ in other applications of the theorem as well.
The result that $a\leq b+\epsilon$ for all $\epsilon>0$ implies $a\le b$ can be useful in showing for example that there is no smallest positive number (in this case, just set $b=0$). It doesn't matter whether we use $a\le b+\epsilon$ for all $\epsilon>0$ or $a<b+\epsilon$ for all $\epsilon>0$, because the two conditions can be shown to be equivalent.