Apostol《数学分析》定理1.1应用疑问：为何假设a≤b+ε而非a<b+ε

Great question—this is a common point of confusion, but it boils down to generality, practicality, and logical equivalence. Let’s unpack it:

• The two conditions are logically equivalent
  You’re right that a < b + ε implies a ≤ b + ε, but the reverse is also true when the condition holds for all ε > 0. If a ≤ b + ε is true for every positive ε, pick any ε and use a smaller value ε' = ε/2. Since a ≤ b + ε' and b + ε' < b + ε, this means a < b + ε. So the statements "for all ε > 0, a ≤ b + ε" and "for all ε > 0, a < b + ε" are actually identical in meaning.

• a ≤ b + ε is more natural in real proofs
  When working through analysis derivations, it’s often easier to end up with a ≤ b + ε than the strict version. For example, if you have a ≤ c and c ≤ b + ε, transitivity gives you a ≤ b + ε directly. Using ≤ in the theorem’s hypothesis lets you apply it immediately without needing an extra step to convert the inequality to strict form.

• It avoids unnecessary edge-case busywork
  If we used a < b + ε instead, the theorem would still be valid, but it might force you to add trivial justifications when you encounter scenarios where equality could hold (even if it doesn’t in the limit). Using ≤ keeps the statement cleaner and more inclusive, covering both strict and non-strict cases in one go.

At the end of the day, either condition would work, but a ≤ b + ε is a more flexible and natural choice for the contexts where this theorem gets applied.

内容的提问来源于stack exchange，提问作者John Van Dyke