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I do not understand the end of Theorem 29.9, on pages 273-4 in Steven Lay's Analysis with an Introduction to Proof (4th edn 2005).

U(f) ≤ U(f,P) < L(f,P) + ε ≤ L(f) + ε

And then it says: "Since ε > 0 is arbitrary, we must have U(f) ≤ L(f)". I don't have any trouble understanding the rest of the proof but I don't understand how we get from the inequality above to U(f) ≤ L(f). Can someone explain to me the reason why we can remove ε and get to U(f) ≤ L(f)?

I've added a picture of the full proof to make it easier to understand my question.

picture of the full proof

Steve Joe
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If you have two (real) quantities $A$ and $B$ such that $$ A\le B+\epsilon $$ for every $\epsilon>0$, then you must have $$ A\le B $$ since otherwise ($A>B$), you can find some $\epsilon_0>0$ such that $$ A>B+\epsilon_0 $$

(For instance, you can take $\epsilon_0=\frac{A-B}{2}$.)

In your example $A=U(f)$ and $B=L(f)$.

  • This has been asked and answered many times before. With >3K reputation you have the privilege to cast close votes, e.g. as a duplicate. Compare https://math.stackexchange.com/help/privileges/close-questions. – Martin R Jan 28 '21 at 18:45
  • MartinR: OP tried to understand a piece of the proof; and the way OP asked the question shows that they don't know the underlying (simple) principle. This particular question was not asked and answered before. Your linked question, which asked for intuition of a known principle, and this one are closely related though. –  Jan 28 '21 at 19:01
  • Maybe I'm being dumb, but if we have A ≤ C < D + ε ≤ B + ε (A, B, C, D being real numbers), wouldn't that give us A < B + ε rather than A ≤ B + ε ? – Steve Joe Jan 28 '21 at 19:23
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    @SteveJoe $A<B+\epsilon$ implies $A\le B+\epsilon$. Remember that $X\le Y$ means "$X<Y$ or $X=Y$". –  Jan 28 '21 at 19:25
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    Ah I see. Thank you for your help – Steve Joe Jan 28 '21 at 19:27
  • @SteveJoe: you are welcome. –  Jan 28 '21 at 19:29