Can the choice of epsilon be arbitrary in epsilon-delta proofs?

Question

I've been reading Spivak's chapter on limits and something that I don't feel I understand entirely is how the epsilon is decided upon. It makes sense to me in the context of $\,|f(x)-L|<\epsilon$ where it appears just on it's own (representing any positive number) but Spivak seems to invoke an arbitrary value of epsilon in some of the given proofs. For example in proving that: $$ \lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}[f(x)]+\lim_{x\to a}[g(x)]$$ he says if: $$\lim_{x\to a}f(x)=l \;\; \text{and} \;\;\lim_{x\to a}g(x)=m$$ then for any $\epsilon>0$ then there are $\delta_1,\delta_2>0$ such that for all $x$: $$ 0<|x-a|<\delta_1 \implies |f(x)-l|<\frac{\epsilon}{2} \\ 0<|x-a|<\delta_2 \implies |g(x)-m|<\frac{\epsilon}{2} $$ He then carries on to show that $|(f+g)(x)-(l+m)|<\epsilon$. I feel like I understand the proof he gives but I just wanted to clarify whether it matters how he defined what the epsilon was. Since for example if he had started with expressions without the $\frac{\epsilon}{2}$ but rather: $$0<|x-a|<\delta_1 \implies |f(x)-l|<\epsilon$$ wouldn't the outcome be $|(f+g)(x)-(l+m)|<2\epsilon$; which I assume still says that it's bounded, and since $\epsilon$ was any positive number it shouldn't matter. Though I'm not certain, so I suspect there's an error in my understanding.

Another example is if you tried proving that:

$$\lim_{x\to 1} 2x-2=0$$ then instead of writing just epsilon, one wrote some positive constant $c$ times epsilon so: $$ |(2x-2)-0|<c\times\epsilon\;\; \rightarrow \;\; |x-1|<\frac{c\times \epsilon}{2}$$ therefore let $\delta =\frac{c\times\epsilon}{2}$ which would still seem to prove it even if we hadn't had the $c$ that is with $\delta=\frac{\epsilon}{2}$ . I know it's not a proper proof but hopefully you see what I'm trying to say (or indeed what I'm doing incorrectly).

Hopefully my question isn't too badly written or that there isn't one asking the same thing, I have found this which is from the same part of the book but I still don't feel sure whether or not the choice of epsilon changes the validity proof.

Answer 1

$\varepsilon$-$\delta$ proofs seem to be the most confusing concept of first-year Calculus (or pre-Calculus, depending on how it's taught).

The $\varepsilon$-$\delta$ definition of a limit is as follows. $\lim_{x\rightarrow a} f(x) = l$ means for all $\varepsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$ then $|f(x) - l| < \varepsilon$. All your answers are contained within this definition so let's break it down and see how all the proofs you reference satisfy the definition.

For the limit to be $l$, we require that every $\epsilon > 0$ has a corresponding $\delta > 0$. A standard way to prove that something holds for all values of a variable in a certain domain is to choose an arbitrary (unspecified) value for that variable in its domain and make a proof that works for that arbitrary choice. Spivak's proof does exactly that: he says "for any $\varepsilon > 0$" and then proceeds to show that there is a $\delta$ corresponding to this specific but arbitrary $\varepsilon$. The unstated conclusion is that there is a $\delta$ for every $\varepsilon$.

Note that the proof of a limit requires that there is a $\delta$ corresponding to every positive "epsilon", but there is no requirement to denote the "epsilon" value as the symbol $\varepsilon$. Instead, we may use an expression such as $c\varepsilon$ for constant $c>0$ so long as this expression can take on all values in the domain of the "epsilon" (namely, all positive numbers). Clearly we may write any positive number as $c\varepsilon$ for some choice of $\varepsilon$. So if we prove that a $\delta$ exists for all $c\varepsilon$ in place of $\varepsilon$, the proof holds for every value that $c\varepsilon$ can take, which means it holds for every positive number, so we are good.

However, if the expression we choose can't take on all possible "epsilon" values, such as $1+\varepsilon$ for $\varepsilon > 0$, then if we prove that a $\delta$ exists for all $1+\varepsilon$ then we haven't proven that a $\delta$ exists for all values for which the definition of limit requires it exist. Such a proof would be invalid.

Finally, you posted a link to another question where a complex expression $\min\left(1,\frac{\epsilon}{2(|m|+1)}\right)$ stands in for the "epsilon". Note that by letting $\epsilon$ vary, this expression can take on all positive values up to but not exceeding 1. Therefore it doesn't take on all values that the definition of limit requires. However, if a $\delta$ works for some "epsilon" then it automatically works for any larger "epsilon" - you can immediately verify this statement in the definition of a limit. So if our expression can take on all positive numbers up to 1, that's actually good enough. (In the most generality, we can prove that there is a $\delta$ corresponding to all "epsilon" in a set of positive numbers containing a sequence going to 0.)

Answer 2

This is not so much a stand-alone answer as an appendix to jef's excellent answer above.

I think of $\varepsilon$-$\delta$ proofs as "challenge-and-response." The challenge is $\varepsilon$: "Can you find a $\delta$ so that ...?" The response is $\delta$: "Yes, I can!"

The proof is a strategy that will guarantee you an answer to the challenge, no matter what (positive) number you're handed.

Answer 3

What Spivak calls an "${\epsilon}/{2}$ argument" is presented in some other books as a "$2\epsilon$ argument" where the sum of two things less than $\epsilon$ is bounded by $2\epsilon$.

It might seem nicer to have the final bound be exactly $\epsilon$, but the second approach is more flexible when you get to more complicated bounds like $\epsilon^5 + 7\sqrt{\epsilon}$ that do not have a simple algebraically described $\delta$ that can be chosen at the start to get $\epsilon$ at the end.