Yesterday, while I was browsing the new questions, I came across this one How is analysis beautiful? -- confusion from an algebraist, which brought up an interesting question, especially for the people who are just starting out with analysis: are the "annoying" epsilons everywhere in analysis, even in the really advanced stuff? Does an actual analyst work with them on a daily basis or are they useful just for establishing the fundamental results in analysis (i.e. what is covered in university)?
I agree with the person that posted the other question that this may be really interesting for some aspiring maths majors (myself included) and that's why I decided to post this question, especially since I have been thinking about quite similar things too (as it can be seen here What should I study next if I enjoy thinking about problems such as these?).
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2Well, many topics in analysis require to use limits. A limit is defined using $\epsilon$. But it doesn't mean you always have to work strictly with the definitions. For example, I don't think anyone really computes Riemann integrals using Riemann sums. As you go farther, you learn more theorems and they give you more tools you can use. Anyway, I don't find all the work with $\epsilon$ annoying. It can be pretty fun sometimes. – Mark Sep 06 '20 at 11:10
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1I'm not an analyst so don't feel qualified to write a full answer, but the impression I got from talking to some/taking different sorts of graduate level analysis courses is that it depends to some degree on the branch/type of analysis. Terry Tao has a blog post about "hard analysis" and "soft analysis" that he references from time to time. "Hard analysis" encompasses the flavors where epsilons come up. – Mark S. Sep 06 '20 at 11:54
1 Answers
Suppose one is working with objects which are hard to pin down. You do not know their nature very well, and it seems impossible to understand them perfectly. What do you do next? Well, the next best thing would be to control the behaviour of the object in some way.
A very simple case of this would be: Suppose you want to solve $f(x) = g(x)$ for some functions $f$ and $g$, and suppose such a problem is very difficult. Well, try to show the next thing which is a little weaker: show that for any degree of accuracy want, the difference between $f$ and $g$ can be fit to this degree of accuracy. More precisely, given $\varepsilon >0$ show that we can find $x$ such that $| f (x) - g(x) | < \varepsilon$.
Seems like a worthwhile pursuit, but maybe that's just me...
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