# Epsilon delta limits if/then language

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1. Jun 8, 2015

### KyleTS

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

This is from the question list at the UC Davis Website epsilon delta exercise list.

In the exercise list we have:

Prove that

Which concludes with:

Thus, if , it follows that [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img13.gif. [Broken] This completes the proof.

My curiosity is here: This concludes with iff-then rhetoric. IFF , THEN [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img13.gif. [Broken] Which is true here as was proven. However, I don't see anything which explicitly demonstrates that for all real ε>0 with ε/3, there is an x in the domain of f(x), such that [PLAIN]https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/img20.gif. [Broken] This seems intuitively obvious, and likely easy to demonstrate in this case, however if the δ = (function of ε) is sufficiently complex, it seems as though verifying 0 < |x-10| < δ may require another epsilon-delta proof itself. That is to say, the proof here does show that IFF, THEN, however I don't see anything which demonstrates the IFF is actually true.

Have I missed something here? Thanks in advance, everyone!

Last edited by a moderator: May 7, 2017
2. Jun 8, 2015

### PeroK

I'm not sure I understand what's confusing you. If $\delta$ was very complex, you might have to justify that $\delta > 0$, but otherwise $f$ must be defined on an open interval containing the limit point (although not necessarily defined at the limit point itself). So, whatever $\delta > 0$ you have there is a set of relevant $x's$.

I was going to say that the "iff" argument is slightly spurious, since you won't always have iff. I would tend to use if. In fact, he runs into trouble on solution 4, as iff is no longer valid. iff will only be valid on linear functions or functions which are 1-1 as far as the limit point is concerned. If that makes sense.

Last edited: Jun 8, 2015
3. Jun 8, 2015

### KyleTS

Okay, now bear with me. Insofar as I understand it, in order to say f is continuously defined on an interval we must establish lim x=>x1 f(x) = f(x1) for all x in the interval. However that means, in this example, 0 < |x-10| < ε/3 is true for all real ε>0 because f(x) is continuous over the interval containing the limit point, however continuity is defined in terms of limits. Hence in order to show f(x) is continuous over the interval containing the limit point, we must use another limit -- but then that limit proof implicitly includes limits in its definitions.

That is, the concluding iff (if?) - δ inequality is established using continuity over the interval, but continuity is defined in terms of limits, hence our limit definition appears to be defined in terms of itself, or recursively. That is when you say "f must be defined on an open interval containing the limit point" in order to prove that we have to use another limit (that is, to show it satisfies the definition of limits and continuity.)

Last edited: Jun 8, 2015
4. Jun 8, 2015

### PeroK

No, I don't understand your argument at all, I'm afraid. The continuity or otherwise of $f$ is not used anywhere in the definition of the limit: only that $f$ be defined in a neighbourhood of the limit point.

5. Jun 9, 2015

### KyleTS

Thank you, I think I'm starting to have the perspective shift I needed.

I came across an article that somewhat touched on what's confusing me.

"Perhaps our students would have less trouble trying to understand the Cauchy-Weierstrass definition if we told them in advance that it was not a formalization of their intuitive conception -- that the mathematician's formal notion of a continuous function is in fact something quite different from the intuitive picture. Indeed, that might help. But if we are getting into the business of open disclosure, we had better go the whole way and point out that the new definition does not explicitly capture continuity at all. That famous -- indeed, infamous -- epsilon-delta statement that causes everyone so much trouble does not eliminate (all) the vagueness inherent in the intuitive notion of continuity. Indeed, it doesn't address continuity at all. Rather, it simply formalizes the notion of "correspondingly" in the relation "correspondingly close." In fact, the Cauchy-Weierstrass definition only manages to provide a definition of continuity of a function by assuming continuity of the real line! "

http://www.cogsci.ucsd.edu/~nunez/web/ESM.PDF

6. Jun 11, 2015

### Fredrik

Staff Emeritus
The definition of "limit" tells us that what it means for a function $f$ to have a limit at $a$, only if $a$ is a limit point of the domain of $f$. The definition of "limit point" says that $a$ is a limit point of a set $E$ if for all $\delta>0$, there's an $x\in E-\{a\}$ such that $|x-a|<\delta$. (Note that $x\in E-\{a\}$ implies $x\neq a$, which implies $0<|x-a|$). The requirement that $a$ must be a limit point of the domain of $f$, is part of the definition of "limit" specifically to deal with the issue that concerns you.

Last edited: Jun 11, 2015
7. Jun 11, 2015

### Fredrik

Staff Emeritus
I find some of these comments pretty odd, in particular the claim that the definition of "continuous" doesn't address continuity at all. It defines what we mean by "continuous", so of course it does.

But there is a funny thing about these definitions. Mathematicians obviously had a pretty good idea what functions they wanted to call "continuous" before the formal definition was invented. So the formal definition is what it is because it ensures that the "right" functions end up being called continuous. Because of this, I would say that a proof of the fact that the f defined by f(x)=3 for all x is continuous isn't really saying anything about f. What the proof tells us is just that the definition that those mathematicians chose for us isn't completely stupid.

Last edited: Jun 11, 2015
8. Jun 11, 2015

### KyleTS

I want to give you a giant hug right now.