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I just watched this youtube video: http://www.youtube.com/watch?v=K4eAyn-oK4M

He lays out his objections against the $ϵ,δ$ definition around 14 min.

Here is the discription of the video:

In this video we aim to give a precise and simpler definition for what it means to say that: a rational polynumber on-sequence p(n) has a limit A, for some rational number A. Our definition is both much simpler and more logical than the usual epsilon -delta definition found in calculus texts. What is required is that we need to find two natural numbers: k called the scale, and m called the start that allow us to bound in a pretty simple way the difference between p(n) and A.

The epsilon-delta definition of a limit is usually considered a high point of logical rigour. Not so. It is also considered too logically involving to be taken seriously as a pedagogical pillar for most undergrads. Hence students may be told about the definition, but are not required to seriously understand it, or be able to use it--unless they are prospective maths majors.

There is a subtle ambiguity in the definition: given an epsilon we are supposed to demonstrate there is a delta (with certain properties) but how are we to do this, since an potential infinity of epsilons are involved? In practice what is required is a correspondence (function/relation etc) between epsilon and delta but the nature of this required correspondence is not clear. We return to our familiar conundrum of using the work``function'' without a proper definition of it.

The key point that makes our simpler more intuitive notion of limit of a sequence work is that we are dealing with very particular and clearly defined on-sequences: those generated by a rationl polynumber. A good example of the benefits of being careful rather than casual when dealing with the foundations of analysis!

My question is: Is this an opinion shared by more mathematicians ? I kind a feel like that this Professor of the University of New South Wales is standing completely alone as it comes to this. I don't really undestand his objections, but I don't think I'm skilled enough to understand if his objections are legit.

Hanul Jeon
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Kasper
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    There is no ambiguity in $\epsilon-\delta$. – Thomas Andrews Apr 18 '13 at 12:51
  • We don't have to find a $\delta$ --- we just have to demonstrate that one exists! – JP McCarthy Apr 18 '13 at 12:52
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    He seems to think that we have to come up with an assignment $\varepsilon \mapsto \delta$... not so. – Clive Newstead Apr 18 '13 at 12:55
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    And with the axiom of choice, such a function exists. – Karl Kroningfeld Apr 18 '13 at 12:55
  • @user1: Yep. But we don't need the axiom of choice to make sense of the $\varepsilon\delta$ definition of continuity. – Clive Newstead Apr 18 '13 at 12:56
  • 19.08 on the clock on the Youtube Clip
    1. A sequence in a set $A$ is a function $s:\mathbb{N}\rightarrow A$??
     – JP McCarthy Apr 18 '13 at 13:00
  • what exactly is the problem in your 2nd-to-last paragraph? In math, we work with "infinities" on a regular basis; it is true that we can not explicitly procure them (as that would take an infinite amount of time) but we certainly may refer to them and describe them. Consider an infinite sequence or sum which we describe by a formula. So too here, $\delta$ will usually depend on $\epsilon$, and so we may describe an infinite amount of $\delta$s by one formula. – Coffee_Table Apr 18 '13 at 13:06
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    That's the same guy who said that infinite sets don't exist. Of course he will have a problem with $\varepsilon$-$\delta$ definitions. – Asaf Karagila Apr 18 '13 at 13:06
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    It is also considered too logically involving to be taken seriously as a pedagogical pillar for most undergrads. It's hard not to read this as "undergrads are too dumb to understand that." I think that is manifestly wrong: they can understand it. The definition is both transparent and accessible. I do not claim that it is obvious or that an untrained person should understand it immediately, I just mean that in a reasonable amount of time anyone can be convinced it captures the intuitive meaning. I would take things this prof says with a grain of salt. – rschwieb Apr 18 '13 at 13:12
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    There seems to be a discussion at xkcd about this individual, if I have not made some gross mistake. – rschwieb Apr 18 '13 at 13:19
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    @rschwieb: This fellow is a bit of a loon, but to many undergraduates the $\epsilon$-$\delta$ definition of continuity is anything but transparent, and in many calculus courses it is discussed briefly or not at all. – Brian M. Scott Apr 18 '13 at 19:37
  • @BrianM.Scott As I said, not obvious or immediately understood, referring to the segment of students grappling with limits (and I gather, the segment you are thinking of.) For a student who "gets" what a limit is, it is hard to imagine a logical definition which is more transparent, (but you might surprise me ?). In case it is still slipping by, I'll reemphasize that I am definitely not saying (and didn't say) that the definition is simple or easy to learn/teach to anybody. – rschwieb Apr 18 '13 at 20:14
  • @rschwieb: You greatly underestimate its difficulty for many students. $\forall\exists\forall$ statements are inherently complicated, and the right-to-left formulation causes more difficulty. That’s one of the reasons people have experimented with other approaches, among them non-standard calculus (Keisler) and nearness (Cameron, Hocking, & Naimpally), that allow an intuitively more natural ‘covariant’ definition. – Brian M. Scott Apr 18 '13 at 20:21
  • @BrianM.Scott I think any reasonable person reading my last two comments will have to concede that I have gone out of my way to say that it is not necessarily easy to learn. I certainly did not understimate the difficulty for students, but I always seem to underestimate my ability to evoke such patronizing comments from you :/ At any rate, thanks for the references to nonstandard approaches... I'll have to take time to look at them. – rschwieb Apr 18 '13 at 20:34
  • Maybe "not easy" is not dramatic enough for some, but for me "not easy=hard". – rschwieb Apr 18 '13 at 20:39
  • @rschwieb: Seriously? For me there’s a long continuum between easy and hard, and not easy definitely does not equate to hard. – Brian M. Scott Apr 18 '13 at 20:40
  • @BrianM.Scott If you put yourself in the shoes of an undergraduate, the "not easy=hard" equation holds. :) For the rest of us, we have enough experience with a range of problems to have more shades. If that was the cause of this prolonged exchange, then I apologize. I just meant to convey the definition isn't trivial to understand, not that it is "slightly harder than easy". – rschwieb Apr 18 '13 at 20:42

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