Cantor Confusion
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From: David Marcus on 17 Jan 2007 16:40 Dik T. Winter wrote: > In article <MPG.20174de97ffb81ac989b5e(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > > Dik T. Winter wrote: > > > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Did you receive and enjoy my book which I sent off on 10. Jan.? > > > > > > Not yet. > > > > You didn't receive it or you didn't enjoy it? > > I didn't receive it and I didn't enjoy it (how could the last happen if > the first did not happen?). It couldn't. Although, I guess you could have enjoyed not receiving it. -- David Marcus
From: Andy Smith on 17 Jan 2007 16:49 MoeBlee writes >> >> I think this is a bit (actually quite a bit) too formal considering >> Andy's mathematical background. You have to walk before you can run. >> >> -- > >Gulp. That IS walking. > >MoeBlee > Thanks MoeBlee ... You are comfortable in the formalism and paradigm that you have been taught, but everything that you know rests on the shoulders of giants. Given a clean slate, could you create infinite set theory and a systemic formalism from the ground up? If not, you can cut me some slack. Here's a thought for you. Possibly you are so locked into your paradigm that you cannot think of any thought not expressed in its terms. So maybe the only way in which you get something new is when some neanderthal like me blunders in and asks some stupid questions. Of course intelligence comes into it and I cheerfully accept that the probability of me making any useful contribution has a lot of 0's following the decimal point. I only discovered usenet a week or so ago, but the sci.math forum is something of a revelation. Snakes on a plane! You have sects, schisms and heretics, massive egos ... no quiet, polite, thoughtful exchanges coming to a common consensus, but ad hominem the norm. All very amusing to an idealist. :) -- Andy Smith
From: David R Tribble on 17 Jan 2007 16:50 MoeBlee wrote: >> f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) >> -> y = z)) > David Marcus wrote: >> I think this is a bit (actually quite a bit) too formal considering >> Andy's mathematical background. You have to walk before you can run. > MoeBlee wrote: > Gulp. That IS walking. Be fair. It's not immediately obvious that f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) -> y = z)) is read as f is a function implies that f is a relation and for all x,y,z it holds that <x y> in f and <x z> in f implies y = z Things like "Axyz" look like words upon first scan. It does take a little experience to know that it is "for all x,y,z" and that "A" is an upside-down "A" (for all). Likewise "xeN" is "x in N", where that little "e" is the non-ASCII "member-of" symbol similar to an epsilon (but isn't). And then there is epsilon, also written as "e". You can see how novices can be overwhelmed by this at first, can you not?
From: David Marcus on 17 Jan 2007 16:52 Andy Smith wrote: > David Marcus writes > >Ah, you are at least a hundred years behind the times. No, a function is > >most definitely not a formula. A function is a rule which assigns, to > >each of certain real numbers, some other real number. For example, the > >rule that assigns to each number a the number 0 if a is irrational and > >the number 1 if a is rational is a function, but you will have a hard > >time coming up with a formula (nor is a formula required). > > Thanks for that, but actually it appears "formula" has a technical > meaning for you. rule and formula are synonyms for me. I don't think "formula" has a precise meaning, but we generally use it to mean some sort of expression, e.g., polynomials, trig functions, exponential functions, a combination of these. If we define a function using cases, e.g., f(x) = 17, if x < 0, -3.14, if x > 0, then we usually don't consider this to be a formula. More formally, a function is a set of ordered pairs such that if (a,b) and (c,d) are both in the function and a = c, then b = d. Have you seen that definition? > >> Well that is undoubtedly the answer to the question. Why doesn't the > >> Taylor expansion work might be a better way of phrasing the question? > > > >An interesting question. It turns out that the behavior of the function > >for complex values of x prevents the Tayor series from working for real > >values. > > OK, thanks, I will look that up I guess. Bit it is odd superficially > in that e(-1/x^2) is super smooth with all its derivatives etc. Yes. But, if you replace x by ix, then you get e^{1/x^2}. So, if you come down the imaginary axis, the function isn't continuous. > >You might enjoy looking at the book Calculus by Michael Spivak. I'm sure > >that it explains Calculus in a way that is very different from how you > >learned it. > > Thanks I might well do that. Besides discussing functions, Spivak also discusses why the Taylor series of e^{-1/x^2} doesn't work. -- David Marcus
From: David Marcus on 17 Jan 2007 17:12
MoeBlee wrote: > David Marcus wrote: > > MoeBlee wrote: > > > Andy Smith wrote: > > > > > > > Doubtless you lot have some legally watertight definition. As far as I > > > > am concerned, a function is a formula that provides a value for a given > > > > input (or inputs, in a multidimensional situation). > > > > > > Dump that definition. It's picture language. Use mathematical > > > definitions instead: > > > > > > {x y} = z <-> At(t in z <-> (t=x v t=y)) > > > > > > {x} = {x x} > > > > > > <x y> = {{x} {x y}} > > > > > > p is an ordered pair <-> Exy p = <x y> > > > > > > r is a relation <-> Ax(x in r -> r is an ordered pair) > > > > > > f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) > > > -> y = z)) > > > > I think this is a bit (actually quite a bit) too formal considering > > Andy's mathematical background. You have to walk before you can run. > > Gulp. That IS walking. Not for an engineer. And, not if you've never taken a math course that does proofs. -- David Marcus |