Possible proof of Gabriel's Theorem?
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From: Robert Israel on 13 Mar 2005 00:31 In article <kff631tnv1np26gkldur1l07bno1gams37(a)4ax.com>, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >On Sat, 12 Mar 2005 11:17:07 -0500, Zbigniew Fiedorowicz ><fiedorow(a)hotmail.com> wrote: >>Given a nonconstant everywhere differentiable function f(x) on the unit >>interval with f'(x)=0 on a dense countable subset D, it is not difficult >>to modify it to one whose derivative vanishes on all the rationals. >>First construct a sequence of C^1 functions g_n(x) from the unit >>interval to itself having the following properties: >>(1) g_n(x) uniformly converge to a function g(x) >(1.5) And g is non-constant. >>(2) g'_n(x) uniformly converge to a function h(x) >>(3) For any fixed rational number r in the unit interval, the >> sequence g_n(r) becomes eventually constant with value in D. >>Then g(x) is easily seen to be C^1 with derivative g'(x)=h(x). >>Then the composite function fg is a nonconstant everywhere >>differentiable function on the unit interval, whose derivative vanishes >>on all the rational numbers. [Note that I am not claiming that (fg)' >>vanishes only on the rationals.] >Well that was pretty simple. You're also not claiming that >g maps the rationals _onto_ D, which is what I was trying >to do for some reason... But that's easy to do too. Let {r_n} and {d_n} be enumerations of the rationals and D respectively. We will also define a sequence of finite subsets A_n of the rationals, where A_0 is empty. g_n will be chosen so that g_n = g_{n-1} on A_{n-1}. If r_n is not in A_{n-1}, then we will ensure that g_n maps r_n to some member of D that is not in g_{n-1}(A_{n-1}). If d_n is not in g_{n-1}(A_{n-1}), then we will ensure that g_n maps some rational q_n to d_n. And we can do this with |g_n - g_{n-1}| + |g'_n - g'_{n-1}| < 2^(-n) everywhere. Then take A_n = A_{n-1} union {r_n, q_n}. Robert Israel israel(a)math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Jesse F. Hughes on 13 Mar 2005 03:52 "William Hughes" <wpihughes(a)hotmail.com> writes: > No. Every epsilon-delta argument has the form: > > Given an epsilon > 0 I can find a delta > 0 > such that if foo < delta, bar < epsilon. Not so! About half of my epsilon-delta arguments have the form Given a delta > 0 I can find an epsilon > 0 such that if foo < epsilon, bar < delta. But this might just be a personal failing. Also, maybe I just thought they were epsilon-delta arguments but they were really delta-epsilon arguments. I can't be sure. -- Jesse F. Hughes "I have put all the information that you need at [a Yahoo! group] where you'll notice a significantly better signal to noise ratio, as I'm just about the only person posting." -- James S. Harris on noise
From: Thomas Nordhaus on 13 Mar 2005 04:50 "Jesse F. Hughes" <jesse(a)phiwumbda.org> schrieb: >"William Hughes" <wpihughes(a)hotmail.com> writes: > >> No. Every epsilon-delta argument has the form: >> >> Given an epsilon > 0 I can find a delta > 0 >> such that if foo < delta, bar < epsilon. > >Not so! About half of my epsilon-delta arguments have the form > > Given a delta > 0 I can find an epsilon > 0 > such that if foo < epsilon, bar < delta. > >But this might just be a personal failing. Also, maybe I just thought >they were epsilon-delta arguments but they were really delta-epsilon >arguments. Boy. Your proofs must be a real pain in the *** to the average mathematician, then. Thomas ;-) > >I can't be sure.
From: David C. Ullrich on 13 Mar 2005 07:10 On 13 Mar 2005 05:31:12 GMT, israel(a)math.ubc.ca (Robert Israel) wrote: >In article <kff631tnv1np26gkldur1l07bno1gams37(a)4ax.com>, >David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >>On Sat, 12 Mar 2005 11:17:07 -0500, Zbigniew Fiedorowicz >><fiedorow(a)hotmail.com> wrote: > >>>Given a nonconstant everywhere differentiable function f(x) on the unit >>>interval with f'(x)=0 on a dense countable subset D, it is not difficult >>>to modify it to one whose derivative vanishes on all the rationals. > >>>First construct a sequence of C^1 functions g_n(x) from the unit >>>interval to itself having the following properties: >>>(1) g_n(x) uniformly converge to a function g(x) > >>(1.5) And g is non-constant. > >>>(2) g'_n(x) uniformly converge to a function h(x) >>>(3) For any fixed rational number r in the unit interval, the >>> sequence g_n(r) becomes eventually constant with value in D. >>>Then g(x) is easily seen to be C^1 with derivative g'(x)=h(x). > >>>Then the composite function fg is a nonconstant everywhere >>>differentiable function on the unit interval, whose derivative vanishes >>>on all the rational numbers. [Note that I am not claiming that (fg)' >>>vanishes only on the rationals.] > >>Well that was pretty simple. You're also not claiming that >>g maps the rationals _onto_ D, which is what I was trying >>to do for some reason... > >But that's easy to do too. Yeah, I realized that a little later. >Let {r_n} and {d_n} be enumerations >of the rationals and D respectively. We will also define a sequence >of finite subsets A_n of the rationals, where A_0 is empty. >g_n will be chosen so that g_n = g_{n-1} on A_{n-1}. >If r_n is not in A_{n-1}, then we will ensure that g_n >maps r_n to some member of D that is not in g_{n-1}(A_{n-1}). >If d_n is not in g_{n-1}(A_{n-1}), then we will ensure that >g_n maps some rational q_n to d_n. >And we can do this with |g_n - g_{n-1}| + |g'_n - g'_{n-1}| < 2^(-n) >everywhere. >Then take A_n = A_{n-1} union {r_n, q_n}. Precisely. Lemme supply a detail you kindly left out, for the benefit of readers who don't see how to ensure the things you say we can ensure (and because I feel kinda stupid, since I didn't see how to do all this a couple weeks ago when I tried, but do see how to do it after Zbigniew's post, even though he didn't say all that much): Start with g_0(x) = 2x. So as long as the increments satisfy the inequalities above we will have g_n' > 1 everywhere, hence g_n maps R onto R. Now given d_n there exists x with g_{n-1}(x) = d_n; hence if q_n is chosen close enough to x we can make g_n(q_n) = d_n and still have |g_n - g_{n-1}| + |g'_n - g'_{n-1}| < 2^(-n). >Robert Israel israel(a)math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich
From: David C. Ullrich on 13 Mar 2005 07:22
On 12 Mar 2005 11:42:17 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Hate to tell you this: This example would apply to the ftoc as well. No, because the hypotheses are correct in ftc. >So >what bullshit are you trying to push here?! > >I can also come up with a number of functioins that don't normally work >by producing some of my own mapings. >Ullrich! You are so damn pathetic, it's just not funny! > >Is this your counter-example? It was. Then Wade gave a simpler one, and now based on a post of Zbigniew Fiedorowicz we have one that's much simpler yet. Maybe not really simpler, but the part that's more complicated is in lots of books: You can find examples in various books, for example Stromberg "Introduction to Classical Real Analysis", of strictly increasing differentiable f with f' = 0 on a dense set. Zbigniew Fiedorowicz has pointed out that it's very easy to modify this to get f'(r) = 0 for all rational r. For such f, if x and w are both rational then the right-hand side of the "theorem" is identically 0. >Gee, no wonder your students are all >idiots!! You say things like this and worse really a lot. Do you keep that in mind when you complain about the personal attacks that you imagine you're getting here? >Jason Wells ************************ David C. Ullrich |