Cantor Confusion
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From: Dave Seaman on 15 Jan 2007 23:12 On Mon, 15 Jan 2007 19:09:41 EST, Andy Smith wrote: > Dave Seaman wrote: > .. >> >> No, sin(pi/x) is undefined at x = 0. >> >> You can define a function f: R -> R by >> >> f(x) = sin(pi/x), if x != 0, >> = 0, if x = 0. >> >> and in that case, it's true that f(x) = 0 (but >> because the definition of >> f says so, not because of antisymmetry. You can >> deduce antisymmetry from >> the definition, not the other way around. >> > If f(x) is such that at any eta, f(eta) = -f(-eta), > this would imply that f(0) = 0? And you would know that f has this property how, exactly? > If f(0) was anything other than 0, the antisymmetry would be destroyed? Assuming it ever existed in the first place. > How could something that is antisymmetric not be other than 0 > at its mirror point - even if it is multivalued, it will retain > antisymmetry in inverting x ? How could a function that is undefined at zero be antisymmetric? By the way, I am assuming that what you call an "antisymmetric" function is what I call an "odd" function, having the property that f(-x) = -f(x). > Alll this is a bit "angels on a pin", but is ultimately important > I think from a philosophical perspective on the nature of infinity. If your definition of an antisymmetric function mentions infinity, then it must be something other than what I thought. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Virgil on 15 Jan 2007 23:26 In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org>, Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > Dave Seaman wrote: > > .. > > > > > > No, sin(pi/x) is undefined at x = 0. > > > > You can define a function f: R -> R by > > > > f(x) = sin(pi/x), if x != 0, > > = 0, if x = 0. > > > > and in that case, it's true that f(x) = 0 (but > > because the definition of > > f says so, not because of antisymmetry. You can > > deduce antisymmetry from > > the definition, not the other way around. > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > this would imply that f(0) = 0? Provided that one also is assured that f(x) is defined for all x. But f(x) = sin(pi/x) does not define a function which has any value at all at x = 0. If one merely defines f(x) = sin(pi/x), then the domain of that f cannot contain 0. One needs an additional statement to extend the definition to include x = 0. > > If f(0) was anything other than 0, the antisymmetry would be destroyed? Not at all. What happens at x = 0 is irrelevant to antisymmetry as long as the function is not given a non-zero value at x = 0. > > How could something that is antisymmetric not be other than 0 > at its mirror point - even if it is multivalued, it will retain > antisymmetry in inverting x ? It need not be defined at all at x = 0 and it still can retain the property of having, for every x in its domain, f(x) = -f(-x). > > Alll this is a bit "angels on a pin", but is ultimately important > I think from a philosophical perspective on the nature of infinity. Since your definition of an "anti-symmetric" function seems to correspond to the more common property of being an /odd/ function, note that f(x) = x^n is an even or odd function according to whether n is an even or odd integer, including negative integers for which the functions are not defined at x = 0.
From: Virgil on 15 Jan 2007 23:46 In article <virgil-D73F49.21261015012007(a)comcast.dca.giganews.com>, Virgil <virgil(a)comcast.net> wrote: > In article > <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org>, > Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > > > Dave Seaman wrote: > > > > .. > > > > > > > > > > No, sin(pi/x) is undefined at x = 0. > > > > > > You can define a function f: R -> R by > > > > > > f(x) = sin(pi/x), if x != 0, > > > = 0, if x = 0. > > > > > > and in that case, it's true that f(x) = 0 (but > > > because the definition of > > > f says so, not because of antisymmetry. You can > > > deduce antisymmetry from > > > the definition, not the other way around. > > > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > > this would imply that f(0) = 0? > > Provided that one also is assured that f(x) is defined for all x. > > But f(x) = sin(pi/x) does not define a function which has any value at > all at x = 0. If one merely defines f(x) = sin(pi/x), then the domain of > that f cannot contain 0. One needs an additional statement to extend the > definition to include x = 0. > > > > If f(0) was anything other than 0, the antisymmetry would be destroyed? > > Not at all. What happens at x = 0 is irrelevant to antisymmetry as long > as the function is not given a non-zero value at x = 0. > > > > How could something that is antisymmetric not be other than 0 > > at its mirror point - even if it is multivalued, it will retain > > antisymmetry in inverting x ? > > It need not be defined at all at x = 0 and it still can retain the > property of having, for every x in its domain, f(x) = -f(-x). > > > > Alll this is a bit "angels on a pin", but is ultimately important > > I think from a philosophical perspective on the nature of infinity. > > Since your definition of an "anti-symmetric" function seems to > correspond to the more common property of being an /odd/ function, note > that f(x) = x^n is an even or odd function according to whether n is an > even or odd integer, including negative integers for which the functions > are not defined at x = 0. It seems that I am being overruled by Wiki on the matter of even and odd functions. When I was young, all f(x) = x^n, with integral n, were regarded as even or odd functions, but nowadays it appears, at least according to Wiki, that it does not hold for negative n.
From: David Marcus on 16 Jan 2007 00:20 Dik T. Winter wrote: > In article <MPG.20160647a0a74784989b3d(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > > Dik T. Winter wrote: > > > In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: > > > ... > > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > > > > this would imply that f(0) = 0? > > > > > > No (assuming any eta > 0). It would only imply that lim{x->0} f(x) = 0. > > > > ?? > > It appears to be clear to me. When the eta are != 0, you can say > nothing about f(0). Sure. But lim{x->0} f(x) doesn't have to equal 0. -- David Marcus
From: David Marcus on 16 Jan 2007 00:23
Virgil wrote: > It seems that I am being overruled by Wiki on the matter of even and odd > functions. When I was young, all f(x) = x^n, with integral n, were > regarded as even or odd functions, but nowadays it appears, at least > according to Wiki, that it does not hold for negative n. I suppose it is a matter of taste. It does point out that it is always good to define your terms. -- David Marcus |