On Jul 2, 4:42 am, Virgil <vir...@ligriv.com> wrote: > In article > <a0f4c494-499c-4628-afa3-08a45d191...@e20g2000vbm.googlegroups.com>, > > > > > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 2 Jul., 01:17, "Mike Terry" > > <news.dead.person.sto...@darjeeling.plus.com> wrote: > > > "WM" <mueck...@rz.fh-augsburg.de> wrote in message > > > >news:email@example.com... > > > > > On 30 Jun., 21:10, "Mike Terry" > > > > <news.dead.person.sto...@darjeeling.plus.com> wrote: > > > > > > The theorem has it's own name: "The > > > > > Intermediate Value Theorem" (IVT). > > > > > Fine that you know it. Now define: f(x) = 1 for x in an interval I_k > > > > and f(x) = -1 for x not in any I_k and > > > > f(x) = 0 for x = y_k_j > > > > Yes, I know it - I learned it in high school, but it seems you never learned > > > it? For IVT to apply we need to have the function f *continuous* on the > > > interval. > > > No, a function like that defined above also obeys that theorem. > > But the English version of the Intermediate Value Theorem, or IVT, says > that a function f continuous on any real interval [a,b] must assume > EVERY value between f(a) and f(b), and your function does NOT assume > every real value between -1 and 1, unless, in WM's world, 0 is the only > real number between -1 and 1. > > Thus WM once again reveals his abysmal ignorance of true mathematics. > --
Of course it does. Just take any interval I_k and let x be the point next to I_k. Then do a linear continuation of x (since there must automatically be a whole raft of other points between x and I_k, even though x is "next"), and voila. Anyone who cannot see that must be a hopeless matheologist, indoctrinated since childhood in the wrong way to do things. All hail WM, the Greatest Mathematician of All Times!