Show  that     A   continuous  open  map  from  R  onto itself  is  monotonic ?
365052
Suppose a continuous open map f: R -> R is not monotonic.  Then we may assume, without loss of generality, that there are points x < y < z in R such that f(y) is the maximum (or minimum) element of f((x,z)).  This implies that f((x,z)) is not open, contradicting the assumption that f is open.
730983
First, define a new function g(x)=f(x)-x. f(x) has a fixed point iff ∃x: g(x)=0. By the fundamental theorem of calculus:\ng(x) = g(0) + [0, x]∫g'(a) da\n\nLet c be sup(|f'(x)|). Then f'(x)≤c<1, and so g'(a)≤c-1<0. So for x>0:\ng(x) < g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\nAnd for x<0:\ng(x) > g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\n\nHowever, [x→∞]lim g(0) + x(c-1) = -∞ (since c-1 is negative), and since g(x) is bounded above by this function for x>0, [x→∞]lim g(x) must also be -∞. Conversely, [x→-∞]lim g(0) + x(c-1) = ∞, and since g(x) is bounded below by this functon for x<0, [x→-∞]lim g(x) is ∞. But this means that g(x) is positive for some negative value of x and negative for some positive value of x, thus by the intermediate value theorem, ∃x: g(x)=0 and therefore f(x) has a fixed point. To see that it's the only one, consider that if g(x) had two zeros then by Rolle's theorem there would exist a point where g'(x) is zero, which contradicts the fact that g'(x) is strictly negative everywhere.\n\nNote that this proof remains valid if c is merely the supremum of f'(x), instead of |f'(x)|.\n\nFor the counterexample with |f'(x)|<1 (but not sup(|f'(x)|)<1), consider f(x)={1/x+x if x≥1, 2 if x<1}. This function is everywhere differentiable ([x→1-]lim f(x) = [x→1+]lim f(x) = 2, and [x→1-]lim f'(x) = [x→1+]lim f'(x) = 0), and ∀x, |f'(x)|<1, but f(x) has no fixed points.
1053820
Stationary in reference to what fixed point?
914601
http://mathworld.wolfram.com/PoincareConjecture.html
473829
The solution set is (-infinity, -7) U (-3/2, infinity)
49883
I don't remember well this kind of things.\n\nAnyway the key is in the fact that between any two rationals there always exist a non-rational: so there are no intervals of rationals which form a continuum\n\nSorry I don't remember exactly what does the term "open" mean
553253
http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
432855
r = a_{n+1}/a_n -> 1 if <a_n> is a sequence of Real values that converges to a non-zero L.\n\na_{n+1}/a_n = (L + (a_{n+1} - L))/(L + (a_n - L)).  But (a_{n+1} - L) -> 0 and (a_n - L) -> 0 since a_n -> L so that (L + (a_{n+1} - L))/(L + (a_n - L)) -> (L+0)/(L+0) = L/L = 1
711884
Homework.\n\nShow it first for intervals. Then extend to simple functions. Then use the fact thats imple functions are dense in L1.
432891
If g(x) is differentiable, then it certainly is continuous, and hence integrable.  You can define H(x) as the following definite integral:\n\nH(x) = 1 + Integral from 0 to x of { g(t) dt }
796566
A function relates each of its inputs to exactly one output. That is for each value of x in the domain, there must be one and only one y.\n1) 2) are OK for each x in R there is exactly one y (and vice versa)\n3) Not OK.  Take y=-2 and y=2. Both have x=4\n4) OK. There are x-values that lead to the same y, but that is allowed.\n6) Not OK. Take x=2, y>4. There are an infinite number of y-values.
511341
I'm not sure if I understood correctly but check out the pigeonhole principle
903277
this is the mean value theorem
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
903397
You might try asking this question under "Mathematics."
653242
Check out http://math.mit.edu/research/index.html
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!