Results 1 to 2 of 2

Math Help - Topology (2)

  1. #1
    Member
    Joined
    Dec 2006
    Posts
    79

    Topology (2)

    Show that the metrics d(f,g) = intergral 0 to 1 abs (f(t) - g(t) dt and
    d'(f,g) = max {abs f(t) - g(t) : t in [0,1] on the set C[0,1] of continuous real valued functions on [0,1] are not topologically equivalent.

    Hint: If d' is topologically equivalent to d, then a sequence {Xn} converges to X in one metric iff it converges to X in the other metric.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by taypez View Post
    Show that the metrics d(f,g) = intergral 0 to 1 abs (f(t) - g(t) dt and
    d'(f,g) = max {abs f(t) - g(t) : t in [0,1] on the set C[0,1] of continuous real valued functions on [0,1] are not topologically equivalent.

    Hint: If d' is topologically equivalent to d, then a sequence {Xn} converges to X in one metric iff it converges to X in the other metric.

    Thanks.
    I never learned topology so I might be way off.

    Consider the sequence x_n = \{ x^n \}, i.e. x_1=x,x_2=x^2,x_3=x^3,.... I claim that the limit of this sequence is 0. Meaning we need to show d(x^n,0)<\epsilon for sufficiently large n. Meaning, \int_0^1 |x^n - 0 | dx =\int_0^1 x^n dx be made sufficiently small. This integral is \frac{1}{n+1} which can be made smaller than any \epsilon >0. We have shown this sequence converges to 0 in the first topology.

    Consider the same sequence in the second topology. I claim the limit is 1. We need to show d'(x^n,1)<\epsilon when n is sufficiently large. But this means \max_{x\in [0,1]} \{ |x^n - 1| \} = \max_{x\in [0,1]} \{ x^n - 1\}. But the maximum value is 0 because 0\leq x^n \leq 1 on [0,1]. This (0) is certainly less than any epsilon thus thus it converges to 1 in the second topology.

    Thue, the two fields are not equivalent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Order topology = discrete topology on a set?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: August 6th 2011, 11:19 AM
  2. a topology such that closed sets form a topology
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 14th 2011, 04:43 AM
  3. Show quotient topology on [0,1] = usual topology on [0,1]
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 5th 2010, 04:44 PM
  4. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 13th 2008, 02:19 PM
  5. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: December 7th 2008, 01:01 PM

Search Tags


/mathhelpforum @mathhelpforum