Results 1 to 6 of 6

Math Help - Implications.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Implications.

    Is at least one of the following two implications valid for every map f:M \rightarrow N of a metric space M to a metric space N?

    a). f is continuous \Rightarrow \ f(U) is open for every U \subset M.

    b). f(U) is open for every open U \subset M \Rightarrow  \ f is continuous.
    I think that a). is valid. The definition of continuity, at a point c \in M, adapted to this case is:

    (\forall \varepsilon >0)(\exists \delta >0)[d_M(c,x)< \delta \ \Rightarrow \ d_N(f(c),f(x))< \varepsilon]

    d_N(f(c),f(x))< \varepsilon is just B_{\varepsilon}(f(c),d_N) valid \forall c,x \in M.

    Hence it does imply that f(U) is open for every U \subset M.

    So, i've answered the question with a yes! However, for the sake of curiosity I will try and figure out b).

    Since a). is true, i'll try and find a counterexample.

    Consider the French Railway Metric. This is open because a ball can be constructed around every point.

    However, f is not continuous.
    Suppose that x=(1,0) and x_n=(1+\frac{1}{n},0).


    As d(x,x_n) \rightarrow 0, d_F(x,x_n) \rightarrow 2 so it cannot be continuous.

    Is this the correct way to think about the problem?
    Last edited by Showcase_22; January 26th 2010 at 04:30 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Showcase_22 View Post
    I think that a). is valid. The definition of continuity, at a point c \in M, adapted to this case is:

    (\forall \varepsilon >0)(\exists \delta >0)[d_M(c,x)< \delta \ \Rightarrow \ d_N(f(c),f(x))< \varepsilon]

    d_N(f(c),f(x))< \varepsilon is just B_{\varepsilon}(f(c),d_N) valid \forall c,x \in M.

    Hence it does imply that f(U) is open for every U \subset M.
    How? To prove that f(U) is open, you need to take a point n \in f(u) and show that there exists an \epsilon_n > 0 such that B_{d_N}(n, \epsilon_n) \subseteq f(U). Have you done that?

    After you understand why your proof is wrong, think of M=N=(\mathbb{R}, d_2) with f(x)=x.

    Consider the French Railway Metric. This is open because a ball can be constructed around every point.

    However, f is not continuous.
    Suppose that x=(1,0) and x_n=(1+\frac{1}{n},0).


    As d(x,x_n) \rightarrow 0, d_F(x,x_n) \rightarrow 2 so it cannot be continuous.

    Is this the correct way to think about the problem?
    What is open? You haven't specified any space, just the metric!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    a) If f:\left(X,d\right)\mapsto\left(\bar{X},\bar{d}\rig  ht) is continuous, then it's open.
    Take X=\bar{X}=\mathbb{R} and the usual metric and consider f(x)=0

    If a function is open, it is continuous.
    It is relatively easy to prove that if f is bijective and open then f^{-1} is continuous. Thus, to state the above would imply that every bijective function with continuous inverse is continuous. Let g:[0,1]\cup\left(2,3\right]\mapsto[0,2] be such that x\mapsto\begin{cases} x & \mbox{if} \quad x\in [0,1] \\ x-1 & \mbox{if} \quad x\in (2,3]\end{cases}. This is a bijection which is continuous but has a non-continuous inverse. So...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Take and the usual metric and consider
    \mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})" alt="f\mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})" />

    f(x)=0 maps every point to 0. The range is just the set \{ 0 \} which is closed.

    f(x)=0 is continuous so we have a counterexample to the question!

    every bijective function with continuous inverse is continuous.
    The function you've written contradicts the above quote!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Showcase_22 View Post
    \mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})" alt="f\mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})" />

    f(x)=0 maps every point to 0. The range is just the set \{ 0 \} which is closed.

    f(x)=0 is continuous so we have a counterexample to the question!



    The function you've written contradicts the above quote!
    Yes. Both are wrong. I was providing counterexamples to my quotes.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Thankyou very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. predicate forms implications
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 22nd 2011, 02:22 AM
  2. Trivially True and Implications
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 7th 2010, 02:28 PM
  3. Help with implications
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 10th 2010, 10:06 AM
  4. Negating implications
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 23rd 2009, 10:32 PM

Search Tags


/mathhelpforum @mathhelpforum