Results 1 to 7 of 7

Thread: Equivalent Metrics

  1. #1
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71

    Equivalent Metrics

    Define $\displaystyle \rho$ on $\displaystyle X \times X$ by

    $\displaystyle \rho(x,y) = min(1,d(x,y)), \ \ \ \ \ \ \ \ \ \ x,y \in X$

    Show that $\displaystyle \rho$ is a metric that is equivalent to d

    My solution:
    Two metrics are equivalent if and only if the convergent sequences in $\displaystyle (X,d)$ are the same as the convergent sequences in $\displaystyle (X,\rho)$

    Let $\displaystyle (x_{n})^\infty_{n=1}$ be convergent in $\displaystyle \rho$ i.e.,

    $\displaystyle \forall \ \epsilon \ \exists \ N_{\epsilon}\ \mbox{such that}\ \forall\ n \geq N_{\epsilon}\ \rho(x_n,x)={min(1,d(x_{n},x))}}<\epsilon $

    Choose $\displaystyle 0 \ < \epsilon \ < 1$. Then $\displaystyle \exists \ N_{\epsilon}$ such that $\displaystyle \forall\ n \ \geq N_{\epsilon}: min (1,d(x_n,x)) < \epsilon < 1$.

    This implies $\displaystyle min (1,d(x_n,x))=d(x_n,x)\ <\ \epsilon$.

    What if $\displaystyle \epsilon > 1$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035
    Irrelevant. The definition of "convergent" says for any$\displaystyle \epsilon> 0 $ but is is small $\displaystyle \epsilon$ that is important.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    And if you still wish to show it for $\displaystyle \epsilon>1$, you can simply note that $\displaystyle min(x,y) \le x $ & $\displaystyle min(x,y) \le y \ \forall x,y \in \mathbb{R}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Defunkt, how would that be relevant ? oO

    I think it's possible to show it for e>1 if you consider the contrapositive. But as Hallsofivy says, it's not useful
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    $\displaystyle min(1, d(x_n,x)) \le 1 < \epsilon$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    But what we want to show is that $\displaystyle d(x_n,x)<\epsilon$. We already know that $\displaystyle \min(1,d(x_n,x))<\epsilon$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976


    Oops
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fun with metrics
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Sep 24th 2011, 07:05 AM
  2. Are equivalent metrics comparable on compact sets?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Aug 18th 2010, 07:23 AM
  3. Equivalent Metrics
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: Feb 16th 2010, 03:24 AM
  4. Challenging Problem!!! (Equivalent Metrics)
    Posted in the Differential Geometry Forum
    Replies: 13
    Last Post: Jan 23rd 2010, 10:23 AM
  5. Metrics
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 29th 2008, 11:38 AM

Search Tags


/mathhelpforum @mathhelpforum