Results 1 to 5 of 5

Thread: Metric Space Isometry

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    41

    Metric Space Isometry

    "Let ($\displaystyle \mathbb{R}, d$) be a metric space where d(x,y) = |y-x|.
    Let $\displaystyle \phi:\mathbb{R}\rightarrow\mathbb{R}$ be an isometry from the metric space to itself.
    Consider the image $\displaystyle x_0 = \phi(0)$. Show that for every x in $\displaystyle \mathbb{R}$, the image $\displaystyle \phi(x)$ equals $\displaystyle x_0 + x$ or $\displaystyle x_0 - x$.
    Next, show that assuming $\displaystyle \phi$ is continuous, either $\displaystyle \phi(x)$ equals $\displaystyle x_0 + x$ for every x, or it equals $\displaystyle x_0 - x$ for every x."

    First, I don't think I understand what "Consider the image $\displaystyle x_0 = \phi(0)$" means. I guess it makes sense that each $\displaystyle \phi(x)$ equals $\displaystyle x_0 + x$ or $\displaystyle x_0 - x$, but I don't know how to prove it.

    For the second statement, I know continuity means that for every $\displaystyle \epsilon > 0, $ there is a $\displaystyle \delta > 0, $ so $\displaystyle d(\phi(x), \phi(x_0)) < \epsilon $ when $\displaystyle d(x, x_0) < \delta $and that it's continuous for each x, but I don't know how to apply this definition to show that it is always either $\displaystyle x_0 + x$ or $\displaystyle x_0 - x$ for every x.

    I really appreciate any help, even (or especially!) if it's just explaining very basic things that I probably don't understand correctly. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Hello Kimberu!

    For the first part, recall that for $\displaystyle \phi:\mathbb{R} \to \mathbb{R}$ to be an isometry, we need to have that $\displaystyle d(\phi(x_1),\phi(x_2)) = d(x_1,x_2) ~ \forall x_1,x_2 \in \mathbb{R}$.

    Now, if we choose $\displaystyle x_1=0$ and let $\displaystyle x\in \mathbb{R}$, then we require that $\displaystyle d(\phi(0), \phi(x)) = d(0,x) = |x|$, that is, $\displaystyle |\phi(x)-\phi(0)| = |x|$. Do you see how to continue?
    Last edited by Defunkt; Sep 10th 2010 at 04:21 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    41
    Oh, now I understand that the question was just defining $\displaystyle x_0 = \phi(0)$!
    So from what you said, from $\displaystyle |\phi(x)-\phi(0)| = |x|$, just solving for $\displaystyle \phi(x)$ would lead to either the case of $\displaystyle x_0 + x$ or $\displaystyle x_0 - x$ by the nature of absolute values if I'm not mistaken.

    Still, I don't know how to apply the definition of continuity though.

    Thank you for your help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

    To prove the second one, assume two values $\displaystyle x,y$ go to $\displaystyle x_0+x$ and $\displaystyle x_0-y$, can the function be an isometry in this case?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    41
    Quote Originally Posted by Jose27 View Post
    Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

    To prove the second one, assume two values $\displaystyle x,y$ go to $\displaystyle x_0+x$ and $\displaystyle x_0-y$, can the function be an isometry in this case?
    So $\displaystyle \phi(x) = x_0+x$ and $\displaystyle \phi(y)=x_0-y$? I assume the answer to your isometry question is 'no', but I don't know why. I tried manipulating the equations for those distances of x and y but I didn't get anywhere with that.

    Sorry to be so dense!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surjectivity of an Isometry given the metric space is complete.
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Mar 6th 2012, 10:23 PM
  2. Replies: 2
    Last Post: Jul 8th 2011, 02:16 PM
  3. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 17th 2010, 02:04 PM
  4. Dual space isometry
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: Jun 1st 2010, 05:08 AM
  5. Isometry between metric spaces
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 13th 2009, 02:24 AM

Search Tags


/mathhelpforum @mathhelpforum