Metric Space Isometry

**kimberu** · September 10th 2010, 03:29 PM

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

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

For the second statement, I know continuity means that for every $\epsilon > 0,$ there is a $\delta > 0,$ so $d(\phi(x), \phi(x_0)) < \epsilon$ when $d(x, x_0) < \delta$ 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 $x_0 + x$ or $x_0 - x$ 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.

**Defunkt** · September 10th 2010, 04:10 PM

Hello Kimberu!

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

Now, if we choose $x_1=0$ and let $x\in \mathbb{R}$ , then we require that $d(\phi(0), \phi(x)) = d(0,x) = |x|$ , that is, $|\phi(x)-\phi(0)| = |x|$ . Do you see how to continue?

**kimberu** · September 11th 2010, 12:32 AM

Oh, now I understand that the question was just defining $x_0 = \phi(0)$ !
So from what you said, from $|\phi(x)-\phi(0)| = |x|$ , just solving for $\phi(x)$ would lead to either the case of $x_0 + x$ or $x_0 - x$ 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!

**Jose27** · September 11th 2010, 09:13 PM

Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

To prove the second one, assume two values $x,y$ go to $x_0+x$ and $x_0-y$ , can the function be an isometry in this case?

**kimberu** · September 12th 2010, 12:49 AM

Originally Posted by Jose27

Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

To prove the second one, assume two values $x,y$ go to $x_0+x$ and $x_0-y$ , can the function be an isometry in this case?

So $\phi(x) = x_0+x$ and $\phi(y)=x_0-y$ ? 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!

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