1. ## 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.

2. 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?

3. 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.

4. 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?

5. 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 $\displaystyle x,y$ 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!