Compact space and isometry

General Topology
Post Reply
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Compact space and isometry

#1

Post by Papapetros Vaggelis »

Let \(\displaystyle{\left(X,d\right)}\) be a compact metric space and \(\displaystyle{f:X\longrightarrow X}\) an isometry.

Prove that the function \(\displaystyle{f}\) is onto \(\displaystyle{X}\) .

Use this result in order to prove that the \(\displaystyle{\mathbb{R}}\) - normed space \(\displaystyle{\left(\ell^2,||\cdot||_{2}\right)}\)

is not compact.
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Re: Compact space and isometry

#2

Post by Tolaso J Kos »

Papapetros Vaggelis wrote:Let \(\displaystyle{\left(X,d\right)}\) be a compact metric space and \(\displaystyle{f:X\longrightarrow X}\) an isometry.

Prove that the function \(\displaystyle{f}\) is onto \(\displaystyle{X}\) .

Use this result in order to prove that the \(\displaystyle{\mathbb{R}}\) - normed space \(\displaystyle{\left(\ell^2,||\cdot||_{2}\right)}\)

is not compact.
Hello Vaggelis,

Suppose that $f$ is not onto. Then there exists an $x \in X$ such that $x \notin f(X)$. Since $f$ is an isometry this means that it will be continuous as a function and since $X$ is compact then $f(X)$ will also be a compact subset of $X$. So, $a= d(x, f(X))>0$.

We pick the sequence $x_0=x, \; x_1=f(x_0), \; \dots, x_n=f(x_{n-1}), \dots$. We note that for this the sequence holds $x_n \in f(X)$ for all $n \geq 1$. Since $f(X)$ is compact, as well as sequently compact, this means that the sequence has a convergent subsequence. Therefore, there exist $m,n $ such that $m>n\geq 1$ and $\rho(x_m, x_n)<a$.

But, $f$ is an isometry meaning that:

$$ \rho\left ( x_m, x_n \right )= \rho \left ( f(x_{m-1}), f\left ( x_{n-1} \right ) \right )=\cdots=\rho\left ( x_{m-n}, x \right )$$

On the other hand $x_{m-n} \in f(X)$ hence

$$d\left ( x, f(X) \right )\leq \rho \left ( x, x_{m-n} \right )=\rho(x_m, x_n)<a$$

a contradiction. Hence $f$ is onto.

In $\ell^2$ we consider the forward shift operator $S_r:\ell^2 \rightarrow \ell^2 $ given by:

$$S_r\left ( x_1, x_2, x_3, \dots \right )= \left ( 0, x_1, x_2, \dots \right )$$

We easily note that it is an isometry but not onto.
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 9 guests