Show that T is surjective

Let $\displaystyle (X,d)$ be a compact metric space, and let $\displaystyle T: X \longrightarrow X$ be a continuous map satisfying the expansion property:

$\displaystyle d(T(x),T(y)) \geq d(x,y)$ for all $\displaystyle x,y \in X$.

Prove that T is surjective.

Hey I've difficulty starting this question. How do I go about doing it?

Thanks in advance.

Re: Show that T is surjective

my thought is this:

choose an open cover of ε-balls for X. since X is compact, it has a finite subcover. consider images under T of this subcover. show they form a cover of X.

(hint: for every U in our sub-cover, we have U contained in T(U) because....?)

Re: Show that T is surjective

Another way would be to prove the result by contradiction. Suppose that the point $\displaystyle u\in X$ is not in the range of T. The range of T is closed (by compactness) so there exists $\displaystyle \delta>0$ such that $\displaystyle d(u,Tx)\geqslant\delta$ for all x in X.

Now consider the sequence $\displaystyle u,Tu,T^2u,T^3u,\ldots$. Use the non-contracting property of T to show that any two points in this sequence are at a distance at least $\displaystyle \delta$ apart. Therefore the sequence cannot have a convergent subsequence, which contradicts the compactness of X.

Re: Show that T is surjective

U is not necessarily contained in T(U). See for example X is the standard sphere, T(p)=-p is the antipodal map.

Quote:

Originally Posted by

**Deveno** my thought is this:

choose an open cover of ε-balls for X. since X is compact, it has a finite subcover. consider images under T of this subcover. show they form a cover of X.

(hint: for every U in our sub-cover, we have U contained in T(U) because....?)

Re: Show that T is surjective

Quote:

Originally Posted by

**xxp9** U is not necessarily contained in T(U). See for example X is the standard sphere, T(p)=-p is the antipodal map.

that's right...for x in U, T(x) might not even be in U. my bad.

Re: Show that T is surjective