Thread: Show that T is surjective

Let be a compact metric space, and let be a continuous map satisfying the expansion property:

for all .

Prove that T is surjective.

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

Thanks in advance.

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....?)

Another way would be to prove the result by contradiction. Suppose that the point is not in the range of T. The range of T is closed (by compactness) so there exists such that for all x in X.

Now consider the sequence . Use the non-contracting property of T to show that any two points in this sequence are at a distance at least apart. Therefore the sequence cannot have a convergent subsequence, which contradicts the compactness of X.

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

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....?)

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.

sorry deleted