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.

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- Nov 19th 2011, 03:46 AMMarkeurShow 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. - Nov 19th 2011, 06:35 AMDevenoRe: 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....?) - Nov 19th 2011, 07:53 AMOpalgRe: Show that T is surjective
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. - Nov 19th 2011, 10:14 AMxxp9Re: Show that T is surjective
- Nov 19th 2011, 10:29 AMDevenoRe: Show that T is surjective
- Nov 19th 2011, 04:50 PMxxp9Re: Show that T is surjective
sorry deleted