Show that a distance preserving map T:X->X is onto

[Ratpigeon]

Homework Statement

I'm trying to show that a distance preserving map is 1:1 and onto. The 1:1 part was easy, but I'm stuck on proving it's onto...

Homework Equations

X is compact
T(X)$\subseteq$X
THere's a hint saying to consider a point y in X\T(X) and consider the minimum distance
between y and x$\in$ T(X) (the infinum of d(x,y) for all x$\in$ T(X) where d is an undefined metric, and call it deltA)
Then it says to consider the sequence
y_n=Tⁿ(y)

The Attempt at a Solution

Because X is compact, and y_n is a subset of X, it must be bounded (I think?) and have a convergent subsequence, and inf(d(T^n(y),T^n(x))=delta for all n (and x_n=T^n(x) will also have a convergent subsequence). Show that the limits of y_n and x_n provide a contradiction?

 

[Dick]

I'm really not sure where that hint is supposed to lead. Here is an alternative hint. You know X can be covered by a finite number of open balls of radius delta/2. Let N be the minimum number of open balls of radius delta/2 you need to cover X. Now throw away a ball containing x. It's doesn't intersect f(X). That means f(X) can be covered by N-1 balls. Do you see a problem looming here? Think about inverse images.

 

[Ratpigeon]

Is the next step something like:
Take a=sup(T(x'),T(x)) and because X is compact, it is closed, therefore using distance preservation of T there exists at least on x in X such that
d(x',x)=a
then lead this to contradiction?

 

[Dick]

Is the next step something like:
Take a=sup(T(x'),T(x)) and because X is compact, it is closed, therefore using distance preservation of T there exists at least on x in X such that
d(x',x)=a
then lead this to contradiction?

I don't see what contradiction that leads to.

 

[Ratpigeon]

Sorry - that one was completely off track, but I think I got it out now, thanks