Continuous expansion is surjective

[boboYO]

Homework Statement

(X,d) a compact metric space, f:X->X cts fn, with
d(f(x),f(y)) >= d(x,y)
for all x, y in X Prove that f is a surjection.

The Attempt at a Solution

Let K be the set of points that are not in f(X). It is a union of open balls because X is closed and hence so is f(X).

Choose a finite open covering of f(X)\K with all the open sets having small radius, call this covering G. Take the pre-image of all these sets, the result will be open balls of smaller or equal radius to the originals, and the centres will be closer to each other. call this set of preimages, H.

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Intuitively it seems like H doesn't cover X and so we have a contradiction (right?), but I don't know how to rigorize it. Hints? Am I on the right track? Missing something obvious?

 

[HallsofIvy]

Yes, I think you are "missing something obvious"! I don't see why you need "compact" or even "continuous". Suppose f were not a surjection. Then there exist $x\ne y$ such that f(x)= f(y). So d(f(x),f(y))= 0. That contradicts "d(f(x),f(y))> d(x,y)" doesn't it?

 

[boboYO]

I don't think that's true, let X=[0,1] on the real line and f(x)=x/2. This isn't a surjection but there is no x,y such that: $$x \neq y$$ and f(x)=f(y). I think you confused surjection and injection ;p