# Continuity, vector function, inverse

1. Dec 19, 2008

### boombaby

1. The problem statement, all variables and given/known data
f:Rn->Rn is continuous and satisfies
|f(x)-f(y)|>=k|x-y|
for all x, y in Rn and some k>0. Show that F has a continuous inverse.

2. Relevant equations

3. The attempt at a solution
It is easy to show that f is injective, but I've no idea how to prove the surjectivity. I was thinking on the R1->R1 case for a while, and guess that I can show that f is unbounded to deduce the surjectivity. But it seems that boundedness is not that useful in Rn->Rn case.
Any hint? THanks!

2. Dec 19, 2008

### Dick

You can still use boundedness, I think. Suppose f(R^n) is not all of R^n. Then pick a point b on the boundary of f(R^n). So you can pick a sequence x_i such that f(x_i) converges to b. The sequence x_i can't be bounded, otherwise it would have a convergent subsequence. So you can pick a subsequence of x_i, y_i such that |y_{i+1}-y_i| is as large as you like, yet f(y_i) must converge to b.

3. Dec 19, 2008

### blerg

i'm working on a similar problem. could you possibly explain why x_i can't be bounded?

4. Dec 19, 2008

### Dick

If x_i were bounded then would be contained in a compact set. So it has a convergent subsequence y_i->y. So f(y_i)->f(y) by continuity. But f(y_i)->b, so b=f(y). But b was assumed to be outside of the range of f.

5. Dec 20, 2008

### boombaby

Thanks, but why b should be outside the range of f? I get it by first proving that injective and continuous f should map open ball to open ball, but not quite clear about the whole reason behind it. Is there some explanation more direct/brilliant/inspiring? Thanks a lot...

6. Dec 20, 2008

### Dick

Good point. Mostly because I drew it that way. :) Better give this some more thought...

7. Dec 20, 2008

### Dick

Have you studied Brouwer's invariance of domain theorem? I think it's necessary. f:R^1->R^2 defined by f(x)=(x,0) satisfies your premises (with unequal dimensions of domain and range) but it's not surjective.

8. Dec 20, 2008

### boombaby

sorry for reply late...last night I was unable to open this forum...
I havn't learned Brouwer's theorem and I've no idea what causes the differences between unequal dimensions and equal dimensions of domain and range (guess it needs a lot of preliminary knowledge that I havn't learned).
but yes, I made a mistake when I tried to prove that "surjective and continuous f maps open ball to open ball". if added f:R^n->R^n, is this right? okay, even if it is true, I guess I cannot prove it, since I am possibly unable to make use of the properties of equal dimensions of domain and range....
so...it seems that the original problem is beyond my knowledge?

Last edited: Dec 20, 2008
9. Dec 20, 2008

### Dick

Ok, if you can prove f maps open balls to open balls then f(R^n) is an open set, then you don't need Brouwer's theorem. If f(R^n) is open then a point on the boundary of f(R^n) is not in f(R^n).

10. Dec 21, 2008

### Dick

You could also use the generalization of Jordan's curve theorem. An injective continuous image of S^(n-1) -> R^n splits the complement of the image into an 'inside' and an 'outside'. It's hard to prove as well, but it's easy to visualize. That should also let you prove f(R^n) is open.

11. Dec 22, 2008

### boombaby

Many things beyond my knowledge:( I'll come back to this problem when I am ready for it. Anyway, thanks very much! I'll keep an eye on Jordan's curve theorem and Brouwer's theorem.