Is there a surjective map $f:X\to X$ on compact metric spcace $(X, d)$ with the following condition?
There is $0<L<1$ such that $d(f(x), f(y))<Ld(x, y)$ for all $x,y\in X$
Asked
Active
Viewed 340 times
2
-
No. See if you can prove it. – GEdgar Apr 09 '17 at 12:38
-
Dear Prof. I can not prove it, I need your help, thanks alot – Ali Barzanouni Apr 09 '17 at 12:42
2 Answers
2
Let $x, y \in X$. Since $f$ is surjective, there are $x_1, y_1 \in X$ such that $f(x_1)=x, f(y_1)=y$, similarly $f(x_2)=x_1, f(y_2)=y_1$, ..., $f(x_{n+1})=x_n, f(y_{n+1})=y_n$. From the assumption, $$d(f(x), f(y))<Ld(x, y)=Ld(f(x_1), f(y_1))<L^2d(x_1, y_1)<\cdots <L^{n+1}d(x_{n}, y_{n})\leqslant L^{n+1} M,$$ for all $n$, where $M:=\sup_{x, y\in X}d(x, y)<\infty$. By letting $n\to \infty$, we obtain from $L^n \to 0$ that $d(f(x), f(y))=0$. Hence, $f$ is constant on $X$. This means $X$ is singleton and there has no such function.
ntt
- 1,143
- 6
- 13
-
For the conclusion a better choice could be that $f$ cannot be surjective and constant at the same time. :) – Yes Apr 09 '17 at 13:03
0
Hints:
Since $f(X)$ is compact, there exists $x,y\in X$ such that $d(f(x), f(y))$ is the diameter of $f(X)$. Then, what can you say about the diameter of $X$ using $$ d(f(x), f(y)) \le Ld(x,y) ? $$
user251257
- 9,229