For a compact metric space $X ,f: X \rightarrow X $ s.t $ d(x,y)\leq d(f(x),f(y))$ is surjective

Question

I want to show that in a compact metrix space $X$ ,the function $f :X \to X$ such that $d(x,y) \le d(f(x),f(y))$ is surjective! I tried to show that f is continuous and injective but i don't think it really have to...

Answer 1

Uhm, I would say it is false. Consider $\mathbb{R}^2$ with the metric (in (in polar coordinates) $d(x,y) = |\rho(x)-\rho(y)| + |\theta(x)-\theta(y)|$, where $\rho(x)$ and $\theta(x)$ are the polar coordinates of $x$. Then consider the function

$$ f(x) = g(\rho,\theta) = (1000 R(\rho), \theta) $$

where $R(\rho)$ is the truncation of $\rho$ to two significant decimal digits. For instance, in scientific notation, $R(\pi)=0.31e02$ and $R(0.000123)=0.12e-03$. Then

$$ d(f(x),f(y)) = d(g(\rho(x),\theta(x)),g(\rho(y),\theta(y)))\\ = 1000|R(\rho(x))-R(\rho(y))| + |\theta(x)-\theta(y)| $$

Now, it is easy to see that $d(f(x),f(y))\geq d(x,y)$. However, the image of $f$ does not contain any point whose distance from the origin is an irrational number, so it cannot be surjective.

Edit: I missed the "compact" in your question. However, we can just consider the closed ball of whatever radius $a$, and modify $f$ to

$$ f(x) = g(\rho,\theta) = (\min\{1000 R(\rho),a\}, \theta) $$

Still, $f$ would miss all the points inside the ball whose distance from the origin is irrational.

Answer 2

Here is a proof for $X = [0,1]$ (and $d$ the Euclidean metric).

Assume that $f$ is not surjective.

Notice that $d(0,1) = 1 = \text{diam}(X)$, and $d(x,y) < 1$ for all other pairs $x,y\in X$.

To satisfy the condition that $d(x,y) \leq d(f(x),f(y))$ for all $x,y \in X$, we must have that $d(f(0),f(1)) \geq d(0,1)$, so $d(f(0),f(1)) = 1$.

Therefore we must have that $f(0) = 0$ and $f(1) = 1$, or $f(0) = 1$ and $f(1) = 0$.

Further, we must have that $d(f(x),f(0)) = d(x,0)$ and $d(f(x),f(1)) = d(x,1)$ for all $x\in X$. Otherwise, either $d(f(x),f(0)) < d(x,0)$ or $d(f(x),f(1)) < d(x,1)$.

Consider that $f$ is not surjective, so there exists $y \in X$ such that $f(x) \neq y$ for all $x\in X$.

In particular, $f(y) \neq y$.

Case 1: $f(0) = 0$ and $f(1) = 1$.

Then it is clear that $d(f(y),f(0)) \neq d(y,0)$ and $d(f(y),f(1)) \neq d(y,1)$.

Case 2: $f(0) = 1$ and $f(1) = 0$.

If $d(y,0) \neq d(f(y),0)$ and $d(y,1) \neq d(f(y),1)$, we are done.

So consider the case where $d(y,0) = d(f(y),0)$ and $d(y,1) = d(f(y),1)$. This means that $f(y) = 1 - y$. Then in order to have that $d(f(1-y),f(0)) = d(1-y,0)$ and $d(f(1-y),f(1)) = d(1-y,1)$, it must be the case that $f(1-y) = y$, but we know already that $f(1-y) \neq y$.

Conclusion: If $f$ is not surjective, then $d(x,y) \leq d(f(x),f(y))$ does not hold for all $x,y \in X$. Considering the contrapositive, if $d(x,y) \leq d(f(x),f(y))$ for all $x,y \in X$, then $f$ must be surjective.

I have a feeling this can be generalized for arbitrary compact metric spaces $X$ using $x,y \in X$ where $d(x,y) = \text{diam}(X)$.