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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...

martini
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user297564
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2 Answers2

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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 $d(x,y) = \text{diam}(X)$.

Adam Francey
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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.

bartgol
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  • Yeah, I noticed, and fixed with an edit. – bartgol Mar 10 '16 at 20:12
  • Probably the other way inequality is correct. – Landon Carter Mar 10 '16 at 20:15
  • @LandonCarter constant functions satisfy the opposite inequality, and they are surely not compact unless your set is one point. – Adam Hughes Mar 10 '16 at 20:15
  • How did you come up with this example? –  Mar 10 '16 at 20:16
  • Look at non-constant functions then! – Landon Carter Mar 10 '16 at 20:17
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    I am having second thoughts about this example. But the idea is to build a function whose range contains only rationals. – bartgol Mar 10 '16 at 20:17
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    @LandonCarter Take a closed ball centered at the origin in $\Bbb R^n$ and shrink its radius by a factor of 1/2--i.e. scale things by 1/2--then the image is a proper subset. – Adam Hughes Mar 10 '16 at 20:29
  • I was trying to recall one similar result where an inequality in reverse direction implies surjectivity in compact metric space. Oh, right! You need equality. – Landon Carter Mar 10 '16 at 20:31
  • Let $\theta_0$ be the argument of $0$, that is $0 = (0, \theta_0)$ in polar coordinates. Then $(\frac 1n, \theta_0 + 1)$ does not have a convergent subsequence. So the space is not compact. – martini Mar 10 '16 at 20:31
  • Uhm. But if the argument is always considered as $\mod 2\pi$, then it does have a convergent subsequence. – bartgol Mar 10 '16 at 20:39
  • It doesn't matter how you define $\theta$ (i.e. what branch of log you choose), it won't be continuous. This example is incorrect. – Patrick Da Silva Mar 10 '16 at 20:41
  • I agree that this metric is not continuous with respect to the Euclidean metric, but why is that relevant? – bartgol Mar 10 '16 at 21:11
  • Your function $f$ does not satisfy the condition that $d(x,y) \leq d(f(x),f(y))$ for all $x,y \in X$.

    Consider $X$ as the closed disk of radius $1$. Pick $x,y\in X$ such that $\theta (x) = \theta (y) = \theta$ for some fixed angle $\theta$, $\rho (x) = 0.5$, and $\rho (y) = 0.75$.

    Then $d(x,y) = 0.25$.

    See for yourself that $f(x) = f(y) = (1, \theta)$, so that $d(f(x),f(y)) = 0 < d(x,y)$.

     – Adam Francey Mar 11 '16 at 19:39
  • Uh, you're right. I still believe the underlying idea would work though. with some modifications. In particular, I would have to modify the coefficient (not 1000) and the number of digits you retain, based on $\rho(x)$, to ensure that if $\rho(x)<1$, then $\rho(f(x))<1$. I have to work out the details, but I believe it's possible. – bartgol Mar 11 '16 at 19:48
  • Nope, it can't be, this question was closed as a duplicate, and the statement was proven to be true. – Adam Francey Mar 11 '16 at 20:26
  • Ok, then I give up. Sorry for wasting everyone's time. – bartgol Mar 11 '16 at 20:27