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Let $(X, d)$ be a metric space. For a subset $A$ of $X$, let $d_A(x) := \inf_{a \in A} d(x, a)$. Let $x,y \in X$. For $b \in A$, we have $$ d_A(x) -d(y, b) = \inf_{a \in A} (d(x, a) - d(y, b)) \le d(x,b)-d(y,b) \le d(x,y). $$

So $d_A(x) -d_A(y) \le d(x,y)$. By symmetry, $|d_A(x) -d_A(y)| \le d(x,y)$ and thus the map $x \mapsto d_A(x)$ is $1$-Lipschitz. In proving this result, I have come across below theorem, i.e.,

Theorem Let $A,B$ be two closed subsets of $X$ such that $A$ is compact and $A \cap B = \emptyset$. Then the map $f:X \to \mathbb R$ defined by $$ f(x) := \frac{d_A(x)}{d_A(x)+d_B(x)} \quad \forall x\in X. $$ is Lipschitz such that $f(A)=1$ and $f(B)=0$.

Could you have a check on my below attempt?

Proof It's clear that $f(A)=1$ and $f(B)=0$. Notice that $d_A(x)+d_B(x) \ge d(A,B) := \inf\{d(a,b) : a\in A, b\in B\}>0$ because $A$ is compact, $B$ is closed, and $A \cap B = \emptyset$. Fix $x,y\in X$. Then $$ \begin{align} |f(x)-f(y)| &= \left | \frac{d_A(x)}{d_A(x)+d_B(x)} - \frac{d_A(y)}{d_A(y)+d_B(y)} \right | \\ &= \frac{|d_A(x) d_B(y) - d_A(y) d_B(x)|}{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} \\ &= \frac{|d_A(x) [d_B(y)-d_B(x)] + d_B(x) [d_A(x)-d_A(y)]|}{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} \\ &\le \frac{d_A(x) |d_B(y)-d_B(x)| }{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} + \frac{d_B(x) | d_A(x)-d_A(y)|}{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} \\ &\le \frac{d_A(x) d(x,y) }{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} + \frac{d_B(x) d(x,y)}{[d_A(x)+d_B(x)][d_A(y)+d_B(y)]} \\ &\le \frac{d(x,y) }{d_A(y)+d_B(y)} + \frac{d(x,y)}{d_A(y)+d_B(y)} \\ &\le \frac{2d(x,y) }{d(A,B)} . \end{align} $$

This completes the proof.

Akira
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  • I'm not exactly sure what $f(A)$ is supposed to be, I guess it's just the image of $A$ under $f$. Anyway, can you please double-check that it's $f(A)=1$, $f(B)=0$ and not the other way around? – Matija Jan 15 '23 at 23:44
  • @Matija You are right! I meant by $f(A)$ the image of $A$ under $f$. Indeed, it should be $f(A)=0$ and $f(B)=1$. – Akira Jan 15 '23 at 23:46
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    The proof is sound. Yes, we need (sequential) compactness to obtain the existence of a limit point in $A$ which is then a limit point to both, yielding a contradiction. I would just change the assertion to $A,B$ such that $d(A,B)>0$. That's intuitive, too, and closedness, compactness are only required to get the positive distance. Clearly, the altered result is significantly stronger, and the Theorem as stated in the question is then an immdiate corollary. But of course, that's a matter of taste. – Matija Jan 16 '23 at 00:01
  • @Matija Thank you so much for your comment. Could you post it as an answer? – Akira Jan 16 '23 at 00:06
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    This has nothing to do with the Urysohn function as such. If $f,g$ are Lipschitz, non negative and $f+g$ uniformly bounded away from zero, then ${f \over f+g}$ is Lipschitz. – copper.hat Jan 16 '23 at 00:17

1 Answers1

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As per request, I post the comment as an answer.

The proof is sound. Yes, we need (sequential) compactness to obtain the existence of a limit point in $A$, which is then a limit point to both, yielding a contradiction. I would just change the assertion to $A,B$ such that $d(A,B)>0$. That's intuitive, too, and closedness, compactness are only required to get the positive distance. Clearly, the altered result is significantly stronger, and the Theorem as stated in the question is then an immdiate corollary. But of course, that's a matter of taste.

Matija
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  • In chess we say "To win you must use all your pieces". The proposer saw the way to use the compactness of $A$ or $B$ by $0<\inf_{x\in X}d_A(x)+d_B(x). $ – DanielWainfleet Jan 16 '23 at 05:33