function valued 1 on an open set and 0 on another

Question

Let $X,Y \subset M$ be closed subsets of a metric space $M$. Assume that $\inf_{x\in X,y\in Y} d(x, y) > 1$. Prove that there exists a 1-Lipschitz function on $M$ which is equal to 1 on $X$ and 0 on $Y$.

Any hints on where to start?

Answer 1

Hint: Think about functions like this:

$$f(x) = \frac{{\rm dist}(x,Y)}{{\rm dist}(x,X)+{\rm dist}(x,Y)},$$

where $\displaystyle {\rm dist}(x,A) = \inf_{y\in A} d(x,y)$.

Answer 2

For $x \in X$, draw $$ D_x := \{p \in M:\; d(p,x) \leq 1\}, $$ and define $$ D := \bigcup_{x \in X} D_x $$ By assumption, $D \cap Y =\varnothing$. Also, define $$ g_x(z) :=\begin{cases} d(z,x),\quad z \in D_x \\ 1,\quad \rm{otherwise} \end{cases} $$ Define $$ f(z) := \inf_{x \in X} g_x(z) $$ Notice $f(y) =1$ for all $y \in Y$.

Consider any two points $u,v \in M$. If $d(u,v) >1$ then we are done. Thus assume $d(u,v) \leq 1$.

Suppose that $u,v \in D$. Let $x_u$, $x_v$ be resp. the center of cone whose shape $f(u),f(v)$ were taken. That is to say, $$ d(x_u,u) =f(u),\quad d(x_v,v) =f(v). $$ And we have, by construction, $$ d(x_u,u) <d(x_v,u),\quad d(x_v,v) <d(x_u,v) $$ Since $\Delta x_u uv$ and $\Delta x_v uv$ both satisfy triangle inequality (as should any metric do), we can draw corresponding quadrilateral $x_u x_v vu$ on the normal Euclidean plane. By A relation between sides of quadrilateral which I just asked, $$ |f(v)-f(u)| =|d(x_u,u) -d(x_v,v)| \leq d(u,v). $$

Then (wlog) consider $u \in D$ but $v \in M-D$, and find $x_u$ the center of the cone on which $f(u)$ is taken. Notice (by assumption) $$ d(x_u,u) \leq 1 \leq d(x_u,v) $$ Thus, similarly, $$ |f(v) -f(u)| =1 -f(u) =1 -d(x_u,u) \leq d(x_u,v) -d(x_u,u) =|d(x_u,v) -d(x_u,u)| \leq d(u,v). $$ ($d(x_u,v) >1$ otherwise $v \in D$. Then triangle inequality is used.)

The case $u,v \in M-D$ is trivial, since $|f(v) -f(u)| =|1-1| =0 \leq |u-v|$.