Urysohn's function on a metric space Let $(X,d)$ be a metric space and $A\subset B\subset X$. $A$ is closed, $B$ is open. If there are developed methods to find at least one (or describe the whole class) of Urysohn's functions for $A$ and $B^c$?
Edited: Martin has already presented a nice example of an Urysohn's function. Now I would like to focus on my second question about the whole class $\mathcal U(A,B^c)$ of Urysohn's functions. I had a function $u$ which is $0$ on $B^c$ and $1$ on A - and I would like to check if it is continuous (so if it is in $\mathcal U(A,B^c)$ class). If there are any sufficient conditions to prove that this function is $\mathcal{U}(A,B^c)$ which allow me pass direct verifying of continuity of $u$, or this question is meaningless and all that I can do - is to verify continuity without theory of Urysohn's functions?
 A: EDIT 2: As pointed out by Theo, my original solution will work only if $d(A,B^c)>0$. (E.g. if $A$ or $B^c$ is  compact.) However, the solution taken from Amman's and Escher's book (see EDIT 1 below) works for arbitrary $A$ and $B$.
Let us denote $\delta=d(A,B^c)$.
Let us define $f(x)=\frac1\delta d(x,A)$. This functions is continuous, $f(a)=0$ for $a\in A$ and $f(b)\ge 1$ for $b\in B^c$.
Similarly define $g(x)=\frac1\delta d(x,B^c)$.
Now define $f'(x)=\min\{1,f(x)\}$ and $g'(x)=\min\{1,g(x)\}$. These functions have similar properties, and additionally they have values in $[0,1]$.
There are several choices for the Urysohn function for $A$ and $B^c$:
$\frac{f'(x)+(1-g'(x))}2$, $\max\{f'(x),1-g'(x)\}$, $\min\{f'(x),1-g'(x)\}$.
I hope I have not overseen some mistake in the above reasoning.
EDIT 1: According to Proposition 4.13 in Amman, Escher: Analysis III, another such function is $\frac{d(x,A)}{d(x,A)+d(x,B^c)}$. I have to admit that this is much more elegant construction than mine. (In my notation it is $\frac{f(x)}{f(x)+g(x)}$.)
A: Consider $f: X \to [0,1]$ given by
$$f(x) = \frac {d(x,A)} {d(x,A) + d(x,B^c)} \ .$$
$f$ is well-defined because the denominator is $0$ if and only if $x \in \bar A \cap \overline {B^c} = A \cap B^c = \emptyset$ (because $A$ is assumed closed and and $B$ assumed open).
We have $f(x) = 0$ if and only if $x \in \bar A = A$, and $f(x) = 1$ if and only if $x \in \overline {B^c} = B^c$.
Finally, $f$ is continuous because the "distance to a closed subset" function is continuous.
