Definition of a function by a metric.

Let (X,p) be a metric space with p metric on X, define for each subset A of X
[tex]p(x,A)=inf_{y\in A}p(x,y)[/tex]
prove that: f_A(x)=p(x,A) f:X->R is continuous.
basically, if U is open in R, we need to show that f^-1_A(U) is open in X, i.e that
{x in X|p(x,A) in U} is open in X, now because U is open in R, then there's an open interval or half open interval contained in U that it contains p(x,A), i.e p(x,A) is bounded, but from this to show that f^-1(U) is open I'm kind of in a stuck.

any hints?

thanks in advance.

 

ok, I have another question, let [tex]S^n[/tex]={[tex]x\in R^n|||x||=1[/tex]}
let x_0=(1,0,...,0) (n entries), prove that S^n-{x_0} is homeomorphic to R^(n+1).
well I thought of dividing by the norm, i.e defining a homeomorphism such as this:
x=(x1,...,x_n+1) in R^n+1 f(x)=(x2,...,x_n-1,1)/||(x2,...,1)|| when (x1,...,xn)=(1,0,...,0)
when (x1,...,x_n)!=(1,0,...,0) then f(x)=(1,x2,...,x_n-1)/||(1,x2,...,x_n-1)||
I know that somewhere I need to divide by the norm in order to keep it that the norm of the vector is 1, but don'y know how, anyone has some hints on this?

thanks in advance.

 

so basically in order to prove it, all you need is to use g(x)=x/||x|| and the fact that S^n-1 is bijective to S^n-1\{x0}, so there's a bijection f between them then gof is the appropiate function.