# Definition of a function by a metric.

Gold Member
Let (X,p) be a metric space with p metric on X, define for each subset A of X
$$p(x,A)=inf_{y\in A}p(x,y)$$
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?

morphism
Homework Helper
It's probably easier to use the good old epsilon-delta definition of continuity here. The following observation will be helpful: p(x,A) <= p(x,z) + p(z,A), for any z in X.

Morphism is right. You have to do what he suggested twice, in order to get the absolute value on the left side. This is assuming that you've already learned that the epsilon-delta definition of continuity applies to metric spaces in general (and it does, calculus clases specialize in R^n, but it can be proven that it applies to metric spaces in general).

Gold Member
ok, I have another question, let $$S^n$$={$$x\in R^n|||x||=1$$}
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?