Is the Characteristic Function Continuous in a Given Metric Space?

[Carl140]

Homework Statement

Let (X,d) be a metric space, A subset of X, x_A: X->R the characteristic
function of A. (R is the set of all real numbers)

Let V_d(x) denote the set of neighbourhoods of x with respect the metric d.
Prove that x_A is continuous in x (x in X) if and only if there
exists an element V in V_d(x) such that V is a subset of A and V is a
subset of X\A.

The Attempt at a Solution

OK, so assume x_A is continuous then for each closed subset of R
the preimage of this closed subset under x_A must be closed.
Take V= {0} then V is closed so (x_A)^(-1) = { y in X: x_A(y) = 0} = X\A.

But I don't see how this helps..I don't know hwo to find such V in V_d(x).

 

[Carl140]

My work:
So assume it is continuous at a point c.
Then put e=1/2 then there exists delta > 0 such that if d(x,c)< delta
then |x_A(x) - x_A(c)|< 1/2.
So we have 2 cases: x and c in A or x,c in X\A.
Then I don't know what to take as V..

 

[morphism]

Am I reading this right? V is a subset of both A and X\A? This is only possible if V is empty, and the empty set is certainly not a neighborhood of anything.