Exploring Metrics on a Set: Examining Conditions of Equality

[darkchild]

Homework Statement

For every set S and every metric d on S, which of the following is a metric on S?
A. 4 + d
B. e^d + 1
C. d - |d|
D. d²
E. square root of d

Homework Equations

none

The Attempt at a Solution

I've ruled out A. because d(x,x) does not equal 0. I've ruled out C. because it's always equal to 0, even when x does not equal y for d(x,y). The correct answer is E. I can't figure out why B. and D. are not metrics.

For B., e^d - 1 is greater than or equal to 0 for all d, is only 0 for d(x,x), d(x,y) = d(y,x), and it's true that d(x,z) is less than or equal to d(x,y)+d(y,z) for all x,y,z in S, and the same is true for answer D...I'm wondering if I'm missing something in the definition of a metric? I know that it is a map to the set of real numbers, so I'm not sure what "every set S" means.

 

[hunt_mat]

E is obviously the correct answer because if $$\rho =\sqrt{d}$$

$$\begin{array}{rcl} \rho^{2}(x,y) & \leqslant & \rho^{2}(x,z)+\rho^{2}(z,y) \\ & \leqslant & \rho^{2}(x,z)+\rho^{2}(z,y)+2\rho (x,z)\rho (z,y) \\ & = & (\rho (x,z)+\rho (z,y))^{2} \end{array}$$

and so:

$$\rho (x,y)\leqslant \rho (x,z)+\rho (z,y)$$

For B and D both the first two axioms, clearly hold, so B(x,x)=0 and B(x,y)=B(y,x), likewise for D, so the way that they fail of for the triangle inequality, squaring the triangle inequality for d, shows that the triangle inequality does not hold for D, as for B, the algebra is so horrible that it clearly can't hold.