1. ## Proof: Metric Space

Let (X,d) be a metric space and let D:X*X-->R be defined by:

D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

Prove that (X,D) is a metric space.

I know that I have to show (X,D) is positive definite, symmetric and satisfies the triangle inequality.

The first I am comfortable with. However, I am not sure of how to go about showing symmetry for the x/=y case...same for the triangle inequality. Anybody have any hints or suggestions?

2. Originally Posted by TexasGirl
Let (X,d) be a metric space and let D:X*X-->R be defined by:

D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

Prove that (X,D) is a metric space.
Do you mean

D(x,y) = (0 if x=y), (min(1,d(x,y)) otherwise)

? What you have written down is essentially the discrete metric.

3. Originally Posted by TexasGirl
Let (X,d) be a metric space and let D:X*X-->R be defined by:

D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

Prove that (X,D) is a metric space.

I know that I have to show (X,D) is positive definite, symmetric and satisfies the triangle inequality.

The first I am comfortable with. However, I am not sure of how to go about showing symmetry for the x/=y case...same for the triangle inequality. Anybody have any hints or suggestions?
SYMMETRY

suppose $\displaystyle x \neq y\$ then

$\displaystyle D(x,y)\ =\ max(1,\ d(x,y))$,

but $\displaystyle d$ is a metric on $\displaystyle X$ so:

$\displaystyle d(x,y)\ =\ d(y,x)$,

so:

$\displaystyle D(x,y)\ =\ max(1,\ d(x,y))\ =\ max(1,d(y,x))\ =\ D(y,x)$.

If $\displaystyle x = y\$ then $\displaystyle D(x,y) = 0 = D(y,x)$

Hence D is symmetric in its arguments (and this works whether $\displaystyle max$ or
$\displaystyle min$ is used in the definition of $\displaystyle D$).

RonL

4. TRIANGLE INEQUALITY

Suppose $\displaystyle x \neq y\$ then

$\displaystyle D(x,y)\ =\ max(1,\ d(x,y))$.

Suppose: $\displaystyle d(x,y)\ \geq\ 1$, then

$\displaystyle D(x,y)\ =\ d(x,y)$

but $\displaystyle d(x,y)$ is a metric on $\displaystyle X$, so:

$\displaystyle d(x,y)\ \leq \ d(x,z)\ +\ d(z,y)$,

for any $\displaystyle z\ \epsilon\ X$. So

$\displaystyle d(x,y)\ \leq \ max(1,d(x,z))\ +\ max(1,d(z,y))$.

Hence:

$\displaystyle D(x,y)\ \leq \ D(x,z)\ +\ D(z,y)$.

Now suppose: $\displaystyle d(x,y)\ <\ 1$, then

$\displaystyle D(x,y)\ =\ 1$,

but $\displaystyle D(u,v)\ \geq \ 1$ for all $\displaystyle u, v\ \epsilon \ X$, so:

$\displaystyle D(x,y)\ \leq \ D(x,z)\ +\ D(z,y)$.

for any $\displaystyle z\ \epsilon\ X$.

When $\displaystyle x\ =\ y,\ D(x,y)=0$ so the triangle inequality
holds trivialy.

RonL