# Proof: Metric Space

• Dec 6th 2005, 12:24 PM
TexasGirl
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?
• Dec 6th 2005, 06:44 PM
hpe
Quote:

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.
• Dec 7th 2005, 09:07 AM
CaptainBlack
Quote:

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 $x \neq y\$ then

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

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

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

so:

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

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

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

RonL
• Dec 8th 2005, 07:59 AM
CaptainBlack
TRIANGLE INEQUALITY

Suppose $x \neq y\$ then

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

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

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

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

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

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

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

Hence:

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

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

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

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

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

for any $z\ \epsilon\ X$.

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

RonL