Thread: 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?

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.

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 then

,

but is a metric on so:

,

so:

.

If then

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

RonL

TRIANGLE INEQUALITY

Suppose then

.

Suppose: , then



but is a metric on , so:

,

for any . So

.

Hence:

.

Now suppose: , then

,

but for all , so:

.

for any .

When so the triangle inequality
holds trivialy.

RonL