metric continuous function

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February 15th 2010, 08:07 AM
raimis

metric continuous function

Suppose $(\mathbb{X},d)$ is a metric space and the corresponding product topology is defined on $\mathbb{X}\times\mathbb{X}$ . Prove that the metric d is continuous.
February 15th 2010, 11:32 AM
Opalg

Quote:

Originally Posted by raimis

Suppose $(\mathbb{X},d)$ is a metric space and the corresponding product topology is defined on $\mathbb{X}\times\mathbb{X}$ . Prove that the metric d is continuous.

You have to show that if $(x,y)$ is close to $(x_0,y_0)$ in $\mathbb{X}\times\mathbb{X}$ , then $d(x,y)$ is close to $d(x_0,y_0)$ . More precisely, you have to show that, given $(x_0,y_0)\in\mathbb{X}\times\mathbb{X}$ and $\varepsilon>0$ , there exists $\delta>0$ such that if $d(x,x_0)<\delta$ and $d(y,y_0)<\delta$ then $|d(x,y) - d(x_0,y_0)|<\varepsilon$ . Use the triangle inequality.
February 15th 2010, 11:48 AM
raimis

product topology

And how do we make use of the product topology here?
February 15th 2010, 01:04 PM
Opalg

Quote:

Originally Posted by raimis

And how do we make use of the product topology here?

The condition $d(x,x_0)<\delta$ and $d(y,y_0)<\delta$ expresses the fact that $(x,y)$ is close to $(x_0,y_0)$ in the product topology.
February 15th 2010, 07:08 PM
Drexel28

Quote:

Originally Posted by raimis

Suppose $(\mathbb{X},d)$ is a metric space and the corresponding product topology is defined on $\mathbb{X}\times\mathbb{X}$ . Prove that the metric d is continuous.

I'm sorry. What is this question actually asking. I see that Opalg has already answered it, but I am just curious. Is it that if $(X,d)$ is a metric space then $d:X^2\mapsto\mathbb{R}$ is continuous?
February 15th 2010, 10:47 PM
raimis

Quote:

Originally Posted by Drexel28

Is it that if $(X,d)$ is a metric space then $d:X^2\mapsto\mathbb{R}$ is continuous?

Yes, the question is as you say. But the problem involves product topology and I am wondering whether it should necessarily be defined on $\mathbb{X}\times\mathbb{X}$ to prove the statement.