Thread: Euclidean metric vs Taxicab metric?

1. Euclidean metric vs Taxicab metric?

Does it matter which metric we use on R^2?

Does it change how sequences converge?

2. Re: Euclidean metric vs Taxicab metric?

Originally Posted by Jame
Does it matter which metric we use on R^2?
Does it change how sequences converge?

Do you know about equivalent metrics?

3. Re: Euclidean metric vs Taxicab metric?

Thank you for answering. I was reading something similar. I was mostly confused since their "open sets" are different.

Open sets under the Euclidean metric being circles, and open sets under the taxicab one being squares.

The second bullet point on the page you linked states they are equivalent if the open sets "nest" (are contained in one another)

I can cleary see why this is true, for you can draw a circle inside of a square and a square inside of a circle (as for the values of r', r'' I'm not sure what they'd be)

So they are equivalent

Is there any reason the Euclidean is preferred? Is it just more natural and generalizes easier to higher dimensions? Must every metric be a geodesic?

4. Re: Euclidean metric vs Taxicab metric?

Originally Posted by Jame
Thank you for answering. I was reading something similar. I was mostly confused since their "open sets" are different.

Open sets under the Euclidean metric being circles, and open sets under the taxicab one being squares.

The second bullet point on the page you linked states they are equivalent if the open sets "nest" (are contained in one another)

I can cleary see why this is true, for you can draw a circle inside of a square and a square inside of a circle (as for the values of r', r'' I'm not sure what they'd be)

So they are equivalent

Is there any reason the Euclidean is preferred? Is it just more natural and generalizes easier to higher dimensions? Must every metric be a geodesic?
It's not really because you can draw one inside each other. It is because a unit circle ( $\mathbb{R}^2$) is homeomorphic to the unit square in $\mathbb{R}^2$. A simplistic way of describing it would be, imagine continously stretching the circle to get the shape of a square.

A metric on $\mathbb{R}^n$ induces a topology, so if you can prove that two different metrics induce the same topology on $\mathbb{R}^n$(find a homeomorphism between the two topological spaces), then those metrics are equivalent.