Proving the Equality of Topologies for the Usual and Taxicab Metrics

[LMKIYHAQ]

Homework Statement

Let d be the usual metric on RxR and let p be the taxicab metric on RxR. Prove that the topology of d = the topology of p.

Homework Equations

The Attempt at a Solution

I am trying to show that the open ball around point (x,y) with E/2 as the radius (in topology d) is a subset of the open ball around point (x,y) with E as the radius (in topology p), which is a subset of the open ball around point(x,y) with E as the radius (in topology d).

I have chosen point (a,b) and assumed it is in the usual metric. I have tried to show then that the point in the taxi-cab metric is less than E, using the usual metric's distance formula less than E/2. I have done some algebra to see that l x-a l + l y-b l < E^{2}. I am not sure how to go from here since I need to show l x-a l + l y-b l < E? Once I get the inequality less than E, then do I show\sqrt{(x-a)^{2} +(y-b)^{2}} < E?


Thanks for the help.

 

[Dick]

What does an open ball of radius r look like in the taxicab metric? Can you picture it? Describe it.

 

[LMKIYHAQ]

I have already pictured it. I thought the "proof by picture" would be a circle (usual top) inside of a rotated square (diamond shape) for taxi-cab top, and that all inside another larger circle. Is this correct? I am still having trouble with the proof even though I have visualized it.

 

[Dick]

I have already pictured it. I thought the "proof by picture" would be a circle (usual top) inside of a rotated square (diamond shape) for taxi-cab top, and that all inside another larger circle. Is this correct? I am still having trouble with the proof even though I have visualized it.

Sure. It's a diamond. It's not that hard to prove. You just figure out that the graph of the boundary |x|-|y|=r consists of four intersecting lines. The 'radius' of the diamond is 1/2 the diagonal. So a diamond of radius r is contained in a circle of radius r. What's the radius of a circle contained in a diamond of radius r?

 

[LMKIYHAQ]

Thanks for the response. I can see how what you have said works pictorially but I still can't see how I can use the formula for the usual topoloy < E/2 to get it to say that the taxicab top formula< E. I know once I do this, I will be able to assume the point (a,b) is in the the taxicab metric and then I will have the taxicab metric formula< E and will have to show somehow that the usual top formula < E? Where should I start?

 

[Dick]

All you have to show is that every ball in the ordinary metric contains a ball in the taxicab metric and vice versa. Isn't that enough to show they induce the same topology? I you know what they look like then just use geometry.

 

[LMKIYHAQ]

What algebra should I use?

 

[Dick]

What algebra should I use?

As I appended to the last message just use the geometry, if you know what the shapes look like. E.g. I have a taxicab ball around x of radius R. If I have an ordinary ball around x of radius r, how small does r have to be relative to R to make sure the circle is contained in the diamond?

 

[LMKIYHAQ]

I guess I am having trouble bc my professor wants me to show in terms of E and E/2 the containment. I know that \sqrt{(x-a)^{2}+(y-b)^{2}}=E for an open ball in the usual and l x-a l +l y-b l=E for the taxicab. I do not know how to show the taxicab is contatined on the usual here.

 

[Dick]

Ok, so you want to do inequalities instead of geometry. Then you want to find constants K and L such that K*(|x-a|+|y-b|)<=sqrt((x-a)^2+(y-b)^2)<=L*(|x-a|+|y-b|). For L just think about the triangle inequality. For K, suppose sqrt((x-a)^2+(y-b)^2)=r. Then |x-a|<=r and |y-b|<=r, right? So |x-a|+|y-b|<=?