Usual and taxicab metric topologies

[Unassuming]

I am trying to show that an "epsilon ball in the taxicab metric" can be contained in a "epsilon ball in the usual metric".

Let "Up" be a taxicab ball, and "Ud" be a usual metric ball.

How can I show that Up( (x,y) , e ) is a subset of Ud( (x,y), e ) ?



I have drawn the picture and I understand how the taxicab ball is inside. It is a diamond that touches the usual ball on each of the diamond's four corners and is less than everywhere else.

How can I show this algebraically?

 

[Unassuming]

I actually figured this out from a previous thread on this forum. Yippee.

I need to show the reverse is true though and I can't it out. I need to show that we can construct a taxicab metric that contains any usual epsilon-ball.

 

[Dick]

Just use geometry again. Now you have a circle inscribed in a diamond instead of vice versa. Neither one is that hard.

 

[Unassuming]

I have to use algebra... for instance, for the first part I said,

Since

$$|x-a| =\sqrt{(x-a)^2} \leq \sqrt{ (x-a)^2 + (y-b)^2 } = \varepsilon$$

and

$$|y-b| = \sqrt{(y-b)^2} \leq \sqrt{ (x-a)^2 + (y-b)^2 } = \varepsilon$$

then

$$|x-a| + |y-b| \leq 2\varepsilon$$

Let

$$k= \frac{1}{2}$$

Then

$$k(|x-a| + |y-b|) \leq \varepsilon = \sqrt{ (x-a)^2 + (y-b)^2 }$$

 

[Unassuming]

Well, I don't know if you can see that latex but this is what I did,

Since |x-a| =\sqrt{(x-a)^2} <= sqrt{ (x-a)^2 + (y-b)^2 } = e

and |y-b| = \sqrt{(y-b)^2} <= \sqrt{ (x-a)^2 + (y-b)^2 } = e

then |x-a| + |y-b| <= 2e

Let k= 1/2. Then k(|x-a| + |y-b|) <= e = \sqrt{ (x-a)^2 + (y-b)^2 }

 

[Dick]

Yes, k=1/2 works fine.

 

[Unassuming]

Is showing the other way similar? I am not getting anywhere. You said use the traingle inequality. Are we still looking for a constant?

 

[Dick]

Just write down the triangle inequality for the three points (a,b), (x,y) and (x,b).