Difference between Metrics

IsThisIt???

Could anyone explain to me the practical difference between the taxi cab metric and the Euclidean metric. In my notes the taxicab is defined using the abs value of the difference of the points while the euclidean is defined using the sqrt of the difference of the points squared.

In practice are these not the same thing. Why do both exist if this is the case?

Thanks for your help

MathsManiac

But they are clearly not the same.

For example, the distance from the point (0, 0) to (3, 4) under the taxicab metric is 3+4=7, while the distance using the ordinary Euclidean metric is sqrt(3^2 + 4^2) = 5.

Furthermore, changing metrics can reverse which of two distances is greater. For example, using the taxicab metric, (3, 4) is farther from the origin than (5, 1), because 7>6, while with the usual Euclidean metric, it is closer, because 5<sqrt(26).

Using the taxicab metric, a "circle" centred at the origin and of radius 5 would look like a square whose vertices are (5,0), (0,5), (-5,0), and (0,-5), as this square (i.e., these four edges) is the set of all points whose distance under this metric is equal to 5.

IsThisIt???

Ah yes, seems clear and obvious now but coming towards the end of 12 hours in the library things get a bit distorted.

Perhaps you could shed some light on the following question I found in my notes as it was this that led me down a thought process that made me question above :

"The metrics d1 and d2 on R2 yield precisely the same open
sets. Why?"

CJC86

IsThisIt??? wrote: »

Ah yes, seems clear and obvious now but coming towards the end of 12 hours in the library things get a bit distorted.

Perhaps you could shed some light on the following question I found in my notes as it was this that led me down a thought process that made me question above :

"The metrics d1 and d2 on R2 yield precisely the same open
sets. Why?"

So, the reason why they yield the exact same open sets is because they are equivalent metrics, hence they are topologically equivalent.

Metrics [latex]d_1[/latex] and [latex]d_2[/latex] are equivalent if there exist [latex]\alpha[/latex] and [latex]\beta[/latex] such that [latex] \alpha d_2(x,y) \leq d_1(x,y) \leq \beta d_2(x,y)[/latex]. I believe that [latex] \alpha=1, \skip \beta = \sqrt{2}[/latex] work in this case.

From this definition we can see that for any open ball in one metric we can find an open ball in the other metric that is contained in the first open ball, hence topological equivalence.

You can read more about it here: http://en.wikipedia.org/wiki/Equivalence_of_metrics