# Definition of a metric

I'm having a hard time understanding what the definition of a metric is. From what I think I understand, it's just a method of measurement between $2$ points in $\mathbb R^n$? Is that somewhere along the lines correct?

Then I have a homework problem that says: Consider the distance function $d:M\times M\to\mathbb R$ and then prove that $d$ is continuous with respect to the natural sum metric defined on $M\times M$, namely $d_{sum}((p,q),(p',q'))=d(p,p')+d(q,q')$.

I just don't understand what it is I'm supposed to prove. On top of that metric spaces seem so foreign and strange to me that I just don't know how to wrap my head around it.

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Look at your script. There should be the definition of a metric somewhere. You have to use this definition. You should try to get some feeling of it, but beware: The imagination in $\mathbb{R}^n$ is not sufficient in general. – azimut Mar 3 '13 at 20:17
What examples of metric spaces have you seen so far? I assume that you’ve seen the usual Euclidean metric in $\Bbb R^n$, at least for $n=1$ and $n=2$. Have you seen the taxicab metric in $\Bbb R^2$, given by $d(\langle a,b\rangle,\langle c,d\rangle)=|a-c|+|b-d|$, or the uniform metric on $\Bbb R^2$, given by $d(\langle a,b\rangle,\langle c,d\rangle)=\max\{|a-c|,|b-d|\}$? The discrete metric on a set? – Brian M. Scott Mar 3 '13 at 20:36

A metric is supposed to quantify distance. Most common type is the so called Euclidean distance that you know as the hypotenuse.

Now consider a cab driver who charges by the miles. That is distance too, but in a non-conventional geometry. He does not drill through buildings, he is restricted to available roads. So you have a "taxi-cab" metric, for the simple case where all roads are perpendicular. On a grid a meaningful distance is then $d((a,b),(x,y))=|a-x|+|b-y|$ which might be associated with $L_1$ norms. Your GPS device also takes you through a road that is shortest but not straight in the conventional sense.

Most important property of a metric is the triangle inequality. (You just won't be happy if the cab does not go through the shortest path.) So in order to go from $A$ to $C$ his chosen path better not be longer than any trip from $A$ to $B$ and then $B$ to $C$.

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Just for clarification: $d$ is a metric on $X$ if and only if it meets the criteria/definition of a metric, which applies to any metric:

• For all $a, b \in X, \;\;d(a, b) \geq 0$
• $d(a, b) = 0 \implies a = b$
• The triangle inequality holds for any three points in $X$:

$d(a, b) + d(b, c) \leq d(a, c)$.

If you are given that $d$ is a metric, then you can use any of the above properties which define a metric to prove things, like continuity, about $d$.

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So, I have a feeling using that $3^{rd}$ criteria will help me prove continuity? – TheHopefulActuary Mar 3 '13 at 21:19
Exactly! That's the one that's key here. – amWhy Mar 3 '13 at 21:24
Have you gotten anywhere on this problem, Kyle? – amWhy Mar 5 '13 at 5:00

A metric is the abstract generalization of distance in analysis. The commonly accepted definition of a metric on a set $X$ is the following:

• $d : X\times X \rightarrow \mathbb{R}$, where $\mathbb{R}$ is the set of real numbers, and it satisfies the following properties $\forall x, y \in X$:
• $d(x,y)\geq 0$
• $d(x,y) = 0 \Leftrightarrow x = y$
• $d(x,y) = d(y,x)$
• $d(x,z) \leq d(x,y) + d(y,z)$ (the triangle inequality)

A set $X$ with a metric $d$ is called a metric space. Also, although $\mathbb{R}^n$ is a metric space that can be equipped with a variety of metrics (such as the taxicab metric, Euclidean norm, etc), we have that we can make any Riemann manifold a metric space by equipping it with the appropriate metric $d$.

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By the triangular inequality for $|\cdot|$ and its reverse form for $d$, we have $$|d(x,y)-d(z,t)|=|d(x,y)-d(x,t)+d(x,t)-d(z,t)\leq |d(x,y)-d(x,t)|+|d(x,t)-d(z,t)|$$ $$\leq d(y,t)+d(x,z)=d_{sum}((x,y),(z,t)).$$ So your function $(x,y)\to d(x,y)$ is $1$-Lipschitz, hence continuous.

The metric $d_{sum}$ is not strange, it is fairly natural.

With $X=\mathbb{R}$ equipped with $d(x,y)=|x-y|$, you get the $\ell^1$ distance $$d_{sum}((x_1,y_1),(x_2,y_2))=|x_1-x_2|+|y_1-y_2|$$ on $\mathbb{R}^2$ which is induced by the $\ell^1$ norm $$\|(x,y)\|_1=|x_1|+|x_2|.$$ So $\mathbb{R}^2$ is even then a normed vector space.

An equivalent way to put a distance on $X\times X$ would be $$d_2((x_1,y_1),(x_2,y_2)):=\sqrt{d(x_1,x_2)^2+d(y_1,y_2)^2}$$ which, in the case of $\mathbb{R}$ would yield the Euclidean distance on $\mathbb{R}^2$.

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