Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $C$.

share|cite|improve this answer

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 $d$.

share|cite|improve this answer
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 $d$.

share|cite|improve this answer

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 $\mathbb{R}^2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.