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.
 A: 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 $d$.
A: 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 $C$.
A: 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 $d$. 
A: 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 $\mathbb{R}^2$.
