Metric spaces Definition

Question

The definition I have been given for a metric space is such:

A metric, or a distance function on $X$ is a function $d : X \times X \to \mathbb{R}$ ........

So my question is: This is essentially saying a mapping from a Cartesian product to the real numbers. So wouldn't this just be some point in your set that maps to a real number, not two points that map to a real number? For example $A=\{1,2,3,4\}$ then $A \times A$ would simply be a pair $(a,b)$ ($a,b \in A$) and how would a pair mapping to a real number create a distance?

Answer 1

Perhaps an example of a metric would help. Suppose you had a distance function $d: \mathbb{R}\times\mathbb{R}\to\mathbb{R}$ defined by $$d(x,y)=|y-x|$$ where $||$ denotes the absolute value of the difference. Then from the set of all ordered pairs $\mathbb{R}\times\mathbb{R}$, you receive a real number as an output, and this number corresponds to a distance, which, in this case, is the difference between two real numbers.

Another sub example, what is the distance between 3 and 1, for instance? Take $d(3,1)=|3-1|=2,$ thus 3 is two away from 1.

Main Idea: The ordered pair $(a,b)$ could be thought of as a point of $\mathbb{R}\times\mathbb{R}$ or as an interval in $\mathbb{R}$.

Answer 2

A metric space generalized distance and is more than merely a mapping from a Cartesian product to the real numbers. First, note that the distance function on $X$ is the cartesian product with itself, not just any random set. So basically we're looking at elements of one set and calculating distances between them.

But the notion of distance has more meaning than simply any function. We have a few other axioms a function must satisfy to be considered a metric.

The first is that $d(x,y) \geq 0$ and $d(x,y)=0$ if and only if $x=y$. So distances are positive numbers and the distance between a point and itself is zero. Secondly we have $d(x,y)=d(y,x)$ which is to say that it doesn't matter which point you measure the distance from because you'll get the same answer.

Finally, and most importantly you have the triangle inequality, $d(x,z) \leq d(x,y) + d(y,z)$ which means that if you are going from $x$ to $z$ and you decide to go through $y$ on the way you can't make your journey any shorter. Either $y$ is on the way or $y$ is out of the way and in either case the distances will be bigger or equal to the distance of going directly.

If you can meet all these requirements you've got a reasonable abstraction of distances. Many proofs only require these three simply properties so they're worth getting to know better.