Introduction to Metric Spaces

Metric Spaces are one of those mathematical topics that everyone intuitively understands. The best example of this is that of three dimensional Euclidean space E^{3}. This serves as the basis for the intuitive concept of a “space”, and our ability to ascribe a distance between to points in three-dimensional space can be described by a distance function d: E\times E \rightarrow E, or a metric. The underlying set E together with the metric d form what is called a metric space (E,d).

To be more mathematically precise, we make the following definition.

Definition. A metric space is a set E together with a rule which associates with pair p,q\in E a real number d(p,q) such that
\displaystyle d(p,q)\geq 0, \forall p,q\in E
\displaystyle d(p,q) = 0 \iff p=q
\displaystyle d(p,q)=d(q,p) \forall p,q\in E
\displaystyle d(p,r) \leq d(p,q)+ d(q,r)

As an example, suppose that the underlying set E = \mathbb{R} and the metric coupled with this set is defined by d(p,q)= |p-q|. To verify that this indeed a metric space we must show that the four axioms are satisfied.

Claim. The mathematical structure (\mathbb{R},d) in which d:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R} is defined by d(p,q)=|p-q| is a metric space.
Proof. Let p,q\in \mathbb{R}. Then by definition of d(p,q) and by definition of the absolute value function, we have that |p-q|\geq 0, so that axiom 1 is satisfied. Suppose now that the points in \mathbb{R} are equal, i.e. that p=q\in \mathbb{R}. Then by definition of d(p,q), we have that |p-q|=|0|=0. Conversely, suppose that |p-q|= 0. By the triangle inequality we have that |p-q|\geq |p|-|q|=0 This implies that |p|= |q|. Thus, condition (2.) is satisfied and hence the distance between two points in \mathbb{R} is zero if and only if the two points are the same. To prove condition (3.), let p,q\in \mathbb{R}, so that d(p,q) = |p-q|. By virtue of the definition of the absolute value, we can say that
\displaystyle d(p,q) = |p-q| = |-(p+q)|= |-1||q-p|= |q-p|= d(q,p).
Thus, we see that the arguments of the proposed distance function is symmetric with respect to its arguments, namely any real numbers p,q\in \mathbb{R}. To prove condition (4.), consider any three points p,q,r\in \mathbb{R}, then the distance function between the points p,r \in \mathbb{R} becomes
\displaystyle d(p,r) = |p-r| = |p-q+q-r|,
wherein we add zero in the form of adding and subtracting the point q (a very common trick in analysis). Then by the properties of the absolute value it follows that
\displaystyle d(p,r) = |p-q+q+r| \leq |p-q|+|q-r| = d(p,q)+d(q,r),
where by the last equality follows from the definition of the distance function. Therefore, all four conditions have been satisfied, and hence by our definition of a metric space, it follows that (\mathbb{R},d) whose distance function d:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R} is defined by d(p,q) = |p-q| is indeed a metric space. \square

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