Metric Space

What is a metric space?

A metric space is a pair $ (X, d) $, where $ X $ is a set and $ d $ is a function (called a metric) that assigns a non-negative real number to every pair of points $ x, y \in X $, denoted $ d(x, y) $, which represents the distance between $ x $ and $ y $. This is usually written as $ (X, d) $. $$ (X,d) $$

The metric must satisfy the following properties:

Non-negativity: $ d(x, y) \geq 0 $ for all $ x, y \in X $, and $ d(x, y) = 0 $ if and only if $ x = y $ (the distance from a point to itself is zero, and the distance between two distinct points is strictly positive).
Symmetry: $ d(x, y) = d(y, x) $ for all $ x, y \in X $ (the distance from $ x $ to $ y $ is the same as from $ y $ to $ x $).
Triangle inequality: $ d(x, y) + d(y, z) \geq d(x, z) $ for all $ x, y, z \in X $ (the direct distance between two points is always less than or equal to the sum of the distances through a third point).

In short, a metric space provides a mathematical structure that allows us to measure the distances between points in a set, enabling us to rigorously explore geometric concepts such as continuity, convergence, and compactness.

Put simply, a metric space is a set X with a distance function d.

This set can be anything from a basic collection of points to a full vector space.

A practical example
The distance function or metric
Distance induced by a norm
Additional Notes

A practical example

A classic example of a metric space is Euclidean space in $ \mathbb{R}^n $, which refers to the set of points in the plane (when $ n = 2 $) or in three-dimensional space (when $ n = 3 $).

Consider $ \mathbb{R}^2 $, the Cartesian plane.

The Euclidean metric $ d $ is defined for two points $ p = (p_1, p_2) $ and $ q = (q_1, q_2) $ as:

$$ d(p, q) = \sqrt{(p_1 - q_1)^2 + (p_2 - q_2)^2} $$

This formula gives the Euclidean distance, which is the straight-line distance between two points $ p $ and $ q $ in the plane.

This definition satisfies the properties of a metric:

Non-negativity: The square root is always non-negative, and $ d(p, q) = 0 $ if and only if $ p_1 = q_1 $ and $ p_2 = q_2 $, meaning $ p = q $.
Symmetry: We have $ d(p, q) = d(q, p) $ because $ (p_1 - q_1)^2 $ is equal to $ (q_1 - p_1)^2 $, meaning the distance is the same no matter the order of the points.
Triangle inequality: The direct distance between two points is always less than or equal to the sum of the distances through a third point, as can be verified using the Pythagorean theorem and other properties of Euclidean geometry.

In conclusion, the space $ (\mathbb{R}^2, d) $, where $ d $ is the Euclidean metric, is an example of a metric space.

The distance function or metric

What is the distance function?

The distance (or metric) is a function $ d(x_1, x_2) $ such that:

$ d(x_1, x_2) \geq 0 $
$ d(x_1, x_2) = 0 $ if and only if $ x_1 = x_2 $
$ d(x_1, x_2) = d(x_2, x_1) $
$ d(x_1, x_2) \leq d(x_1, x_3) + d(x_3, x_2) $

for all $ x_1, x_2, x_3 \in X $.

Types of distances

There are several different distance functions, not just one.

Euclidean distance

$$ d_2(x, y) := \sqrt{ \sum{(x_i - y_i)^2 } } $$

This is the most common distance because it's the foundation of Euclidean geometry.

Manhattan distance

This distance is fundamental in taxi geometry. It’s called "Manhattan distance" because, just like navigating a city grid, you can't move diagonally between buildings.

$$ d_1(x_1, x_2) := \sum{ |x_i - y_i| } $$

Discrete distance

In this case, the distance is always 1 unless the two points $ x $ and $ y $ are the same, in which case the distance is 0.

$$ d(x, y) := \begin{cases} 0 \:\:\: if \: x = y \\ 1 \:\:\: if \: x \ne y \end{cases} $$

Distance induced by a norm

The norm always induces a distance function.

In such cases, the distance is known as the induced distance.

$$ ||v|| := d(v, 0_V) $$

The length of a vector is simply the distance from the origin.

Thus, if a vector space has a norm, it is also a metric space.

Note. However, the reverse is not always true. A distance function may not necessarily be associated with a norm.

Properties of the induced distance

A distance is said to be induced if it satisfies the following conditions:

$ d(v_1 + v_3, v_2 + v_3) = d(v_1, v_2) $
$ d(k \cdot v_1, k \cdot v_2 ) = |k| \cdot d(v_1, v_2) $

Where $ v_1, v_2, v_3 $ are vectors in vector space $ V $ and $ k $ is a scalar $ k \in K $.

Example

Using this definition, it's easy to demonstrate that the Euclidean norm induces the Euclidean distance because it meets the conditions above.

Let's take three vectors $ v_1, v_2, v_3 $ in Euclidean space

$$ v_1 = (6,8) \\ v_2 = (3,4) \\ v_3 = (3,0) $$

The vectors have the following norms:

$$ ||v_1||_2 = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 $$ $$ ||v_2||_2 = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 $$ $$ ||v_3||_2 = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 $$

Therefore, they have the following induced distances:

$$ ||v_1||_2 = d(v_1, 0_v) = 10 $$ $$ ||v_2||_2 = d(v_2, 0_v) = 5 $$ $$ ||v_3||_2 = d(v_3, 0_v) = 3 $$

According to the definition, $$ ||v|| = d(v, 0_v) $$ holds if the following conditions are satisfied:
1] $ d(v_1 + v_3, v_2 + v_3) = d(v_1, v_2) $
2] $ d(k \cdot v_1, k \cdot v_2) = |k| d(v_1, v_2) $

Let’s verify these two conditions.

First condition

$$ d(v_1 + v_3, v_2 + v_3) = d(v_1, v_2) $$ $$ d(10 + 3, 5 + 3) = d(10, 5) $$ $$ d(13, 8) = d(10, 5) $$

The distance on the left-hand side is

$$ d(13, 8) = \sqrt{(13 - 8)^2} = \sqrt{5^2} = \sqrt{25} = 5 $$

The distance on the right-hand side is

$$ d(10, 5) = \sqrt{(10 - 5)^2} = \sqrt{5^2} = \sqrt{25} = 5 $$

Therefore

$$ d(13, 8) = d(10, 5) = 5 $$

The first condition is satisfied.

Second condition

$$ d(k \cdot v_1, k \cdot v_2) = |k| \cdot d(v_1, v_2) $$ $$ d(k \cdot 10, k \cdot 5) = |k| \cdot d(10, 5) $$

Let's take $ k = 2 $.

$$ d(2 \cdot 10, 2 \cdot 5) = |2| \cdot d(10, 5) $$ $$ d(20, 10) = |2| \cdot d(10, 5) $$

Where

$$ d(20, 10) = \sqrt{(20 - 10)^2} = \sqrt{10^2} = 10 $$

and

$$ |2| \cdot d(10, 5) = 2 \cdot \sqrt{(10 - 5)^2} = 2 \cdot 5 = 10 $$

Thus

$$ d(k \cdot 10, k \cdot 5) = |k| \cdot d(10, 5) = 10 \:\:\: with \:\: k = 2 $$

The second condition is also satisfied.

We’ve now shown that in Euclidean space, the distance is induced by the norm.

Additional Notes

Some additional observations and remarks on metric spaces:

Bounded Set in a Metric Space
Let $(X, d)$ be a metric space, where $d$ is the metric defined on $X$. A subset $A \subseteq X$ is said to be bounded if there exists a positive real number $\mu > 0$ and a fixed point $x_0 \in X$ such that: $$ d(x, x_0) \leq \mu \quad \text{for all } x \in A $$ In simple terms, all points in $A$ lie within an open or closed ball of radius $\mu$, centered at $x_0$. In this context, a set is considered "bounded" if it can be fully enclosed within a ball (a region of space determined by the metric) with a finite radius.
In the topology induced by the metric $d$, the concept of a bounded set is not inherently linked to whether the set is open or closed. It solely concerns the distances between points in the set.
Bounded Metric
In a metric space $(X, d)$, if the entire set $X$ is bounded, then the metric $d$ is referred to as a bounded metric.
The basis theorem for the topology induced by a metric
In a metric space $(X, d)$, the collection of open balls $ \mathcal{B} $ serves as a basis for a topology on $X$. $$ \mathcal{B} = \{B_d(x, \varepsilon) \mid x \in X, \varepsilon > 0\} $$
Continuity Theorem in Metric Spaces
A function $f : X \to Y$ between two metric spaces $(X, d_X)$ and $(Y, d_Y)$ is continuous if, for every point $x \in X$ and any $\varepsilon > 0$, there exists a $\delta > 0$ such that whenever two points $x, x' \in X$ are less than $\delta$ apart $$ d_X(x, x') < \delta $$ their images $f(x)$ and $f(x')$ in $Y$ will be less than $\varepsilon$ apart $$ d_Y(f(x), f(x')) < \varepsilon $$
Metric Spaces are Hausdorff Spaces
Every metric space is Hausdorff. Conversely, if a topological space is not Hausdorff, it cannot arise from a metric.
Note: A topological space is considered Hausdorff if any two distinct points can be separated by disjoint open neighborhoods.

And so forth.