Euclidean Space

Introduction to Euclidean Spaces. Metric Spaces

German Sanmi June 16, 2026 en

0. Index

Presentation and Definition. $\mathbb{R}^k$
Metric Spaces.
- 2.1. Metric definition. Metric spaces.
- 2.2. Euclidean metric.
- 2.3. $\mathbb{R}^k$ is a metric space with the euclidean metric.
Main properties of the norm in $\mathbb{R}^k$.
Summary.

1. Presentation and Definition. $\mathbb{R}^k$

A Euclidean space is an affine space over the reals such that the associated vector space is a euclidean vector space; a finite-dimensional inner product space over the real numbers.

Let's consider the affine space $(\mathbb{R}^k,V,+)$ where the associated $\mathbb{R}$-vector space is $V= (\mathbb{R},(\mathbb{R}^k,+),·)$, with the operations as follows:

\[(x_1,...,x_k)+(y_1,...,y_k)=(x_1+y_1,...,x_k+y_k)\] \[\alpha · (x_1,...,x_k) = (\alpha x_1,..., \alpha x_k)\]

Being $+: V \times \mathbb{R}^k \to \mathbb{R}^k $ a simply transitive action over $\mathbb{R}^k$

Over $V$, we define the following inner product, which we remember from the Complex Numbers post, on $6.3.1$.

An inner product is the object which injects geometry by defining how a vector expresses its own direction (norm), introducing subsequently the orthogonality. In this particular presentation, this inner product receives the particular name dot product.

\[\quad \quad \langle x,y \rangle = x · y =\sum_{i=1}^k x_iy_i\]

The defined dot product gives rise to the norm; for each vector $x \in \mathbb{R}^k$ it is defined:

\[\|x\| = \sqrt{ x·x}= \sqrt{\sum_{i=1}^k x_i^2}\]

Which gives us a notion of "length" of the vector. It is worth saying that this is specifically true in $\mathbb{R}^k$ where intuitive geometry matches the formal geometry instituted through the dot product.

Thus, a euclidean space is the result of injecting geometry into an affine space through a real inner product space; a real vector space with geometry.

As a consequence of the geometry injected by the inner product, we can consider a notion of proximity (metric) between the points in the affine space which allows us to introduce the notion of limit and develop the analysis.

2. Metric spaces.

2.1. Metric definition. Metric spaces.

A metric abstracts the notion of distance. This is an important concept since from this all the analysis departs. From the distance notion comes proximity and then limits gain sense.

A distance tries to capture how far two points are from each other in a non-empty set $X$. Observe that this is not an algebraic structure or algebraic property; contrary to the norm, the distance doesn't need to sum or scale.

Let $X$ be a non-empty space, then a metric space is a pair $(X,d)$ where: $d : X \times X \to \mathbb{R}$ is an application called distance, verifying:

$M1$: Symmetry, the distance is indifferent to the point of measurement:
\[d(x,y)=d(y,x) \quad \forall x,y \in X\]
$M2$: Gets null on the diagonal:
\[d(x,y)=0 \iff x = y \quad \forall x,y \in X\]
$M3$: Triangular inequality:
\[d(x,z) \leq d(x,y) + d(y,z) \quad \forall x,y,z \in X\]

Observe some immediate consequences of the three axioms:

\[d(x,x) = 0 \leq d(x,y) + d(y,x) = 2d(x,y) \implies 0 \leq d(x,y) \quad \forall x,y \in X\]

The distance is always positive as a consequence of the triangular inequality and the symmetry.

2.2. Euclidean metric.

Let's consider again the affine space $\mathbb{R}^k$ as defined above with the dot product. Then, $V$ is an inner product vector space and, consequently, a normed vector space. Let's see how we can develop a metric in $\mathbb{R}^k$ to state it as a metric space.

First, at this point, we will develop the metric in $\mathbb{R}^k$ through its norm (depending on the inner product) although we established this is not entirely necessary.

We define the metric in $\mathbb{R}^k$ as, for $x,y \in \mathbb{R}^k$, we have that:

\[d(x,y) = \|\overrightarrow{xy}\| = \|y - x\|\]

This is, the distance between two points on $\mathbb{R}^k$ is the length of the vector that connects both points. Remember from our knowledge of affine space that $x + \overrightarrow{xy} = y$

2.3. $\mathbb{R}^k$ is a metric space with the euclidean metric.

Let's see that $(\mathbb{R}^k,d)$ as defined is a metric space:

Symmetry:
\[d(x,y) = \|x - y\| = \sqrt{\sum_i^k (x_i -y_i)^2} = \sqrt{\sum_i^k (y_i - x_i)^2} = \| y - x \| = d(y,x)\]
Zero distance along the same point:
\[d(x,y) = \|x - y\| = \sqrt{\sum_i^k (x_i -y_i)^2}\]
Since it is a sum of positive terms, to be zero all the operands have to be zero, this is:
\[d(x,y) = 0 \iff x_i =y_i \quad i=1,2,...,k \iff x = y \quad \forall x,y \in \mathbb{R}^k\]
Triangle inequality: This result comes immediately from the inner product. Take $x,y \in \mathbb{R}^k$, then:
\[\|x+y\|^2 = \|x\|^2 + \|y\|^2 + 2 \langle x,y \rangle\]
Let's see that, by Cauchy-Schwarz: $|\langle x,y \rangle| \leq ||x|| ||y||$, thus
\[\|x+y\|^2 = \|x\|^2 + \|y\|^2 + 2 \langle x,y \rangle \leq \|x\|^2 + \|y\|^2 + 2\|x\| \|y\| = (\|x\| + \|y\|)^2\]
Since $||x+y|| \geq 0$ and $||x|| + ||y|| \geq 0$ is:
\[\|x+y\|^2 \leq (\|x\| + \|y\|)^2 \implies \|x+y\| \leq \|x\| + \|y\|\]
Giving place to the triangular inequality (which is always true in a real inner product space).

Now, observe that:
\[d(x,z) = \|x -z\| = \|x -y + y-z\| \leq \|x-y\| + \|y-z\| = d(x,y)+d(y,z)\]
The inequality is guaranteed by the triangle inequality in the norm, preceded by Cauchy-Schwarz.

3. Main properties of the norm in $\mathbb{R}^k$

Let's evaluate some main properties of the norm of $\mathbb{R}^k$.

$||x|| \geq 0$, is immediate from the fact that is the square of a real number.
$||x|| = 0 \iff x = 0$, immediate from norm definition.
$||\alpha x || = (\langle \alpha x, \alpha x \rangle)^\frac{1}{2} = (\alpha^2 \langle x,x \rangle)^\frac{1}{2} = |\alpha| ||x||$
$||x · y|| \leq ||x|| ||y||$ Immediate from Cauchy-Schwarz.
$||x+y|| \leq ||x|| + ||y||$ Triangle inequality previously demonstrated.
$||x-z|| \leq ||x-y|| + ||y-z||$ An extension of the triangle inequality relative to the euclidean metric also demonstrated above.

4. Summary.

Thus, we've seen that $\mathbb{R}^k$ as an affine space with $\mathbb{R}^k$ as the associated vector space and the real inner product as the dot product; $x · y$ conforms a Euclidean space; an affine space with a geometry induced by the vector space which crystallizes in the metric $d(x,y) = ||\overrightarrow{xy}||$

5. Exercises.

5.1.

Suppose $k \geq 3$, $x,y \in \mathbb{R}^k :

x-y

= d > 0$ and $r>0$. Prove:

If $2r > d$ there are infinitely many $z \in \mathbb{R}^k$ such:
\[|z - x| = |z - y| = r\]
Observe that by the triangle inequation is:
\[|x -y| \leq |x-z| + |z-y|\]
Thus, the points requested by the exercise are those points satisfying:
\[z \in \mathbb{R}^k: \begin{cases}|x-z| = |z-y|\\ d = |x -y| < |x-z| + |z-y| = 2r\end{cases}\]
Note that the first condition give us a general caracterization of the points $z$, those whose coordinates satisfies the equation:
\[\left(\sum_{i=1}^k(x_i - z_i)^2\right)^{1/2} = \left(\sum_{i=1}^k(y_i - z_i)^2\right)^{1/2} \iff \sum_{i=1}^k(x_i - z_i)^2 = \sum_{i=1}^k(y_i - z_i)^2\]
Note that for each square is:
\[(x_i - z_i)^2 = x^2_i -2x_iz_i + z^2_i = y^2_i -2y_iz_i + z^2_i = (y_i - z_i)^2\]
The $z^2_i$ cancels from both sides and sorting:
\[2z_i(y_i - x_i)=y^2_i - x^2_i\]
Adding now the summatories we have:
\[z·(y-x) = \frac{1}{2}(|y|^2 - |x|^2)\]

Maths