2. Euclidean distance; balls and spheres

1. Euclidean distance in $\mathbf{R}^n$

In conformity with the general definitions the Euclidean distance between two points $x = (x_i)$ and $y = (y_i)$ is the number

We recall its principal properties. The relation $d(x, y) = 0$ is equivalent to $x = y$. We have $d(x, y) = d(y, x)$; for all scalars $t \in \mathbf{R}$, $d(tx, ty) = |t| d(x, y)$; for all $z \in \mathbf{R}^n$, $d(x + z, y + z) = d(x, y)$; in other words, the distance between two points is invariant under translation. The distance $d(o, x)$ from the origin $o$ to a point $x$ is denoted also by $\|x\|$ and is called the Euclidean norm of $x$ (or simply the norm of $x$, when there is no likelihood of confusion; cf. Chapter IX, § 3, no. 3). Then $d(x, y) = \|y - x\|$.

For $n = 1$, the Euclidean distance between the points $x, y$ of $\mathbf{R}$ reduces to the length $|y - x|$ of the intervals with $x$ and $y$ as end-points. For any $n$, we say that $d(x, y) = \|y - x\|$ is the length of the segments with $x$ and $y$ as end-points.

The Euclidean distance satisfies the triangle inequality

(I)

We recall that the proof of (I) reduces to that of the inequality

this in turn is equivalent to the Cauchy-Schwarz inequality

which is an immediate consequence of Lagrange’s identity

This proof shows at the same time that the two sides of (1) can be equal only if $z$ is a point of the segment with $x$ and $y$ as end-points.

From (1) we deduce the inequality

Finally, if $x = (x_i), y = (y_i)$, we have

Hence a subset $A$ of $\mathbf{R}^n$ is bounded (§ 1, no. 1) if and only if

2. Displacements

We recall again that the affine transformations $f$ of $\mathbf{R}^n$ onto itself which leave invariant the distance between any two points [that is to say, such that $d(f(x), f(y)) = d(x, y)$ for all $x, y$] are called Euclidean displacements (or simply displacements) (*); they form a group, called the group of displacements of $\mathbf{R}^n$. This group operates transitively on $\mathbf{R}^n$; more generally, if $V$ and $V'$ are any two affine linear varieties of the same dimension in $\mathbf{R}^n$, there exists a displacement which transforms $V$ into $V'$. The displacements which leave the origin fixed, called orthogonal transformations, form a subgroup of the group of all displacements. This subgroup is called the orthogonal group on $n$ real variables; the linear mappings which belong to this group are characterized by the fact that they leave invariant the norm $\|x\|$ of every point $x \in \mathbf{R}^n$, or, equivalently, the quadratic

(*) If $f$ is subjected only to the condition $d(f(x), f(y)) = d(x, y)$ for all $x, y$, then in fact $f$ must be affine and linear, and therefore a displacement.

form $\|x\|^2 = \sum_{i=1}^n x_i^2$. The scalar product of two vectors $x = (x_i)$ and $y = (y_i)$ of $\mathbf{R}^n$ is the value $\sum_{i=1}^n x_i y_i$ of the bilinear form associated with the quadratic form $\sum_{i=1}^n x_i^2$; it is denoted by $(x|y)$, or simply by $xy$ if there is no likelihood of confusion. Every orthogonal transformation leaves invariant the scalar product of any two vectors. Two vectors $x, y$ are said to be orthogonal if $(x|y) = 0$; two vector subspaces $V, V'$ of $\mathbf{R}^n$ are said to be orthogonal if each $x \in V$ is orthogonal to each $y \in V'$; and two affine linear varieties $P, P'$ are said to be orthogonal if the vector subspaces parallel respectively to $P$ and $P'$ are orthogonal.

3. Euclidean balls and spheres

For each integer $p > 0$, let $U_p$ denote the set of all pairs $(x, y)$ of points of $\mathbf{R}^n$ such that $d(x, y) < 1/p$; the inequalities (3) show that the sets $U_p$ form a fundamental system of entourages of the uniformity of $\mathbf{R}^n$ (cf. Chapter IX, § 2).

From this fact and from the inequality

When there is no risk of confusion we say simply "ball" (resp. "sphere") for "Euclidean ball" (resp. "Euclidean sphere"). When $n = 2$, a ball of two dimensions is called a disc, and a sphere of one dimension is called a circle. When $n = 1$, the open (resp. closed) ball with centre $x_0$ and radius $r$ is the interval $]x_0 - r, x_0 + r[$ (resp. $[x_0 - r, x_0 + r]$); the sphere with centre $x_0$ and radius $r$ is the set consisting of the two end-points $x_0 - r, x_0 + r$ of these intervals.

From what has been said, the balls (open or closed) with centre $x_0$ (or just those with radii $1/p$, where $p$ runs through the set of integers $> 0$) form a fundamental system of neighbourhoods of the point $x_0$.

4. Stereophic projection

Consider the point $e_n = (0, \ldots, 0, 1)$ of $S_{n-1}$, and the hyperplane $H$ with equation $x_n = 0$, orthogonal to the vector $e_n$. To every point $x = (x_i)$ of $S_{n-1}$, other than $e_n$, let us make correspond the point $y$ where the line through $e_n$ and $x$ meets the hyperplane $H$ (Fig. 3). It is easily verified that we have

and

!Diagram showing stereographic projection

Figure 3.

If we denote by $A$ the complement of $\{ e_n \}$ in $S_{n-1}$, these formulas show that we have thus defined a homeomorphism of $A$ onto the hyperplane $H$. This homeomorphism is called the stereographic projection of $A$ onto $H$, or (by abuse of language) the stereographic projection of $S_{n-1}$ onto $H$; $e_n$ is the vertex of the projection, $H$ the hyperplane of projection. More generally, if $H'$ is any hyperplane passing through $o$ (a diametral hyperplane of $B_n$) and if $a$ is one of the points of intersection of $S_{n-1}$ and the line orthogonal to $H'$ passing through $o$, we can define in the same way the stereographic projection with vertex $a$ onto the hyperplane of projection $H'$; in any case this projection can be brought back to the preceding one by an orthogonal transformation which transforms $H'$ into $H$ and $a$ into $e_n$.

Exercises

See the exercises for § 2.