9.1 Points and Coordinates

In school the plane comes first as a picture, a sheet with two axes drawn on it, and pairs of numbers are labels attached to its points afterwards. Analysis turns this around. A point is its list of coordinates, and every statement about points, lines, lengths or angles is a statement about real numbers. Chapter 3 has already done this once: there a complex number was by definition a pair of reals, and the plane of complex numbers was the set $\mathbb{R}^2$ . This section does the same for lists of any length. It defines $\mathbb{R}^n$ , describes subsets of it by conditions on coordinates, and introduces the simplest such subsets, the boxes.

Standing setting. Throughout the chapter, $\mathbb{R}$ is the complete ordered field of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Chapter 5 (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2). Every rule proved there for ordered fields holds in $\mathbb{R}$ , and no proof below looks inside a real number. The letters $n$ and $m$ stand for natural numbers with $n, m \ge 1$ , and $k$ and $j$ for indices in $\{1, \dots, n\}$ , unless we say otherwise.

9.1.1. A point is its list of coordinates

Coordinates entered geometry so that a curve could become an equation and a construction a computation; a proof about the plane then needs the laws of $\mathbb{R}$ and nothing else. We take that step for every number of coordinates.

For example, the pair $(3, -1)$ is a point of the plane, with first coordinate $3$ and second coordinate $-1$ . The triple $(2, 0, 5)$ is a point of space. A list of four reals, such as $(1, -2, 0, 3)$ , is just as good a point, though we can no longer draw it. In each case the point is nothing but the list.

A point is its list of coordinates.

Definition 9.1.1 (The Space

\mathbb{R}^n

)

Let $n \in \mathbb{N}$ with $n \ge 2$ . The set $\mathbb{R}^n$ is the set of all $n$ -tuples $x = (x_1, \dots, x_n) \qquad \text{with } x_k \in \mathbb{R} \text{ for every } k \in \{1, \dots, n\} ;$ it is the set $X^n$ of Sets and Functions, Definition 1.4.2 with $X = \mathbb{R}$ . The real number $x_k$ is the $k$ -th coordinate of $x$ . For $n = 1$ we define $\mathbb{R}^1 \mathrel{:=} \mathbb{R}$ , and read a real number $x$ as a list of length one whose first coordinate is $x_1 \mathrel{:=} x$ . The elements of $\mathbb{R}^n$ are called points or vectors; the two words mean the same thing.

In words: a point of $\mathbb{R}^n$ is an ordered list of $n$ real numbers, and nothing more. Two points are equal if and only if they agree in every place: $x = y \iff x_k = y_k \text{ for every } k \in \{1, \dots, n\} .$ Sets and Functions, Section 1.4, defined this equality of $n$ -tuples, just after ordered pairs (Sets and Functions, Definition 1.4.1). For $n = 1$ it says only that $x_1 = y_1$ . So one equation between points of $\mathbb{R}^n$ is $n$ equations between real numbers. For $n = 2$ the definition gives $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ . This is also the underlying set of the complex numbers: $\mathbb{C} \mathrel{:=} \mathbb{R}^2$ in Definition 3.1.1. A complex number and a point of the plane are the same object; Section 9.4 compares what the two readings do with it.

The case $n = 1$ is a convention, not a theorem. Sets and Functions defines tuples only of length at least $2$ , and a real number is not a tuple. We declare $\mathbb{R}^1$ to be $\mathbb{R}$ so that every statement “for all $n \ge 1$ ” in this chapter includes the line, with $x_1$ meaning $x$ itself.

Examples. The point $(1, -2, 0, 3)$ lies in $\mathbb{R}^4$ ; its third coordinate is $0$ . The point $(0, 0, 0)$ lies in $\mathbb{R}^3$ . It is not the real number $0$ : it is a triple, and Sets and Functions agreed that a tuple is never a number (Sets and Functions, Definition 1.4.1). And the smallest case: $\mathbb{R}^1$ is the line $\mathbb{R}$ , where a point has one coordinate, and the equality rule above reads $x = y$ .

A list is not a set. Replace the list by a set. The set $\{1, 2\}$ holds the same two numbers as the point $(1, 2)$ , but it forgets their order, $\{1, 2\} = \{2, 1\}$ , and it forgets repetitions, $\{1, 1\} = \{1\}$ . The points $(1, 2)$ , $(2, 1)$ and $(1, 1)$ are three different points of $\mathbb{R}^2$ . What a coordinate system needs is exactly what sets lose: which number sits in which place.

Why this definition. A point defined as “a position in the plane” cannot be reasoned about: there is no rule for positions, only pictures of them. A point defined as a list of reals carries all the rules of the field $\mathbb{R}$ with it, one coordinate at a time. The price is that every geometric word now needs a definition in coordinates. The rest of this chapter supplies these definitions, each by a formula in the coordinates:

the sum of two points (Section 9.2);
the dot product (Section 9.3);
length and distance (Section 9.4);
the cosine of an angle (Section 9.6).

Remark. The familiar model of the plane, with an origin, a unit of length and two perpendicular axes, is how we draw a point of $\mathbb{R}^2$ and choose what to prove. It is motivation only. In this chapter a picture is never a reason: every claim is proved from the definitions and the laws of $\mathbb{R}$ .

Warning. Order and nesting matter. The points $(1, 2)$ and $(2, 1)$ of $\mathbb{R}^2$ are different, and the point $(1, 2)$ is not the set $\{1, 2\}$ . Nor is $\mathbb{R}^2 \times \mathbb{R}$ literally $\mathbb{R}^3$ . Its elements are pairs $((x_1, x_2), x_3)$ whose first entry is itself a pair. The elements of $\mathbb{R}^3$ are triples, and a triple is not a pair (Sets and Functions, Section 1.4).

Second readingGluing two lists into one

Although $\mathbb{R}^m \times \mathbb{R}^n$ and $\mathbb{R}^{m+n}$ are different sets, nothing is lost in passing between them. Let $g \colon \mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^{m+n}$ glue two lists, $g\bigl((x_1, \dots, x_m), (y_1, \dots, y_n)\bigr) \mathrel{:=} (x_1, \dots, x_m, y_1, \dots, y_n) ,$ with a real number read as a list of length one when $m = 1$ or $n = 1$ . Then $g$ is a bijection (Sets and Functions, Definition 2.2.3). It is injective. Suppose two pairs have the same image. The images agree in all $m + n$ places, so the first lists agree in every place and so do the second lists. The two pairs are therefore equal by Sets and Functions, Definition 1.4.1. It is surjective: a list $(z_1, \dots, z_{m+n})$ is the image of the pair $\bigl((z_1, \dots, z_m), (z_{m+1}, \dots, z_{m+n})\bigr)$ . We use this identification only in words, as in “a point of $\mathbb{R}^3$ is a point of the plane together with a height”.

Quick check. Which of $(0, 0, 0)$ , $(1, 2)$ and $((1, 2), 3)$ lie in $\mathbb{R}^3$ ? How many points has $\{0, 1\}^3$ ?

Solution. Only $(0, 0, 0)$ . The point $(1, 2)$ is a pair, a list of length two. The object $((1, 2), 3)$ is also a pair, whose first entry is the pair $(1, 2)$ ; it lies in $\mathbb{R}^2 \times \mathbb{R}$ , and a pair is not a triple.

The set $\{0, 1\}^3$ has eight points: each of the three coordinates is $0$ or $1$ , and the choices are independent, so there are $2 \cdot 2 \cdot 2 = 8$ lists. Written out, they are $(0, 0, 0)$ , $(0, 0, 1)$ , $(0, 1, 0)$ , $(0, 1, 1)$ , $(1, 0, 0)$ , $(1, 0, 1)$ , $(1, 1, 0)$ and $(1, 1, 1)$ .

9.1.2. Sets described by conditions

With points as lists, a subset of $\mathbb{R}^n$ is described by a condition on coordinates: the set of all $x \in \mathbb{R}^n$ whose coordinates satisfy it. Deciding whether a given point belongs is then a computation with real numbers. Here are the subsets of the plane that return most often.

Example (Subsets of the Plane by Conditions)

For each of the following subsets of $\mathbb{R}^2$ , name one point that lies in it and one that does not, and check both against the condition.

(a) The first axis $A \mathrel{:=} \{ x \in \mathbb{R}^2 : x_2 = 0 \}$ .
(b) The closed first quadrant $Q \mathrel{:=} \{ x \in \mathbb{R}^2 : x_1 \ge 0 \text{ and } x_2 \ge 0 \}$ .
(c) The diagonal $D \mathrel{:=} \{ x \in \mathbb{R}^2 : x_1 = x_2 \}$ .
(d) The graph $G \mathrel{:=} \{ (t, t^2) : t \in \mathbb{R} \}$ of the squaring function.
(e) The set $E \mathrel{:=} \{ x \in \mathbb{R}^2 : x_1^2 + x_2^2 \le 1 \}$ .

Solution.

(a) The point $(5, 0)$ is in $A$ , since its second coordinate is $0$ . The point $(0, 5)$ is not, since $5 \ne 0$ .
(b) The point $(0, 3)$ is in $Q$ : $0 \ge 0$ and $3 \ge 0$ . “Closed” means the axes are included. The point $(-1, 2)$ is not in $Q$ , since $-1 \ge 0$ fails; one failed condition is enough, because the two conditions are joined by “and”. Compare the condition $x_1 x_2 \ge 0$ , a single condition that looks similar: the point $(-1, -1)$ satisfies it, as $(-1)(-1) = 1 \ge 0$ , but $(-1, -1) \notin Q$ . So $\{ x : x_1 x_2 \ge 0 \}$ is a larger set than $Q$ : it contains the closed third quadrant as well.
(c) The point $(-2, -2)$ is in $D$ , and $(1, 2)$ is not, since $1 \ne 2$ .
(d) Here the set is described by listing its points, one for each real $t$ , rather than by a condition. It has a description by a condition too: $G = \{ x \in \mathbb{R}^2 : x_2 = x_1^2 \}$ . Indeed $(t, t^2)$ satisfies $x_2 = x_1^2$ ; and if $x_2 = x_1^2$ , then $x = (t, t^2)$ with $t \mathrel{:=} x_1$ . So $(3, 9) \in G$ , while $(9, 3) \notin G$ , because $3 \ne 81 = 9^2$ . Section 8.1 treats the graph of any real function in the same way.
(e) The point $(3/5, 4/5)$ is in $E$ : $9/25 + 16/25 = 1 \le 1$ . The point $(1, 1)$ is not: $1 + 1 = 2 > 1$ . In the picture $E$ is a disk, but the picture is not the definition; once Section 9.4 has defined length, the condition says that the length of $x$ is at most $1$ .

9.1.3. Boxes

Some conditions treat the coordinates separately: one condition on $x_1$ , one on $x_2$ , and so on, with no condition linking two coordinates. The quadrant $Q$ of Example (Subsets of the Plane by Conditions) is of this kind; the diagonal $D$ is not, since $x_1 = x_2$ ties the two coordinates together. The simplest separate conditions bound each coordinate between two numbers. On the line that gives a closed interval $[a, b]$ , and in the plane a rectangle $[a_1, b_1] \times [a_2, b_2]$ with sides parallel to the axes. Later work in analysis traps points inside such sets in $\mathbb{R}^n$ , so we give them names.

For example, take $a = (0, 1)$ and $b = (2, 3)$ . The points $x$ of $\mathbb{R}^2$ with $0 \le x_1 \le 2$ and $1 \le x_2 \le 3$ form the rectangle $[0, 2] \times [1, 3]$ , with corners $(0, 1)$ and $(2, 3)$ . The point $(1, 2)$ lies in it. The point $(3, 2)$ does not, since its first coordinate $3$ is larger than $2$ .

Definition 9.1.2 (Boxes)

Let $a, b \in \mathbb{R}^n$ with $a_k \le b_k$ for every $k$ . The closed box with corners $a$ and $b$ is $[a_1, b_1] \times \dots \times [a_n, b_n] \mathrel{:=} \{ x \in \mathbb{R}^n : a_k \le x_k \le b_k \text{ for every } k \} .$ If $a_k < b_k$ for every $k$ , the open box with corners $a$ and $b$ is $(a_1, b_1) \times \dots \times (a_n, b_n) \mathrel{:=} \{ x \in \mathbb{R}^n : a_k < x_k < b_k \text{ for every } k \} .$

The notation is the product of sets from Sets and Functions, Section 4.2, applied to intervals. A point lies in the box exactly when its $k$ -th coordinate lies in the $k$ -th interval, for every $k$ . In the open box each $(a_k, b_k)$ is an interval, not a pair. Sets and Functions, Section 1.4 lets context decide between the two readings of $(a, b)$ . Inside a product of this kind it always means the interval.

Examples. For $n = 1$ a closed box is a closed interval $[a_1, b_1]$ and an open box an open interval. For $n = 2$ the closed box $[0, 1] \times [0, 1]$ is the unit square. In $\mathbb{R}^3$ , the closed box $[-1, 1] \times [0, 1] \times [2, 2]$ is flat: its third interval is $[2, 2] = \{2\}$ , so every point of it has third coordinate $2$ . The definition allows this, since it asks only $a_3 \le b_3$ . At the extreme, $a = b$ gives a box with exactly one point, $a$ itself.

Every box is non-empty, and we can name a point in it. The corner $a$ lies in the closed box, since $a_k \le a_k \le b_k$ for every $k$ ; so does $b$ . The open box contains the center $c$ with $c_k \mathrel{:=} (a_k + b_k)/2$ , because $a_k < b_k$ gives $a_k < (a_k + b_k)/2 < b_k$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (b)). This is why the definition demands $a_k \le b_k$ , and $a_k < b_k$ for the open box. Without it, $[1, 0] \times [0, 1] = \{ x \in \mathbb{R}^2 : 1 \le x_1 \le 0 \text{ and } 0 \le x_2 \le 1 \}$ would be “a box”, an empty one, and its “corners” would mean nothing. Exercise C1 shows that with the condition in place, a closed box determines its corners.

A set that is not a box. Let one bound depend on another coordinate: $T \mathrel{:=} \{ x \in \mathbb{R}^2 : 0 \le x_2 \le x_1 \le 1 \} ,$ a triangle in the picture. Each coordinate still lies in $[0, 1]$ , but the bound $x_2 \le x_1$ links the two. That one change makes $T$ no box. Suppose $T = [p_1, q_1] \times [p_2, q_2]$ . The points $(0, 0)$ and $(1, 1)$ lie in $T$ , so $p_1 \le 0$ , $p_2 \le 0$ , $0 \le q_1$ and $1 \le q_2$ . Then $(0, 1)$ satisfies $p_1 \le 0 \le q_1$ and $p_2 \le 1 \le q_2$ , so it lies in the box. But $(0, 1) \notin T$ , since $1 \le 0$ fails. An open box fails the same way: if $T = (p_1, q_1) \times (p_2, q_2)$ , the points $(0, 0)$ and $(1, 1)$ give $p_1, p_2 < 0$ and $q_1, q_2 > 1$ , so again $(0, 1)$ lies in the box but not in $T$ . To show a set is not a box, read bounds off points of the set, then build a point the box must contain.

9.1.4. Exercises

A. Check your understanding

Exercise (A1)

(a) Find all $s, t \in \mathbb{R}$ with $(s + t, \; s - t) = (4, 2)$ .
(b) Is $(1, 1/2)$ in $[0, 1] \times [1/2, 1]$ ? Is $(1/2, 0)$ ?
(c) Describe the set $\{ x \in \mathbb{R}^2 : x_1 = x_2 \text{ and } x_1 + x_2 = 1 \}$ .
(d) Is $(0, 2)$ in the open box $(0, 1) \times (1, 3)$ ?
(e) Write the closed box with corners $(-1, 0, 2)$ and $(1, 1, 2)$ , and count its points whose coordinates are all integers.

Solution.

(a) By the equality of pairs, $s + t = 4$ and $s - t = 2$ . Adding gives $2s = 6$ , so $s = 3$ ; then $t = 1$ . Conversely $(3 + 1, \; 3 - 1) = (4, 2)$ . So $s = 3$ , $t = 1$ is the only solution.
(b) Yes: $0 \le 1 \le 1$ and $1/2 \le 1/2 \le 1$ . No: the second coordinate $0$ fails $1/2 \le 0$ .
(c) A point in the set has $x_1 = x_2$ , so $2x_1 = 1$ and $x_1 = x_2 = 1/2$ . Conversely $(1/2, 1/2)$ satisfies both conditions. The set is $\{ (1/2, 1/2) \}$ , a single point.
(d) No. The first coordinate would need $0 < 0 < 1$ , and $0 < 0$ is false. The second coordinate passes ( $1 < 2 < 3$ ), but every coordinate must pass.
(e) The corners satisfy $-1 \le 1$ , $0 \le 1$ and $2 \le 2$ , so the box is $[-1, 1] \times [0, 1] \times [2, 2]$ . A point of it with integer coordinates has $x_1 \in \{-1, 0, 1\}$ , $x_2 \in \{0, 1\}$ and $x_3 = 2$ , and every such choice gives one. So there are $3 \cdot 2 \cdot 1 = 6$ of them.

B. Practice

Exercise (B1: When one box contains another)

Let $P = [a_1, b_1] \times \dots \times [a_n, b_n]$ and $Q = [c_1, d_1] \times \dots \times [c_n, d_n]$ be closed boxes in $\mathbb{R}^n$ . Prove that $P \subseteq Q$ if and only if $c_k \le a_k$ and $b_k \le d_k$ for every $k$ .

Solution. $(\Leftarrow)$ Let $x \in P$ and fix $k$ . Then $c_k \le a_k \le x_k \le b_k \le d_k$ , so $c_k \le x_k \le d_k$ by transitivity. As $k$ was arbitrary, $x \in Q$ . Hence $P \subseteq Q$ .

$(\Rightarrow)$ Suppose $P \subseteq Q$ . The corner $a$ lies in $P$ , because $a_k \le a_k \le b_k$ for every $k$ , so $a \in Q$ : reading coordinate $k$ , $c_k \le a_k$ . Likewise $b \in P$ , so $b \in Q$ and $b_k \le d_k$ . This holds for every $k$ .

The hypothesis $a_k \le b_k$ of Definition 9.1.2 was used in $(\Rightarrow)$ , to put the corners in $P$ . Without it the claim fails: the “box” $\{ x \in \mathbb{R} : 1 \le x \le 0 \}$ is empty, so it lies in every $Q$ , including $Q = [5, 6]$ , although $5 \le 1$ is false.

Exercise (B2: Intersections and unions of boxes)

Let $P = [a_1, b_1] \times \dots \times [a_n, b_n]$ and $Q = [c_1, d_1] \times \dots \times [c_n, d_n]$ be closed boxes in $\mathbb{R}^n$ .

(a) Prove that $P \cap Q = \{ x \in \mathbb{R}^n : \max\{a_k, c_k\} \le x_k \le \min\{b_k, d_k\} \text{ for every } k \}$ . Deduce that $P \cap Q$ is a closed box if $\max\{a_k, c_k\} \le \min\{b_k, d_k\}$ for every $k$ , and is empty otherwise.
(b) Prove that $([0, 1] \times [0, 1]) \cup ([2, 3] \times [0, 1])$ is not a closed box.

Solution.

(a) Fix $k$ and a real $t$ . Since $\le$ is a total order on $\mathbb{R}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (i)), $\max\{a_k, c_k\}$ is one of $a_k$ , $c_k$ and is $\ge$ both, so $a_k \le t$ and $c_k \le t$ hold together exactly when $\max\{a_k, c_k\} \le t$ . In the same way $t \le b_k$ and $t \le d_k$ hold together exactly when $t \le \min\{b_k, d_k\}$ . Now $x \in P \cap Q$ means $a_k \le x_k \le b_k$ and $c_k \le x_k \le d_k$ for every $k$ , which by the two equivalences, with $t = x_k$ , is the displayed condition.
If $\max\{a_k, c_k\} \le \min\{b_k, d_k\}$ for every $k$ , the displayed set is the closed box with corners $u$ and $v$ , where $u_k \mathrel{:=} \max\{a_k, c_k\}$ and $v_k \mathrel{:=} \min\{b_k, d_k\}$ ; the hypothesis is the condition $u_k \le v_k$ that Definition 9.1.2 requires. If $\max\{a_k, c_k\} > \min\{b_k, d_k\}$ for some $k$ , no real $x_k$ satisfies $\max\{a_k, c_k\} \le x_k \le \min\{b_k, d_k\}$ , by transitivity, so $P \cap Q = \varnothing$ .
(b) Call the union $U$ and suppose $U = [p_1, q_1] \times [p_2, q_2]$ . The points $(0, 0)$ and $(3, 0)$ lie in $U$ , so $p_1 \le 0$ and $3 \le q_1$ , and $p_2 \le 0 \le q_2$ . Then $(3/2, 0)$ lies in the box, since $p_1 \le 0 \le 3/2 \le 3 \le q_1$ and $p_2 \le 0 \le q_2$ . But $(3/2, 0)$ lies in neither square: $3/2 > 1$ and $3/2 < 2$ . So $(3/2, 0) \notin U$ , a contradiction. Hence $U$ is not a closed box. The argument is the one used above to show that the triangle $T$ is not a box.

C. Going deeper

Exercise (C1: A box determines its corners)

Let $P = [a_1, b_1] \times \dots \times [a_n, b_n]$ and $Q = [c_1, d_1] \times \dots \times [c_n, d_n]$ be closed boxes in $\mathbb{R}^n$ . Prove that if $P = Q$ , then $a_k = c_k$ and $b_k = d_k$ for every $k$ .

Solution (Hint 1). Use Exercise B1, twice.

Solution (Hint 2). From $P \subseteq Q$ and $Q \subseteq P$ , write down four inequalities for each $k$ .

Solution. Suppose $P = Q$ and fix $k$ . Since $P \subseteq Q$ , Exercise B1 gives $c_k \le a_k$ and $b_k \le d_k$ . Since $Q \subseteq P$ , the same exercise with the roles of the boxes exchanged gives $a_k \le c_k$ and $d_k \le b_k$ . The order of $\mathbb{R}$ is antisymmetric, being a total order (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (i)), so $a_k = c_k$ and $b_k = d_k$ . As $k$ was arbitrary, this proves the claim.

So a closed box, as a set, determines the numbers $a_k$ and $b_k$ . This is what makes “the box $[a_1, b_1] \times \dots \times [a_n, b_n]$ ” an unambiguous name, and “the corners of a box” meaningful. Exercise B1 needed $a_k \le b_k$ for the box on the left. This result applies B1 both ways, so it needs both $a_k \le b_k$ and $c_k \le d_k$ : the empty sets $\{ x \in \mathbb{R} : 1 \le x \le 0 \}$ and $\{ x \in \mathbb{R} : 5 \le x \le 2 \}$ are equal, with different “corners”.

What a complete answer needs. (10 points) - 2: fixes $k$ and states the goal $a_k = c_k$ , $b_k = d_k$ . - 3: obtains $c_k \le a_k$ and $b_k \le d_k$ from $P \subseteq Q$ (by B1, or directly from the corners $a, b \in P$ ). - 3: obtains the reverse inequalities from $Q \subseteq P$ . - 2: concludes by antisymmetry, for every $k$ . Slips that lose the step: using one inclusion only; reading the corners off the notation without proof (that is the claim).