Real Topology

Introduction

This article develops the topology of the real algebra $\mathbb{R}$: the order topology, the metric topology and their coincidence, the completeness and local compactness of the line, its connectedness and interval structure, its contractibility, its one-point and two-point compactifications, and its two ends. It is the base case of the topology series of the corpus.

The subject is elementary, and its place in this blog is that of the base case. The topology is developed here for the one-dimensional algebra $\mathbb{R}$, in the companion articles Complex Topology for the two-dimensional definite algebra $\mathbb{C}$ and Biquaternion Topology for the eight-dimensional algebra $\mathbb{B}$, and in the companion articles on the topology of the other members of the family. The whole structure is one connected line with two ends: every homotopy group vanishes, the unit set is the finite sign group, and the compactification that appears is the circle. The properties that the higher members acquire — a compact unit sphere, a null cone with a link, a nontrivial fundamental group — are all absent here, and their absence is the mathematical content of the one-dimensionality.

Conventions. The basis is $e_0 = 1$; a real number is $a = a e_0$ with its single real coordinate $a$; the norm is $N(a) = a^2$ and the modulus and metric are $|a| = \sqrt{N(a)} = \sqrt{a^2}$; the group of units is $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$. No physics is invoked.

The Algebra as a Topological Space

The map

$$ a \longmapsto a $$

is a linear isometry of $(\mathbb{R}, |\cdot|)$ onto the Euclidean line $\mathbb{R}^1$. The topology of $\mathbb{R}$ is therefore the Euclidean topology of the line; multiplication is bilinear, hence continuous, so $\mathbb{R}$ is a topological algebra over $\mathbb{R}$; and inversion is continuous on the units, so $\mathbb{R}^\times$ is a topological group.

The two distinguished structures of the line are its order and its metric, and each generates a topology; the first result of the article is that the two are the same.

The Order Topology and the Metric Topology

Definition. The order topology on $\mathbb{R}$ is generated by the open intervals

$$ (a, b) = \{c : a < c < b\}, \qquad (-\infty, b) = \{c : c < b\}, \qquad (a, \infty) = \{c : c > a\}, $$

together with $\mathbb{R}$ and $\varnothing$. The metric topology is generated by the open balls

$$ B(a_0, \varepsilon) = \{c : |c - a_0| < \varepsilon\}, \qquad \varepsilon > 0, $$

of the metric $d(a, b) = |a - b|$.

Theorem. The order topology and the metric topology on $\mathbb{R}$ coincide.

Proof. An open ball is the interval $(a_0-\varepsilon, a_0+\varepsilon)$, which is order-open, so every metric-open set is order-open. Conversely an open interval $(u,v)$ is the union of the balls $B\bigl(\tfrac{u+v}{2}, r\bigr)$ over $r < \tfrac{v-u}{2}$ centred in the interval, and a ray is a union of balls, so every order-open set is metric-open. The two topologies therefore have the same open sets.

The coincidence is the statement that the order and the metric on the line determine the same notion of nearness; the metric is itself order-defined, $d(a,b) = |a-b|$ with $|a|$ the absolute value, so the order generates the metric, and conversely. In the complex case there is no order, and the topology comes from the Euclidean metric alone; in the real case the two descriptions are available and agree.

Completeness and Local Compactness

Theorem (completeness). Every Cauchy sequence of real numbers converges to a real limit; equivalently, $\mathbb{R}$ is a complete metric space.

Proof. This is the completeness axiom of Real Algebra in its metric form. A Cauchy sequence is bounded, so its limit inferior

$$ L = \sup_{n} \inf_{k \ge n} a_k $$

exists in $\mathbb{R}$ by completeness; because the sequence is Cauchy, its oscillation over the tail falls below every $\varepsilon > 0$ eventually, so $|a_n - L| < \varepsilon$ for all large $n$, and $a_n \to L$. The completeness of $\mathbb{R}$ is what the construction of $\mathbb{R}$ from $\mathbb{Q}$ by Dedekind cuts or Cauchy sequences supplies, and it is the property that fails for $\mathbb{Q}$.

Theorem (local compactness). $\mathbb{R}$ is locally compact: every point has a compact neighbourhood.

Proof. The closed bounded interval $[a_0-1, a_0+1]$ is a neighbourhood of $a_0$, and by the Heine–Borel theorem it is compact: every open cover has a finite subcover. Equivalently, every sequence in a closed bounded interval has a convergent subsequence, by Bolzano–Weierstrass.

Local compactness is the property that makes the one-point compactification available in the next section. The line is $\sigma$-compact, being the union of the intervals $[-n, n]$, and it is second countable, since the rational-ended intervals form a base.

Connectedness and the Interval Structure

Theorem. $\mathbb{R}$ is connected, and its connected subsets are exactly the intervals.

Proof. Suppose $\mathbb{R} = A \sqcup B$ with $A$, $B$ non-empty open. Pick $u \in A$ and $v \in B$ with $u < v$, and let $c = \sup\{a \in A : a < v\}$; by completeness $c$ exists, and the openness of $A$ and $B$ gives a contradiction at $c$. Hence no such separation exists, so $\mathbb{R}$ is connected. A subset is connected exactly when it contains every point between any two of its points, which is the interval property.

Thus the connected subsets of $\mathbb{R}$ are the intervals — the sets $[a,b]$, $(a,b)$, $[a,b)$, $(a,b]$ and the rays — and every one of them is connected, giving the interval structure of the line. Every interval is connected and locally connected, and the line is path-connected, indeed every two points are joined by the interval between them.

The punctured line. Removing a point disconnects the line into exactly two components:

$$ \mathbb{R} \setminus \{a_0\} = (-\infty, a_0) \sqcup (a_0, \infty), \qquad \pi_0(\mathbb{R}\setminus\{a_0\}) \cong \{\pm1\}. $$

In particular the group of units $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$ has two components, the positive and negative rays, as recorded in Real Norm and Invertibility and Real Exponential and Lie Group Structure. In the complex case the punctured plane is connected, because a plane minus a point can be walked around; the disconnectedness of the punctured line is the one-dimensional exception, and it is why the real unit group is disconnected.

The Sign Group and the Group of Units

Definition. The unit set of $\mathbb{R}$ is the norm-one level set

$$ \{a : |a| = 1\} = \{a : N(a) = 1\} = \{\pm1\} = O(1). $$

The two descriptions coincide because the norm is positive definite and equals the square of the modulus, $N(a) = |a|^2$. The unit set is therefore the two-point set, a finite discrete topological group.

Proposition. $\{\pm1\}$ is a compact, disconnected, abelian topological group of dimension $0$, and it is the maximal compact subgroup of $\mathbb{R}^\times$.

Proof. It is the kernel of the norm on $\mathbb{R}^\times$, hence a subgroup; it is finite, hence compact and discrete, of dimension $0$; it is abelian and has the two components $\{+1\}$ and $\{-1\}$. A compact subgroup of the one-dimensional group $\mathbb{R}^\times$ lies in a compact interval around $1$, and the only such subgroups are $\{1\}$ and $\{\pm1\}$; since $\{1\}$ is not maximal, the maximal compact subgroup is $\{\pm1\}$.

The unit set is the group $O(1)$, and it is the base case of the theorem that the unit sphere of a real normed division algebra is a Lie group. The cases are the finite list $S^0 = \{\pm1\}$, $S^1$, $S^3$, and the real member is $S^0$. It is disconnected, in contrast with the connected circle $S^1$ of the complex case; the disconnectedness is the sign, and it is the topological form of the two-component unit group.

The group of units. $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$ is an open, dense, disconnected, non-compact topological group with two contractible components, the positive and negative rays. Its identity component is $\mathbb{R}_{>0}$, and its group of components is the sign group:

$$ \mathbb{R}^\times \cong \mathbb{R}_{>0} \times \{\pm1\}, \qquad \pi_0(\mathbb{R}^\times) \cong \{\pm1\}. $$

Contractibility and the Homotopy Groups

Theorem (contractibility). $\mathbb{R}$ is contractible, hence path-connected and simply connected, with $\pi_n(\mathbb{R}) = 0$ for all $n \ge 1$.

Proof. The straight-line homotopy

$$ H(t, a) = (1-t) a, \qquad t \in [0,1], $$

is continuous with $H(0,a) = a$ and $H(1,a) = 0$, so the identity is homotopic to the constant map at $0$.

Every map into $\mathbb{R}$ is null-homotopic, and the same homotopy contracts each of the two rays and each interval, since they are closed under the contraction toward any base point. The single distinguished subspace, the whole line, therefore carries no topology beyond that of the contractible line; the content on the norm of Real Algebra lies in the quadratic form, not in the topology.

The homotopy groups of the unit group. The two components of $\mathbb{R}^\times$ are contractible, so

$$ \mathbb{R}^\times \simeq \{\pm1\}, \qquad \pi_1(\mathbb{R}^\times) = 0, \qquad \pi_n(\mathbb{R}^\times) = 0 \ \ (n \geq 1). $$

The real unit group is thus homotopy equivalent to a two-point space, and every homotopy group vanishes. This is the base case of the topology of the unit groups of the family: $\mathbb{C}^\times \simeq S^1$ has $\pi_1 \cong \mathbb{Z}$ from the winding of the phase, and $\mathbb{B}^\times \simeq U(2) \simeq S^1\times S^3$ has $\pi_1 \cong \mathbb{Z}$ and $\pi_3 \cong \mathbb{Z}$. The real case has no compact connected factor and therefore no homotopy to wind.

The One-Point and Two-Point Compactifications, and the Ends

The one-point compactification. $\mathbb{R}$ is locally compact and Hausdorff but not compact, so it has an Alexandrov one-point compactification

$$ \mathbb{R} \cup \{\infty\} \cong S^1, $$

the circle, obtained by adjoining the single point $\infty$ that closes up the two unbounded directions. The construction is the quotient of the two-point compactification below that identifies the two ends, and it is the standard instance of the one-point compactification of a locally compact space.

The two-point compactification. Adjoining the two order-ends separately gives the extended real line

$$ \overline{\mathbb{R}} = [-\infty, +\infty] \cong [0,1], $$

a compact ordered space in which every subset has a supremum and an infimum. The extra points are the two ends: the one-point compactification identifies them, the two-point compactification keeps them apart. In the complex case the one-point compactification is the Riemann sphere $S^2$; here it is the circle, and the circle appears because the two ends of the line, once joined, form a loop.

The ends. The line has exactly two ends, the directions $a \to +\infty$ and $a \to -\infty$; both are ends of the contractible line, and the end compactification is $\overline{\mathbb{R}}$, which adds one point for each. The group of units $\mathbb{R}^\times$ has the same two ends, since removing the origin does not merge or split the unbounded directions. The number of ends is the topological invariant that distinguishes the line from the plane (which has one end) and from the biquaternion algebra (also one end, since dimension eight is greater than one); the two ends of $\mathbb{R}$ are the topological form of the two signs.

Comparison with the Complex and Biquaternion Topology

Structure	$\mathbb{R}$	$\mathbb{C}$	$\mathbb{B}$
Algebra as a space	$\mathbb{R}$, contractible, two ends	$\mathbb{R}^2$, contractible, one end	$\mathbb{R}^8$, contractible, one end
Unit sphere	$\{\pm1\} = S^0$, not connected	$S^1$, a connected group	$S^7$, not a group
Norm-one level set	$\{\pm1\} = O(1)$, finite	$S^1 = U(1)$, compact	$SL(2,\mathbb{C})$, non-compact
Null cone $\{N = 0\}$	$\{0\}$, a point	$\{0\}$, a point	real dimension $6$
Link of the null cone	none	none	$S^1$-bundle over $S^2\times S^2$
Group of units	two contractible components	$\mathbb{C}^\times \simeq S^1$, connected	$\mathbb{B}^\times \simeq U(2)$
Homotopy of the units	all $\pi_n = 0$	$\pi_1 \cong \mathbb{Z}$	$\pi_1 \cong \mathbb{Z}$, $\pi_3 \cong \mathbb{Z}$
One-point compactification	$S^1$	$S^2$	(dimension $8$, not a group)
Ends	two	one	one

The three algebras share the contractibility of the ambient space. They differ in the unit sphere, which is the finite sign group for $\mathbb{R}$, a connected circle for $\mathbb{C}$, and a non-group seven-sphere for $\mathbb{B}$; in the unit group, which has two contractible components for $\mathbb{R}$ and a compact connected core for the other two; and in the ends, which are two for the line and one for the higher-dimensional spaces. Each difference traces to a single cause: the dimension of the real algebra, which makes the unit set finite, a circle or a sphere, and the definiteness of the norm, which makes the zero set of $N$ a single point rather than a cone.

Summary

The real algebra $\mathbb{R}$ is contractible, hence path-connected and simply connected, with $\pi_n(\mathbb{R}) = 0$ for all $n \ge 1$; its order topology and its metric topology coincide, it is complete and locally compact, and its connected subsets are exactly the intervals. The unit set $\{\pm1\} = O(1)$ is a compact, disconnected, abelian topological group of dimension $0$, the base case $S^0$ of the unit spheres of the normed division algebras; it is the maximal compact subgroup of the unit group.

The group of units is $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$, an open, dense, disconnected, non-compact topological group with the two contractible components $\mathbb{R}_{>0}$ and $\mathbb{R}_{<0}$, homeomorphic to $\mathbb{R}_{>0}\times\{\pm1\}$. It is homotopy equivalent to a two-point space, so all of its homotopy groups vanish, in contrast with the winding of the complex and biquaternion unit groups. The unit group is the disjoint union of its two contractible rays; there is no null cone and no link, because the norm is definite and the zero-divisor class is empty.

The line has two ends, and the two compactifications are the two-point compactification $\overline{\mathbb{R}} = [-\infty,+\infty] \cong [0,1]$, which keeps the ends apart, and the one-point compactification $\mathbb{R}\cup\{\infty\} \cong S^1$, which joins them into the circle. Compared with the complex and biquaternion algebras, the real case loses the connected unit circle, the seven-sphere, the null cone and its link, and the non-vanishing homotopy groups, and keeps a contractible line whose compactification is the simplest closed curve.

Summary of Notation

symbol	meaning
$\mathbb{R}$	the real algebra as a topological space; contractible, two ends
$a$	a real number with real coordinate $a$
$N(a) = a^2$, $\lvert a\rvert = \sqrt{N(a)}$	the norm and the modulus
$d(a,b) = \lvert a-b\rvert$	the metric generating the topology
$B(a_0,\varepsilon) = (a_0-\varepsilon, a_0+\varepsilon)$	the open ball, an open interval
$\{\pm1\} = O(1) = S^0$	the unit set, a discrete compact group
$\mathbb{R}^\times = \mathbb{R}_{>0}\sqcup\mathbb{R}_{<0}$	the group of units, two contractible components
$\pi_0(\mathbb{R}^\times) \cong \{\pm1\}$	the group of components of the units
$\mathbb{R}\cup\{\infty\} \cong S^1$	the one-point compactification
$\overline{\mathbb{R}} = [-\infty,+\infty] \cong [0,1]$	the two-point compactification, the extended line
$\pi_n(\mathbb{R}) = 0$, $\pi_n(\mathbb{R}^\times) = 0$	vanishing homotopy groups