The Extended Real Number System

What is the supremum of the set of all natural numbers? Think of it as a subset of R\mathbb{R}. By the Completeness Axiom, every nonempty subset of R\mathbb{R} bounded above has a least upper bound in R\mathbb{R}. But N\mathbb{N} is not bounded above. So no real number serves as its supremum. The definition of supremum simply does not apply. We are forced to say "sup⁡N\sup \mathbb{N} does not exist in R\mathbb{R}." This is a mild but persistent inconvenience. Theorems about suprema, limits, and Bounds of a Set constantly need an exception clause. This exception handles the unbounded case. It clutters otherwise clean statements. The extended real number system repairs this. It adjoins two new symbols, +∞+\infty and −∞-\infty, to R\mathbb{R}. It also adds an order and a partial arithmetic. These make the symbols behave the way limits and suprema already suggest they should. The result is denoted R‾\overline{\mathbb{R}} or [−∞,+∞][-\infty, +\infty]. It is not a field. There is no consistent way to make ∞−∞\infty - \infty or 0⋅∞0 \cdot \infty meaningful. But it is a complete, totally ordered set. In it, every subset has a supremum and an infimum, with no exceptions. This single fact is the entire payoff. Statements like "every monotone sequence converges in R‾\overline{\mathbb{R}}" become unconditionally true. So does "sup⁡A\sup A exists for every A⊆RA \subseteq \mathbb{R}." The idea matters for an important reason. It is the quiet infrastructure behind measure theory. It also underlies the study of lim sup⁡\limsup and lim inf⁡\liminf, improper integrals, and extended real-valued functions. These functions are used throughout analysis. Upper and lower Lebesgue integrals are literally elements of R‾\overline{\mathbb{R}}. Once R‾\overline{\mathbb{R}} is built rigorously, one benefit follows. One never again has to write "if the set is bounded above" as a caveat. This caveat previously qualified theorems about suprema. This article gives the precise construction of R‾\overline{\mathbb{R}}. It states and proves the theorem that makes it worthwhile. That theorem is completeness of order without exception. The article also carefully documents which arithmetic operations extend. It notes which operations are deliberately left undefined. Treating ∞\infty as an ordinary number is the most common source of error.

Key takeaways
  • R‾:=R∪{+∞,−∞}\overline{\mathbb{R}} := \mathbb{R} \cup \{+\infty, -\infty\}, with −∞<x<+∞-\infty < x < +\infty for every x∈Rx \in \mathbb{R}.
  • Every subset of R‾\overline{\mathbb{R}} has a supremum and an infimum in R‾\overline{\mathbb{R}} — no exceptions, unlike in R\mathbb{R}.
  • Arithmetic extends partially: x+∞=∞x + \infty = \infty, x⋅∞=∞x \cdot \infty = \infty for x>0x>0, but ∞−∞\infty - \infty and 0⋅∞0 \cdot \infty are left undefined.
  • R‾\overline{\mathbb{R}} is not a field and not a metric space under the usual metric, only an ordered set (though it is metrizable via a bounded homeomorphism to [−1,1][-1,1]).
  • The convention sup⁡∅=−∞\sup \varnothing = -\infty and inf⁡∅=+∞\inf \varnothing = +\infty is forced by consistency, not arbitrary.
  • Key idea: adjoining two ideal endpoints turns "bounded-above subsets have suprema" into "all subsets have suprema," at the cost of losing full arithmetic.

Where this lives today

In measure theory, the Lebesgue outer measure and the integral of a nonnegative measurable function both take values in [0,+∞][0,+\infty]; the monotone convergence theorem and Fatou's lemma are stated cleanly only because +∞+\infty is an admissible value, avoiding endless case splits on finiteness. In probability theory, extended real-valued random variables (e.g. hitting times of a Markov chain, which may be +∞+\infty if the target is never reached) are modeled directly as measurable functions into R‾\overline{\mathbb{R}}, so that "the event never happens" is a single well-defined value rather than an undefined expression. In real analysis proper, lim sup⁡n→∞an\limsup_{n\to\infty} a_n and lim inf⁡n→∞an\liminf_{n\to\infty} a_n for an arbitrary real sequence (an)(a_n) are defined as

lim sup⁡n→∞an:=inf⁡n≥1sup⁡k≥nak∈R‾,\limsup_{n\to\infty} a_n := \inf_{n\ge1} \sup_{k \ge n} a_k \in \overline{\mathbb{R}},

and this definition is unconditional — it needs no boundedness hypothesis on (an)(a_n) — precisely because both the inner sup⁡\sup and outer inf⁡\inf are guaranteed to exist in R‾\overline{\mathbb{R}} by the completeness theorem proved above. In optimization and convex analysis, extended-real-valued functions f:X→R‾f: X \to \overline{\mathbb{R}} (allowing f(x)=+∞f(x)=+\infty to encode constraints, as in the indicator function of a convex set) are the standard language of convex analysis (Rockafellar's framework), where infima of such functions are always well-defined thanks to the same completeness property. In computer science and numerical computing, IEEE 754 floating-point arithmetic implements a finite-precision shadow of R‾\overline{\mathbb{R}}, with explicit `+Infinity`, `-Infinity`, and `NaN` (Not-a-Number) values; `NaN` is the machine's realization of exactly the "undefined" operations ∞−∞\infty - \infty and 0×∞0\times\infty identified above, and its propagation rules mirror the mathematical restriction that these operations carry no consistent value.

Ask the tutor: How does the order topology on R‾\overline{\mathbb{R}} relate to the usual topology on The Real Line, and why does that make R‾\overline{\mathbb{R}} compact?

DefinitionThe Extended Real Number System

Let R\mathbb{R} be the real numbers with its usual order <<. Introduce two distinct objects +∞+\infty and −∞-\infty, neither of which is a real number, and define

R‾:=R∪{−∞,+∞}.\overline{\mathbb{R}} := \mathbb{R} \cup \{-\infty, +\infty\}.

Extend the order << on R\mathbb{R} to a total order << on R‾\overline{\mathbb{R}} by declaring, for every x∈Rx \in \mathbb{R},

−∞<x<+∞,−∞<+∞,-\infty < x < +\infty, \qquad -\infty < +\infty,

and retaining the usual order among elements of R\mathbb{R}. Write ≤\le for "<< or equal," as usual.

Arithmetic (partial extension). For x∈Rx \in \mathbb{R} define:

  • x+(+∞)=(+∞)+x=+∞x + (+\infty) = (+\infty) + x = +\infty, and x+(−∞)=(−∞)+x=−∞x + (-\infty) = (-\infty) + x = -\infty;
  • (+∞)+(+∞)=+∞(+\infty) + (+\infty) = +\infty, (−∞)+(−∞)=−∞(-\infty) + (-\infty) = -\infty;
  • for x>0x > 0: x⋅(+∞)=+∞x \cdot (+\infty) = +\infty, x⋅(−∞)=−∞x\cdot(-\infty) = -\infty; for x<0x<0: x⋅(+∞)=−∞x\cdot(+\infty) = -\infty, x⋅(−∞)=+∞x \cdot (-\infty) = +\infty;
  • 0⋅(+∞)=0⋅(−∞)=00 \cdot (+\infty) = 0 \cdot (-\infty) = 0 by convention in measure theory only (this convention is not used in the order/limit theory below and is flagged explicitly whenever invoked);
  • −(+∞)=−∞-(+\infty) = -\infty, −(−∞)=+∞-(-\infty) = +\infty;
  • (+∞)−(−∞)=+∞(+\infty) - (-\infty) = +\infty, (−∞)−(+∞)=−∞(-\infty) - (+\infty) = -\infty.

The expressions

(+∞)+(−∞),(−∞)+(+∞),0⋅(±∞) (outside the measure-theoretic convention)(+\infty) + (-\infty), \quad (-\infty) + (+\infty), \quad 0\cdot(\pm\infty) \text{ (outside the measure-theoretic convention)}

are left undefined. Division by ±∞\pm\infty of a real number gives 00; division by 00 remains undefined; ±∞/±∞\pm\infty / \pm\infty is undefined.

Bounds. For A⊆R‾A \subseteq \overline{\mathbb{R}}, an element u∈R‾u \in \overline{\mathbb{R}} is an upper bound of AA if a≤ua \le u for all a∈Aa \in A, and sup⁡A\sup A is the least upper bound of AA in R‾\overline{\mathbb{R}} (an element ss with ss an upper bound and s≤us \le u for every upper bound uu); lower bound and inf⁡A\inf A are defined dually. By convention,

sup⁡∅:=−∞,inf⁡∅:=+∞.\sup \varnothing := -\infty, \qquad \inf \varnothing := +\infty.

Intuition

Think of the completed real line as a closed interval that has been given two extra endpoints, one at each "end at infinity." Every real number sits strictly between −∞-\infty and +∞+\infty, and the order is total: any two elements of R‾\overline{\mathbb{R}} are comparable. The picture to keep in mind is R‾\overline{\mathbb{R}} drawn as a compactified line segment [−1,1][-1,1], with 0↦00 \mapsto 0, +∞↦1+\infty \mapsto 1, −∞↦−1-\infty \mapsto -1, via an order-preserving bijection such as x↦x1+∣x∣x \mapsto \frac{x}{1+|x|} extended by ±∞↦±1\pm\infty \mapsto \pm 1. This map is not arithmetic-preserving — it is purely a device for visualizing the order and the topology. The reason completeness becomes exception-free is simple set-theoretic bookkeeping. In R\mathbb{R}, an unbounded-above set has no supremum because there is no candidate real number large enough. In R‾\overline{\mathbb{R}}, the element +∞+\infty is by definition larger than every real number, so it is automatically an upper bound for any subset of R\mathbb{R}, and it becomes the least one exactly when no real number works. The empty set is the boundary case in the other direction: vacuously, every element of R‾\overline{\mathbb{R}} is an upper bound of ∅\varnothing (there is nothing to violate the bound), so the least upper bound is the smallest element of the whole space, namely −∞-\infty. This is why sup⁡∅=−∞\sup \varnothing = -\infty is forced, not a matter of taste.

Figure
The extended real line as a compactified interval, with a bounded-above-but-unbounded set
-∞ +∞ 0 1 2 3 4 N accumulates toward +∞ sup(ℕ) = +∞ in R̄ (does not exist in R)

The central theorem

TheoremOrder-Completeness of the Extended Reals

Let R‾=R∪{−∞,+∞}\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty,+\infty\} with the order defined above. Then: 1. R‾\overline{\mathbb{R}} is totally ordered by <<: for all x,y∈R‾x,y \in \overline{\mathbb{R}}, exactly one of x<yx<y, x=yx=y, y<xy<x holds. 2. (Unconditional completeness.) Every subset A⊆R‾A \subseteq \overline{\mathbb{R}}, including A=∅A = \varnothing and unbounded AA, has a supremum and an infimum in R‾\overline{\mathbb{R}}. 3. For A⊆RA \subseteq \mathbb{R} nonempty and bounded above in R\mathbb{R}, the supremum of AA in R‾\overline{\mathbb{R}} coincides with the supremum of AA in R\mathbb{R} furnished by the Completeness Axiom. If AA is nonempty and not bounded above in R\mathbb{R}, then sup⁡A=+∞\sup A = +\infty in R‾\overline{\mathbb{R}}.

Proof

Technique. We reduce everything to the Completeness Axiom for R\mathbb{R} by a case analysis on whether a given subset of R‾\overline{\mathbb{R}} contains ±∞\pm\infty or is bounded, and we verify the supremum property directly from its definition (least upper bound) in each case. No circularity arises because we never assume completeness of R‾\overline{\mathbb{R}} to prove itself; we only use completeness of R\mathbb{R}, which is a hypothesis available by prerequisite.

Part 1: Total order. First we check the extended relation << is well defined and total. For x,y∈Rx,y\in\mathbb{R}, trichotomy holds by the order axioms of R\mathbb{R}. For x∈Rx \in \mathbb{R} compared to +∞+\infty: by definition x<+∞x < +\infty always, x≠+∞x \neq +\infty since +∞∉R+\infty \notin \mathbb{R}, and +∞<x+\infty < x is declared false; exactly one of the three trichotomy relations holds. The symmetric argument handles xx versus −∞-\infty. Finally −∞<+∞-\infty < +\infty is declared and is the only relation among the two symbols since they are distinct and no other relation was declared. Transitivity: it suffices to check all cases where at least one of x,y,zx,y,z is ±∞\pm\infty, since transitivity among real triples is inherited from R\mathbb{R}.

  • If z=+∞z = +\infty and x<yx < y with y∈Ry\in\mathbb R or y=+∞y=+\infty... more simply, we check the two nontrivial patterns directly. Suppose x<yx<y and y<+∞y<+\infty. If y∈Ry\in\mathbb R, x∈R∪{−∞}x\in\mathbb{R}\cup\{-\infty\}, then in either case x<+∞x<+\infty by definition. Suppose −∞<x-\infty<x and x<yx<y; if x∈Rx\in\mathbb{R} or x=x= anything with y∈R∪{+∞}y \in \mathbb R\cup\{+\infty\}, then −∞<y-\infty<y by definition of the order at −∞-\infty (every element other than −∞-\infty itself is >−∞>-\infty). These exhaust the cases where an endpoint is involved, so << is transitive on R‾\overline{\mathbb{R}}.

Hence (R‾,<)(\overline{\mathbb{R}}, <) is a total (linear) order.

Part 2: Every subset has a supremum (the argument for infimum is symmetric, replacing sup⁡\sup/upper bound by inf⁡\inf/lower bound throughout, and is omitted after the remark following the proof). Let A⊆R‾A \subseteq \overline{\mathbb{R}} be an arbitrary subset. We produce sup⁡A\sup A by cases. Case (a): A=∅A = \varnothing. Every element u∈R‾u \in \overline{\mathbb{R}} satisfies the defining condition "a≤ua \le u for all a∈Aa \in A" vacuously, since there is no a∈Aa \in A to check. So every element of R‾\overline{\mathbb{R}} is an upper bound of ∅\varnothing. The least such element is the minimum of R‾\overline{\mathbb{R}}, which is −∞-\infty (since −∞≤x-\infty \le x for every x∈R‾x \in \overline{\mathbb{R}}, directly from the order definition: −∞<x-\infty < x for all x≠−∞x \ne -\infty, and −∞≤−∞-\infty \le -\infty trivially). Hence sup⁡∅=−∞\sup \varnothing = -\infty, matching the stated convention. Case (b): +∞∈A+\infty \in A. Then +∞+\infty is the only possible upper bound, because any u<+∞u < +\infty fails +∞≤u+\infty \le u. And +∞+\infty is an upper bound of AA, since +∞+\infty is the maximum of R‾\overline{\mathbb{R}} (by definition, x<+∞x < +\infty or x=+∞x = +\infty for every x∈R‾x\in\overline{\mathbb{R}}). So sup⁡A=+∞\sup A = +\infty. Case (c): A≠∅A \ne \varnothing, +∞∉A+\infty \notin A, and AA is not bounded above in R\mathbb{R} (i.e., A⊆R∪{−∞}A \subseteq \mathbb{R}\cup\{-\infty\}, A∩R≠∅A\cap\mathbb R\ne\varnothing, and A∩RA \cap \mathbb{R} has no upper bound in R\mathbb{R}). We claim sup⁡A=+∞\sup A = +\infty. Since +∞+\infty is the maximum of R‾\overline{\mathbb{R}}, it is an upper bound of AA. Suppose u∈R‾u \in \overline{\mathbb{R}}, u<+∞u < +\infty, were also an upper bound of AA; then u∈R∪{−∞}u \in \mathbb{R} \cup \{-\infty\}. If u=−∞u = -\infty, then since A∩R≠∅A \cap \mathbb{R} \neq \varnothing, pick a0∈A∩Ra_0 \in A \cap \mathbb{R}; then a0≤u=−∞a_0 \le u = -\infty forces a0=−∞a_0 = -\infty (as −∞-\infty is the minimum), contradicting a0∈Ra_0 \in \mathbb{R}. If u∈Ru \in \mathbb{R}, then uu is a real upper bound of the real subset A∩RA \cap \mathbb{R} (every element of A∩RA \cap \mathbb{R} is ≤u\le u since uu bounds all of AA), contradicting the assumption that A∩RA\cap\mathbb R has no real upper bound. In both subcases we reach a contradiction, so no u<+∞u < +\infty is an upper bound of AA. Hence +∞+\infty is the least upper bound: sup⁡A=+∞\sup A = +\infty. Case (d): A≠∅A \ne \varnothing, +∞∉A+\infty \notin A, and A∩RA \cap \mathbb{R} is bounded above in R\mathbb{R} (including the subcase A={−∞}A = \{-\infty\}). Sub-case (d1): A∩R=∅A \cap \mathbb{R} = \varnothing, so A={−∞}A = \{-\infty\}. Then −∞-\infty is an upper bound of AA (trivially, −∞≤−∞-\infty \le -\infty), and it is the least upper bound since −∞-\infty is the minimum of R‾\overline{\mathbb{R}}, hence certainly the least among upper bounds. So sup⁡A=−∞\sup A = -\infty. Sub-case (d2): B:=A∩RB := A \cap \mathbb{R} is a nonempty subset of R\mathbb{R} bounded above in R\mathbb{R}. By the Completeness Axiom, s:=sup⁡RBs := \sup_{\mathbb{R}} B exists in R\mathbb{R}: ss is a real upper bound of BB, and s≤ts \le t for every real upper bound tt of BB. We claim s=sup⁡R‾As = \sup_{\overline{\mathbb R}} A. First, ss is an upper bound of AA in R‾\overline{\mathbb{R}}: every element of AA is either in BB (so ≤s\le s by choice of ss) or equals −∞-\infty (so ≤s\le s automatically, as −∞-\infty is the minimum of R‾\overline{\mathbb{R}} and s≠−∞s \ne -\infty since s∈Rs\in\mathbb R). Since +∞∉A+\infty \notin A by hypothesis, these are the only two possibilities, so ss bounds AA above. Second, ss is least among upper bounds of AA in R‾\overline{\mathbb{R}}. Let uu be any upper bound of AA. Since B⊆AB \subseteq A, uu is in particular an upper bound of BB. If u=+∞u = +\infty, then trivially s≤us \le u. If u=−∞u = -\infty, then since BB is nonempty (say b0∈Bb_0 \in B), b0≤u=−∞b_0 \le u = -\infty forces b0=−∞b_0 = -\infty, contradicting b0∈Rb_0 \in \mathbb{R}; so this case cannot occur. If u∈Ru \in \mathbb{R}, then uu is a real upper bound of BB, so s≤us \le u by the least-upper-bound property of ss in R\mathbb{R}. In every possible case, s≤us \le u. Hence ss is the least upper bound of AA in R‾\overline{\mathbb{R}}: sup⁡A=s=sup⁡RB\sup A = s = \sup_{\mathbb{R}} B. Cases (a)–(d) exhaust all subsets A⊆R‾A \subseteq \overline{\mathbb{R}} (partitioned by whether A=∅A=\varnothing, +∞∈A+\infty \in A, or not, and in the last case by boundedness of A∩RA \cap \mathbb R), and in each case we exhibited sup⁡A∈R‾\sup A \in \overline{\mathbb{R}} satisfying the definition of least upper bound. This proves every subset of R‾\overline{\mathbb{R}} has a supremum in R‾\overline{\mathbb{R}}. The infimum statement follows by the identical argument applied to lower bounds (or via the order-reversing bijection x↦−xx \mapsto -x on R‾\overline{\mathbb{R}}, which sends +∞↔−∞+\infty \leftrightarrow -\infty and reverses <<, converting the supremum argument into the infimum argument); we omit the repetition.

Part 3. This is precisely what was shown in Case (c) (unbounded case, giving sup⁡A=+∞\sup A = +\infty) and Case (d2) (bounded case, giving sup⁡A=sup⁡RB\sup A = \sup_{\mathbb R} B where here B=AB = A since A⊆RA \subseteq \mathbb R), applied to AA itself viewed as a subset of R‾\overline{\mathbb R}. ■\blacksquare

∎

Worked examples

Example 1 (a genuinely unbounded, non-monotone set). Let

A={ (−1)n n  :  n∈N }∪{1k:k∈N}⊆R.A = \{\, (-1)^n \, n \;:\; n \in \mathbb{N} \,\} \cup \left\{ \frac{1}{k} : k \in \mathbb{N}\right\} \subseteq \mathbb{R}.

So A={−1,2,−3,4,−5,6,… }∪{1,12,13,… }A = \{-1, 2, -3, 4, -5, 6, \dots\} \cup \{1, \tfrac12, \tfrac13, \dots\}. Compute sup⁡A\sup A and inf⁡A\inf A in R‾\overline{\mathbb{R}}. Solution. The even-index terms (−1)nn(-1)^n n for even n=2,4,6,…n = 2,4,6,\dots give 2,4,6,…2,4,6,\dots, which is unbounded above in R\mathbb{R} (for any real MM, choosing an even n>Mn > M gives an element of AA exceeding MM, by the Archimedean property of R\mathbb{R}). So AA is not bounded above in R\mathbb{R}; by Theorem Part 3 (unbounded case), sup⁡A=+∞\sup A = +\infty in R‾\overline{\mathbb{R}}. For the infimum: the odd-index terms give −1,−3,−5,…-1, -3, -5, \dots, unbounded below in R\mathbb{R} by the same Archimedean argument (for any real mm, an odd nn with n>−mn > -m gives (−1)nn=−n<m(-1)^n n = -n < m). So AA has no lower bound in R\mathbb{R}, and by the infimum-analogue of Part 3, inf⁡A=−∞\inf A = -\infty. Note the reciprocal terms 1/k1/k play no role in either computation — they are bounded (0<1/k≤10 < 1/k \le 1) and hence irrelevant to whether AA is unbounded overall; the unboundedness is driven entirely by the (−1)nn(-1)^n n part. This illustrates that sup⁡\sup and inf⁡\inf in R‾\overline{\mathbb{R}} genuinely depend on global behavior, not on any single "typical" element.

Example 2 (a family of nested sets and the role of sup⁡∅\sup \varnothing). For n∈Nn \in \mathbb{N}, define

An={x∈R:x>n and x2<n}.A_n = \left\{ x \in \mathbb{R} : x > n \text{ and } x^2 < n \right\}.

Determine sup⁡An\sup A_n for each nn, and determine lim⁡n→∞sup⁡An\lim_{n\to\infty} \sup A_n if it exists in R‾\overline{\mathbb{R}} (using the ordinary ε\varepsilon-definition adapted to R‾\overline{\mathbb R}, or just by inspection of eventual behavior). Solution. Fix n∈Nn \in \mathbb{N}. The condition "x>nx > n and x2<nx^2 < n" requires x>n≥1x > n \ge 1, hence x2>n2≥nx^2 > n^2 \ge n whenever n≥1n \ge 1 (since n2≥nn^2 \ge n for n≥1n\ge1), which contradicts x2<nx^2 < n. So no real xx satisfies both conditions simultaneously, for any n∈Nn \in \mathbb{N}: An=∅A_n = \varnothing for every nn. By the convention established in the definition and justified in Case (a) of the theorem's proof,

sup⁡An=−∞for every n∈N.\sup A_n = -\infty \quad \text{for every } n \in \mathbb{N}.

Consequently the sequence (sup⁡An)n(\sup A_n)_{n} is the constant sequence −∞,−∞,−∞,…-\infty, -\infty, -\infty, \dots, which trivially "converges" to −∞-\infty in R‾\overline{\mathbb{R}} (in the sense that it is eventually, indeed always, equal to −∞-\infty). This example is a useful sanity check: it shows the empty-set convention is not a pathological edge case invented for aesthetics, but arises naturally from ordinary-looking algebraic conditions, and the theorem handles it uniformly with no special pleading.

CounterexampleWhy we cannot make R‾\overline{\mathbb{R}} a field: dropping "leave 0⋅∞0\cdot\infty and ∞−∞\infty-\infty undefined"

Suppose, contrary to the definition above, we tried to force a value for ∞−∞\infty - \infty in order to make R‾\overline{\mathbb{R}} a field extending R\mathbb{R}'s arithmetic — say by declaring ∞−∞:=0\infty - \infty := 0, matching the naive idea "infinity minus itself is zero." We show this breaks associativity of addition, so (R‾,+)(\overline{\mathbb{R}}, +) cannot be a group (hence not a field) under any such extension. Take three elements a=1∈Ra = 1 \in \mathbb{R}, b=+∞b = +\infty, c=−∞c = -\infty. Compute two ways:

(a+b)+c=(1+∞)+(−∞)=∞+(−∞)=0(by the forced convention),(a+b) + c = (1 + \infty) + (-\infty) = \infty + (-\infty) = 0 \quad \text{(by the forced convention)},
a+(b+c)=1+(∞+(−∞))=1+0=1.a + (b+c) = 1 + (\infty + (-\infty)) = 1 + 0 = 1.

We get (a+b)+c=0(a+b)+c = 0 but a+(b+c)=1a+(b+c) = 1, so associativity fails: (a+b)+c≠a+(b+c)(a+b)+c \ne a+(b+c). Since associativity of ++ is required for (R‾,+)(\overline{\mathbb R},+) to be a group, and being a field requires (R‾,+)(\overline{\mathbb R},+) to be an abelian group, no assignment of a value to ∞−∞\infty - \infty (or to 0⋅∞0\cdot\infty, by an analogous computation using distributivity, e.g. 0=0⋅∞=(1−1)⋅∞0 = 0\cdot\infty = (1-1)\cdot\infty versus 1⋅∞−1⋅∞=∞−∞1\cdot\infty - 1\cdot\infty = \infty - \infty) can make R‾\overline{\mathbb{R}} a field extending the arithmetic of R\mathbb{R}. This is exactly why the definition above stops short of defining ∞−∞\infty-\infty and 0⋅∞0\cdot\infty: the order-completeness theorem (suprema always exist) survives without full arithmetic, but full arithmetic is genuinely impossible to add consistently.

Scope, edge cases & common traps

SituationWhat holds in R‾\overline{\mathbb{R}}What fails / must be checked
sup⁡\sup of a nonempty, bounded-above A⊆RA \subseteq \mathbb{R}equals ordinary sup⁡RA\sup_{\mathbb R} Anone — fully consistent extension
sup⁡\sup of unbounded-above A⊆RA \subseteq \mathbb{R}sup⁡A=+∞\sup A = +\inftydoes not exist in R\mathbb{R} alone
sup⁡∅\sup \varnothing=−∞= -\infty (forced by the definition of least upper bound)easy to misremember as "undefined"
x+∞x + \infty, x⋅∞x\cdot \infty (xx real, x≠0x\ne0)always defined, matches limit intuitionsign of xx matters for multiplication
∞−∞\infty - \infty, 0⋅∞0\cdot\inftyundefined in the order/limit frameworksome measure-theoretic texts set 0⋅∞=00\cdot\infty=0 by fiat, only for integration theory — do not import this into limit computations
R‾\overline{\mathbb{R}} as an algebraic structuretotally ordered set, complete latticenot a field, not a vector space, not even a group under ++
Limits of sequences with terms →±∞\to \pm\inftycan be defined meaningfully using R‾\overline{\mathbb{R}}'s orderstill need separate careful definitions of convergence "to ∞\infty"; not automatic from field structure since there is none

The mathematician behind it

The extended real numbers, in the form used throughout modern analysis, do not trace to a single named theorem but crystallized as completeness itself was being made rigorous in the late nineteenth century, particularly through the work of Richard Dedekind and Georg Cantor on the construction of R\mathbb{R}, and through the subsequent generation (Carathéodory, Lebesgue, Hausdorff) who needed +∞+\infty as a genuine value — not a mere symbol of divergence — for measures and integrals to behave uniformly.

Origins · 1 minConstantin Carathéodory and how this was discovered
CC
Constantin Carathéodory
1873–1950 · measure theory and the outer measure

Carathéodory's 1914 construction of outer measure assigns to every subset of Rn\mathbb{R}^n a value in [0,+∞][0,+\infty], treating +∞+\infty as a bona fide element of the value space on exactly the same footing as any finite number — precisely the extended real line's order-completeness at work, since every set has some outer measure, finite or not, with no exceptional case to handle separately.

History
- 1858–1872 — Dedekind and later Cantor/Weierstrass give rigorous constructions of R\mathbb{R} via cuts and Cauchy sequences, establishing the Completeness Axiom as the defining property of the reals.
- 1902–1904 — Lebesgue's theory of measure and integration requires "sets of infinite measure" to be treated as ordinary objects of the theory, pushing +∞+\infty into routine use as a value rather than a divergence symbol.
- 1914 — Carathéodory formalizes outer measure with values in [0,∞][0,\infty], making the extended nonnegative reals standard technical apparatus in measure theory.
- 1930s — Textbook treatments (notably in the tradition leading to Rudin's Principles of Mathematical Analysis) codify R‾=[−∞,+∞]\overline{\mathbb{R}} = [-\infty,+\infty] with its order and partial arithmetic as standard preliminary material before defining lim sup⁡\limsup and lim inf⁡\liminf.
- Present — R‾\overline{\mathbb{R}} and its higher-dimensional analogues (the extended nonnegative reals [0,∞][0,\infty], the two-point compactification) are standard in probability theory, functional analysis, and optimization, wherever suprema over unbounded families must always exist.
Is R‾\overline{\mathbb{R}} a metric space under the usual absolute-value distance?

Not under the ordinary metric d(x,y)=∣x−y∣d(x,y) = |x-y|, since ∣+∞−x∣|{+\infty} - x| is not defined for real xx (subtraction involving ∞\infty from a finite number is fine, e.g. +∞−x=+∞+\infty - x = +\infty, but then "∣+∞∣|+\infty|" is not a real number, so dd does not map into [0,∞)[0,\infty) as a genuine metric requires). However, R‾\overline{\mathbb{R}} can be metrized: the map φ(x)=x/(1+∣x∣)\varphi(x) = x/(1+|x|) for x∈Rx\in\mathbb R, extended by φ(±∞)=±1\varphi(\pm\infty)=\pm1, is an order-preserving bijection onto [−1,1][-1,1], and pulling back the usual metric on [−1,1][-1,1] via φ\varphi gives a genuine metric on R‾\overline{\mathbb{R}} compatible with its order topology. This is how one rigorously discusses convergence "to ∞\infty" topologically.

Why do we need both +∞+\infty and −∞-\infty, rather than a single point at infinity?

Because R‾\overline{\mathbb{R}} is an ordered completion, not merely a topological one. The one-point compactification of R\mathbb{R} (used in some contexts, analogous to the Riemann sphere for C\mathbb{C}) adds a single point ∞\infty that both "ends" wrap around to, but this destroys the linear order: you cannot consistently say whether ∞\infty is greater or less than every real number. Since the entire purpose here is to make sup⁡\sup and inf⁡\inf total operations respecting the order, we need one ideal point at each end — a top element +∞+\infty for suprema of unbounded-above sets, and a bottom element −∞-\infty for infima of unbounded-below sets (and for sup⁡∅\sup\varnothing).

Does the theorem's proof use the Completeness Axiom in a circular way?

No. The proof only invokes completeness of R\mathbb{R} (the hypothesis that nonempty, bounded-above subsets of R\mathbb{R} have a real least upper bound) as an already-established prerequisite. It never assumes the conclusion — completeness of R‾\overline{\mathbb{R}} — while proving that conclusion. Each case in the proof (empty set, set containing +∞+\infty, unbounded set, bounded nonempty real set) is handled either by direct verification from the definition of least upper bound, or by reduction to the real Completeness Axiom applied to a genuinely bounded real subset. This is a standard non-circular "case-exhaustion" argument.

Can I use 0⋅∞=00 \cdot \infty = 0 freely once I've seen it in a measure theory course?

Only within the specific convention of that context (typically, defining the Lebesgue integral of a nonnegative function that is zero on a set of infinite measure) and it must be stated explicitly as a convention, not derived from the arithmetic of R‾\overline{\mathbb{R}} described here. Outside of measure theory — for instance in ordinary limit computations, or in the algebra used in the counterexample above — treating 0⋅∞0\cdot\infty as a defined quantity leads to contradictions (as shown via distributivity), so it should never be used silently. Always name the convention explicitly when it is in force.

How does R‾\overline{\mathbb{R}} interact with lim sup⁡\limsup and lim inf⁡\liminf of sequences?

It is what makes those definitions total. For any real sequence (an)(a_n), whether or not (an)(a_n) is bounded, the quantities bn:=sup⁡k≥nakb_n := \sup_{k\ge n} a_k and lim inf⁡\liminf's analogous cn:=inf⁡k≥nakc_n := \inf_{k \ge n} a_k are well-defined elements of R‾\overline{\mathbb{R}} by the completeness theorem, and (bn)(b_n) is a non-increasing sequence in R‾\overline{\mathbb{R}}, so lim sup⁡an:=inf⁡nbn\limsup a_n := \inf_n b_n is again guaranteed to exist in R‾\overline{\mathbb{R}}. Without the extended reals, one would need separate definitions and case splits depending on boundedness of (an)(a_n); with them, lim sup⁡\limsup and lim inf⁡\liminf are always defined, and ana_n converges in the extended sense iff the two agree.

Is the empty-set convention sup⁡∅=−∞\sup\varnothing = -\infty just a notational trick, or does it have real content?

It has real content: it is forced by applying the actual definition of supremum (least upper bound) to the empty set, as shown in the proof's Case (a) — every element of R‾\overline{\mathbb{R}} vacuously satisfies "is an upper bound of ∅\varnothing," and the least element overall is −∞-\infty. It is not an arbitrary convenience; changing it would break the theorem that every subset has a supremum equal to the least element among its upper bounds. It also matches intuition from lattice theory: in a complete lattice, sup⁡\sup of the empty set is always the bottom element, and inf⁡\inf of the empty set is the top element.

Does adjoining ±∞\pm\infty change any true theorem about ordinary real numbers?

No — Part 3 of the theorem guarantees the extension is conservative: for A⊆RA\subseteq\mathbb R nonempty and bounded above, sup⁡A\sup A computed in R‾\overline{\mathbb R} agrees exactly with sup⁡A\sup A computed in R\mathbb R via the ordinary Completeness Axiom. So every classical theorem phrased only about bounded, nonempty subsets of R\mathbb{R} remains word-for-word true; R‾\overline{\mathbb{R}} only adds new statements (about unbounded and empty sets) rather than altering old ones.

Before this