The rational numbers feel solid until you try to measure the diagonal of a unit square. You ask for a number whose square is 2, and the rationals have nothing to offer: there is a gap exactly where you reach. Real analysis begins by repairing that gap, and the repair is a single structural assumption called completeness. Everything else in the course, from convergent sequences to power series used in option pricing, rests on it.
Ordered fields
The real numbers form an ordered field. The field axioms say we can add, subtract, multiply, and divide (except by zero) with the usual rules. The order axioms add a relation that is total, transitive, compatible with addition (if a is less than b then a+c is less than b+), and compatible with multiplication by positives. The rationals satisfy all of these. So do the reals. The order axioms alone cannot distinguish them, which is exactly why we need one more axiom.
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Lesson close · Real Analysis for Quants
Consolidate The Real Number System & Completeness
Use this checkpoint to recall the lesson before marking it complete. The concepts below come from this lesson's authored sections.
Key concepts covered
01Ordered fields
02Bounds, supremum, infimum
03The completeness axiom
04The Archimedean property
05Density of the rationals
What you should know now
□Explain The Real Number System & Completeness in your own words and state the assumptions.
□Reproduce the main argument, derivation, or implementation without copying the lesson.
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Progress
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Check one example against units, boundary conditions, or edge cases.
□Name one modeling or implementation limitation and when it matters.
Sequence
Continue to Sequences & Cauchy Convergence, the next prerequisite step in this course.
Let S be a nonempty set of real numbers. A number u is an upper bound for S if every element s of S satisfies s at most u. The set is bounded above if such a u exists. The supremum, written supS, is the least upper bound: an upper bound that is at most every other upper bound. Symmetrically the infimuminfS is the greatest lower bound.
The defining property of the supremum has two halves. First, supS is an upper bound. Second, nothing smaller works: for any ε greater than 0, there is an element s of S with s greater than supS−ε. That second half is the workhorse in proofs.
The completeness axiom
Completeness: every nonempty subset of R that is bounded above has a supremum in R.
This is the axiom Q fails. Consider the set of rationals whose square is less than 2. It is bounded above (by 2, say) but its least upper bound would have to be 2, which is not rational. Inside R the supremum exists; inside Q it does not. The reals are, up to isomorphism, the unique complete ordered field.
The Archimedean property
Claim. For any real x there is a natural number n with n greater than x.
Proof sketch. Suppose not. Then the natural numbers N are bounded above by x. By completeness N has a supremum α. Since α is the least upper bound, α−1 is not an upper bound, so some natural number m satisfies m greater than α−1. But then m+1 is a natural number with m+1 greater than α, contradicting that α bounds N. Hence no such x exists, proving the claim. ■
An immediate consequence: for every ε greater than 0 there is an n with 1/n less than ε. Apply the property to x=1/ε.
Density of the rationals
Claim. Between any two reals a less than b there is a rational.
Derivation. Since b−a is positive, the Archimedean property gives an n with 1/n less than b−a, that is, n(b−a) greater than 1. Now choose the integer m to be the smallest integer greater than na. Then m−1 is at most na, so m is at most na+1, which is less than na+n(b−a)=nb. Combining, na less than m less than nb, hence a less than m/n less than b. The rational m/n lands strictly between a and b. ■
Worked example
Let S be the set of all numbers 1−1/n for natural n, namely 0,1/2,2/3,3/4,…. We claim supS=1. First, 1 is an upper bound: each 1−1/n is less than 1. Second, take any ε greater than 0. By the Archimedean corollary pick n with 1/n less than ε. Then the element 1−1/n exceeds 1−ε. So no number below 1 can be an upper bound, and 1 is the least one. Notice 1 itself is not an element of S: a supremum need not be attained.
Why quants care
Every convergence result a quant relies on is downstream of completeness. The fixed point that defines an implied volatility, the limiting price of a binomial tree as steps go to infinity, the value of a perpetual annuity as an infinite sum, and the existence of an optimal portfolio on a closed bounded constraint set all depend on suprema and limits actually existing. Completeness is the guarantee that when a Cauchy iteration or a monotone bisection search ought to converge, there is a genuine real number waiting at the end rather than a hole. Without it, numerical methods would chase limits that may not exist.
Adapted from MIT OpenCourseWare (18.100 Real Analysis), CC BY-NC-SA.
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Knowledge check
Q1.Why does the set of rationals with square less than 2 fail to have a supremum within the rational numbers?
Q2.The Archimedean property is most directly used to prove which statement?
Q3.If u is the supremum of a nonempty set S, which statement must hold?
Problems
P1
Let S be the set of values 3 + 2/n for natural numbers n = 1, 2, 3, ... . Compute the infimum of S. Give a decimal.
P2
Prove that between any two distinct real numbers there exists an irrational number.
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