Why Completeness Matters

Completeness Supplies Missing Boundary Points

The greatest lower bound property gives a sharp lower boundary to every nonempty real set that is bounded below. Its order counterpart, the least upper bound property, guarantees a sharp upper boundary for every nonempty real set that is bounded above. Together, these properties express a central feature of the real numbers: bounded sets cannot have a missing boundary that ought to be a real number.

The field and order rules alone do not ensure this. Earlier in the course, the set of nonnegative rational numbers whose squares are less than $2$ was shown to be bounded above in $\mathbb{Q}$ but to have no supremum in $\mathbb{Q}$. The real numbers fill this kind of gap. Completeness matters because many basic results about boundaries, intervals, and limits depend on having no such gaps.

Definition (Completeness). The real numbers are complete if every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$. Equivalently, they are complete if every nonempty subset bounded below has an infimum in $\mathbb{R}$. The Least Upper Bound Property and the Greatest Lower Bound Property express these equivalent forms of completeness.

This definition is not an additional algebraic rule. It is an existence guarantee about sets: when a nonempty set has an upper bound, its least upper bound is a real number. The guarantee is useful even when the supremum is not in the set. It lets us locate a boundary by considering all elements of a set at once, rather than requiring that the boundary be attained.

A Common Point in Nested Intervals

One consequence of completeness concerns intervals that become more and more restricted. Suppose each interval is closed and bounded, and every interval after the first lies inside its predecessor. The intervals may narrow indefinitely without any one of them becoming empty. Completeness ensures that they still have a point in common.

Theorem (Nested Closed Interval Theorem). Let $I_n=[a_n,b_n]$, for each positive integer $n$, be a sequence of nonempty closed bounded intervals such that $I_{n+1}\subseteq I_n$. Then $$ \bigcap_{n=1}^{\infty} I_n\ne\varnothing. $$ If, in addition, the lengths $b_n-a_n$ tend to $0$, the intersection contains exactly one point.

Proof. Since $I_{n+1}\subseteq I_n$, the left endpoints do not decrease: $a_n\leq a_{n+1}$. Also, every left endpoint is at most $b_1$, because every interval lies inside $I_1=[a_1,b_1]$. Thus the set $A=\{a_n:n\geq1\}$ is nonempty and bounded above. By the Least Upper Bound Property, $\alpha=\sup A$ is a real number.

Fix any positive integer $n$. We show that $b_n$ is an upper bound for $A$. If $m\geq n$, then $a_m\in I_m\subseteq I_n$, so $a_m\leq b_n$. If $m<n$, the nesting gives $a_m\leq a_n$, and $a_n\leq b_n$. Therefore $a_m\leq b_n$ in either case. Since $\alpha$ is the least upper bound of $A$, it follows that $\alpha\leq b_n$. Also $a_n\leq\alpha$, because $\alpha$ is an upper bound for $A$. Hence $a_n\leq\alpha\leq b_n$, so $\alpha\in I_n$. This holds for every $n$, proving that the intersection is nonempty.

Now suppose the lengths tend to $0$. If two distinct points $x$ and $y$ belonged to every interval, assume without loss of generality that $x<y$. Set $\varepsilon=y-x>0$. By the assumption on the lengths, some $n$ satisfies $b_n-a_n<\varepsilon$. But both points belong to $[a_n,b_n]$, so $y-x\leq b_n-a_n$, contradicting $b_n-a_n<\varepsilon=y-x$. Thus there cannot be two distinct common points. The intersection contains exactly one point. $\square$

Worked Example: Nested Intervals with More Than One Common Point

Define $I_1=[1,4]$, $I_2=[2,4]$, and $I_n=[2,3]$ for every $n\geq3$. These are nonempty closed bounded intervals. The first inclusion holds because $[2,4]\subseteq[1,4]$; the next holds because $[2,3]\subseteq[2,4]$; and all later intervals are equal. A point belongs to every interval exactly when it lies in $[2,3]$. Therefore $$ \bigcap_{n=1}^{\infty}I_n=[2,3]. $$ The theorem guarantees a common point, not necessarily a unique one. Here the lengths do not tend to $0$: from $n=3$ onward, each length is $3-2=1$.

Worked Example: Nested Intervals with One Common Point

Let $I_1=[-2,5]$, $I_2=[1,4]$, and $I_n=[3,3]$ for every $n\geq3$. The intervals are nested: $[1,4]\subseteq[-2,5]$, and $[3,3]\subseteq[1,4]$. Every interval from the third onward contains only $3$, and $3$ also belongs to the first two intervals. Thus $$ \bigcap_{n=1}^{\infty}I_n=\{3\}. $$ The lengths are $7$, $3$, and then $0$ for every $n\geq3$, so they tend to $0$, as required by the uniqueness part of the theorem.

The proof shows exactly where completeness enters: the set of left endpoints has a supremum, and that supremum belongs to every interval. Without an appropriate completeness principle, nested intervals could keep imposing tighter bounds without leaving any point in the number system that satisfies all of them.

Bounded Monotone Sequences Have Limits

The same idea applies to an ordered list of real numbers. A nondecreasing sequence that is bounded above has a least upper bound. Its terms approach that boundary: once one term is sufficiently close to the supremum, all later terms are at least as large and remain below the supremum. The decreasing case follows from the corresponding infimum argument.

For clarity, a sequence $(x_n)$ converges to $L$ if, for every $\varepsilon>0$, there is an index $N$ such that $|x_n-L|<\varepsilon$ whenever $n\geq N$. A sequence is nondecreasing if $x_n\leq x_{n+1}$ for every $n$, and nonincreasing if $x_{n+1}\leq x_n$ for every $n$.

Theorem (Bounded Monotone Convergence). Every nondecreasing real sequence that is bounded above converges to the supremum of its set of terms. Every nonincreasing real sequence that is bounded below converges to the infimum of its set of terms.

Proof. First let $(x_n)$ be nondecreasing and bounded above. Its set of terms $S=\{x_n:n\geq1\}$ is nonempty and bounded above, so completeness gives $\alpha=\sup S$. Let $\varepsilon>0$. The number $\alpha-\varepsilon$ is strictly less than $\alpha$, so it cannot be an upper bound for $S$; otherwise $\alpha$ would not be the least upper bound. Consequently, there is an index $N$ for which $\alpha-\varepsilon<x_N$. For every $n\geq N$, monotonicity and the upper-bound property of $\alpha$ give $$ \alpha-\varepsilon<x_N\leq x_n\leq\alpha. $$ Therefore $0\leq\alpha-x_n<\varepsilon$, which means $|x_n-\alpha|<\varepsilon$. This proves that $x_n$ converges to $\alpha=\sup S$.

Now let $(x_n)$ be nonincreasing and bounded below. The set $S=\{x_n:n\geq1\}$ is nonempty and bounded below, so the Greatest Lower Bound Property gives $\beta=\inf S$. For any $\varepsilon>0$, the number $\beta+\varepsilon$ cannot be a lower bound for $S$, since it is greater than the greatest lower bound. Thus some index $N$ satisfies $x_N<\beta+\varepsilon$. For every $n\geq N$, nonincreasing monotonicity and the lower-bound property of $\beta$ give $$ \beta\leq x_n\leq x_N<\beta+\varepsilon. $$ Hence $|x_n-\beta|<\varepsilon$, so the sequence converges to $\beta=\inf S$. $\square$

Worked Example: A Bounded Nondecreasing Sequence

For each positive integer $n$, let $x_n=\min\{n,4\}$. The first terms are $1,2,3,4,4,\ldots$, so the sequence is nondecreasing and bounded above by $4$. Its set of terms contains $4$, and no term exceeds $4$, so $\sup\{x_n:n\geq1\}=4$. The theorem gives convergence to $4$; directly, $x_n=4$ for every $n\geq4$. Thus for any $\varepsilon>0$, choosing $N=4$ gives $|x_n-4|=0<\varepsilon$ whenever $n\geq N$.

Why These Consequences Matter

The nested interval and bounded monotone convergence theorems are different ways to use the same completeness guarantee. For nested intervals, the left endpoints form a bounded set whose supremum supplies a point satisfying every interval constraint. For a bounded nondecreasing sequence, the supremum of the terms is the limit. In both arguments, the existence of a sharp boundary turns an indefinite process of tightening bounds into an actual real number.

The hypotheses cannot be discarded. Nested intervals must be nonempty and nested; otherwise the intersection need not exist. A nondecreasing sequence must be bounded above for the stated theorem to apply. Completeness does not say that every sequence converges, nor that every collection of intervals has a common point. It guarantees these conclusions only when the relevant bounds and structure are present.

The contrast with the rational numbers is instructive. The earlier example of a bounded subset of $\mathbb{Q}$ with no rational supremum shows that the least upper bound guarantee does not follow merely from being an ordered field. Completeness distinguishes $\mathbb{R}$ by ruling out these missing boundaries. That is why it supports so many later arguments in analysis: proofs about limits, approximation, and continuity often need to know that a boundary approached by real numbers is itself real.

Key takeaway. Completeness is an existence principle for sharp boundaries. It guarantees common points for nested nonempty closed bounded intervals and limits for bounded monotone real sequences. These are concrete reasons the completeness of $\mathbb{R}$ matters.

Check Your Understanding

Use the completeness definition and the two proved theorems to answer the following questions.

Which set is used in the nested interval proof to produce a common point, and what completeness property guarantees its supremum?
Why does the Nested Closed Interval Theorem guarantee at least one common point even when the interval lengths do not tend to $0$?
What additional condition in the theorem ensures that nested intervals have only one common point?
Why can $\alpha-\varepsilon$ not be an upper bound for the terms of a sequence whose supremum is $\alpha$?
Which completeness property gives the limit of a nonincreasing sequence bounded below?

What You'll Learn