4.3 Completeness and gaps in Q

Completeness is an existence property

The previous note introduced supremum and infimum. Completeness asks whether those extremal bounds actually exist whenever they should exist.

This is the first point where $Q$ and $R$ separate decisively. Algebraically, $Q$ already looks very strong: it is a field, and it comes with a familiar total order. But order-theoretically, $Q$ still misses some boundary points.

The definition

Definition

Complete ordered set

An ordered set $X$ is complete if:

• every nonempty subset $Y\subseteq X$ that is bounded above has a supremum in $X$;
• every nonempty subset $Y\subseteq X$ that is bounded below has an infimum in $X$.

So completeness is not about having many elements. It is about never losing the least upper bound or greatest lower bound when those bounds ought to be there.

Worked example

Finite ordered sets are complete

If $X$ is a finite totally ordered set, every nonempty subset already has a largest and smallest element. Those elements automatically serve as supremum and infimum.

So finite totally ordered sets are complete for a simple reason: the relevant extremes are already attained inside the set.

The interesting question is what happens for infinite totally ordered sets such as $Q$ and $R$.

The rational counterexample

The standard counterexample is

$$S=\{x\in Q\mid x^2\lt 2\}.$$

This set is not empty, because $1\in S$. It is also bounded above in $Q$; for instance $2$ is an upper bound.

What would happen if $Q$ were complete? Then $S$ would have a rational supremum. The whole argument below shows that no rational number can play that role.

Theorem

Q is not complete

The ordered set $(Q,\le)$ is not complete.

Why no rational number can be sup(S)

Any upper bound $s$ must satisfy $s\ge1$, since $1\in S$. In particular, $s\le0$ cannot be an upper bound. We may therefore restrict the candidate supremum to positive rational numbers before comparing squares.

The two rational perturbations

Suppose $s\in Q$ and $s\gt 0$. If $s^2\lt 2$, let $M=2-s^2$ and choose

$$h=\frac12\min\left\{1,\frac{M}{2s+1}\right\}\gt 0.$$

This is an explicit rational choice. Since $h\lt 1$ and $h\lt M/(2s+1)$,

$$(s+h)^2=s^2+h(2s+h)\lt s^2+M=2.$$

Thus $s+h\in S$ and $s+h\gt s$, so $s$ is not an upper bound.

If $s^2\gt 2$, let $M=s^2-2$ and choose

$$h=\frac12\min\left\{s,\frac{M}{2s}\right\}\gt 0.$$

Now $r=s-h\gt 0$, and

$$r^2=s^2-2sh+h^2\gt s^2-2sh\gt 2.$$

Every $x\in S$ satisfies $x\lt r$: otherwise $x\ge r\gt 0$ would imply $x^2\ge r^2\gt 2$. Consequently $r$ is an upper bound smaller than $s$, contradicting leastness. Positivity of $r$ is essential; a negative rational whose square exceeds $2$ is not an upper bound of $S$.

Finally, $s^2=2$ is impossible for rational $s$ by the earlier parity proof. These cases exhaust the positive candidates, so $S$ has no supremum in $Q$. All quantities used here are rational; no real square root or completeness principle was assumed.

Completeness is not the same as maximality

The word “maximum” describes membership, whereas completeness describes an existence principle for bounds. A maximum is an element of the subset; a supremum may be an endpoint in the ambient set that the subset never reaches. Conversely, a supremum can exist for a set even when the ambient ordered set is not complete. Completeness says that this happens for every nonempty bounded set in the ambient universe.

Theorem

Finite total order and completeness

Every finite totally ordered set is complete because each nonempty subset has a maximum and a minimum. This finite argument does not extend to arbitrary infinite ordered sets: an infinite set may have a sharp boundary without attaining it, or may have a bounded subset whose boundary is absent from the ambient set.

Worked example

An explicit finite total-order calculation

Let $X=\{-2,0,3\}$ with its usual order and let $Y=\{-2,3\}$. In the ambient set $X$, the only upper bound of $Y$ is $3$, so $\sup_X(Y)=3$; the only lower bound is $-2$, so $\inf_X(Y)=-2$. Both are attained because they belong to $Y$. If instead $Z=\{0,3\}\subseteq X$, then its upper bounds in $X$ are again $\{3\}$, while its lower bounds are $\{-2,0\}$. The ambient set controls the list of bounds, but the maximum and minimum are still determined by membership in the subset.

Worked example

An infimum without a minimum

The positive rational set $Q_{>0}$ has no minimum: for every $q>0$, the rational $q/2$ is positive and smaller. Nevertheless $0$ is its infimum in $Q$. Indeed, $0$ is a lower bound, while a lower bound $\ell>0$ would have to satisfy $\ell\le\ell/2$, which is impossible. This proves $\inf_Q(Q_{>0})=0$, even though the minimum does not exist. It is the same membership distinction that later lets $S$ have a real supremum without containing $\sqrt 2$.

Density is not the same as completeness

It is tempting to object:

"But there are infinitely many rationals between any two rationals. Surely the gap can be filled by approximations."

That objection confuses two ideas.

• Density says that between two distinct rationals, another rational can be found.
• Completeness says that every bounded nonempty set has the correct least upper bound and greatest lower bound inside the ambient ordered set.

The set $S$ shows that you can have endless rational approximations to a missing boundary and still fail completeness.

Common mistake

Approximation does not equal possession

$Q$ contains rationals arbitrarily close to $\sqrt{2}$. That does not mean $Q$ contains the least upper bound of $S$. A sequence of better and better approximations is weaker than actually having the boundary point.

Reading the perturbation proof as a completeness test

The two perturbations above are more than a trick for this one polynomial. They are a general way to test a proposed least upper bound. A candidate $s$ can fail in two different directions. If there is room below the target, the upward perturbation $s+h$ remains in the set and defeats $s$ as an upper bound. If the candidate is too high, the downward perturbation $s-h$ remains an upper bound and defeats $s$ as the least upper bound. A genuine boundary can survive both tests only when it sits exactly at the transition between the two cases.

For the square condition, the transition is $s^2=2$. The argument first uses $1\in S$ to show every upper-bound candidate is positive. That preliminary sign step is necessary: the map $x\mapsto x^2$ is not increasing on all of $Q$, so one cannot compare squares until the candidate and the relevant witness are known to be nonnegative. In the second case, one also proves $r=s-h>0$; without that fact, $x\ge r$ would not imply $x^2\ge r^2$.

The argument is therefore a model of proof order:

1. establish nonemptiness and an explicit bound;
2. restrict candidates using a simple element of the set;
3. split all remaining candidates into exhaustive algebraic cases;
4. construct a strict witness in the first case or a smaller bound in the second;
5. use the rational parity argument to eliminate the equality case.

Skipping any one of these steps leaves a logical gap. In particular, “the values approach $2$” is not a substitute for the second case, because it never shows that every smaller rational is defeated.

Worked example

A bounded rational set whose endpoint is present

Let $T=\{q\in Q:0\lt q\le1\}$. The element $1$ belongs to $T$ and is an upper bound, so it is immediately the maximum and therefore $\sup_Q(T)=1$. The element $0$ is a lower bound. If $\ell>0$ were a larger lower-bound candidate, then $\ell/2$ would be positive; when $\ell/2\le1$, it lies in $T$ and is smaller than $\ell$, contradicting the lower-bound condition. If $\ell/2>1$, then already $1\in T$ contradicts $\ell\le1$. Thus $\inf_Q(T)=0$, and $T$ has no minimum. This example is deliberately close to the $\sqrt{2}$ set: the same language of bounds is used, but the endpoint $1$ is rational and present, so no completeness failure occurs.

The contrast identifies the exact claim made by the completeness axiom. It does not say that every subset has a maximum or minimum. It says that when a nonempty subset is bounded in its chosen ambient ordered set, a best upper or lower bound exists there. For $S$, the real boundary exists in $R$ but the rational boundary does not exist in $Q$; density supplies approximating points, while completeness supplies the missing ambient boundary.

The ambient qualification is also visible in a simple change of notation. The set $T$ has $\sup_Q(T)=1$, and the same value is $\sup_R(T)$, because $1$ is rational and the upper-bound comparison is unchanged when the ambient universe is enlarged. For the square-root set, however, writing $\sup_Q(S)$ is a claim about a rational number and is false, while writing $\sup_R(S)$ is a claim about the completed real line and is true. Supremum notation is therefore never just a decoration on a set; it records where the proposed boundary is required to live.

This viewpoint also prevents a common converse error. If a particular set has a supremum, one cannot conclude that its ambient ordered set is complete. The geometric-sum set in the previous note has a rational supremum even though $Q$ is not complete. To establish completeness, one must quantify over every nonempty bounded subset, including subsets whose boundary is not described by an obvious formula. The $\sqrt{2}$ cut is valuable precisely because it shows that one such universal claim fails.

The example also separates two meanings of “gap.” There is no gap between two distinct rationals in the density sense: another rational always lies between them. The gap in $S$ is an endpoint gap, detected only after quantifying over all upper bounds. Rational points can occupy every interval around the missing boundary, but the boundary itself can still be absent. A proof of completeness must therefore ask where the best bound lives, which set supplies the witnesses, and which theorem guarantees existence in that ambient set.

The practical lesson is to keep three questions separate whenever a boundary is discussed. First ask whether the set is nonempty. Next ask whether a bound exists in the chosen ambient set. Finally ask whether a best bound exists there. For a finite total order the last question is answered by the largest or smallest member. For the positive rationals the answer is supplied by zero, which is outside the set. For the square-root cut the answer depends on moving from the rational ambient set to the complete real ambient set. The same words therefore describe three different mechanisms, and the proof must identify which mechanism is being used.

This is the boundary question that completeness answers. It is an existence statement about the ambient ordered set.

Quick checks

Checkpoint

Why is 2 an upper bound of $S=\{x\in Q:x^2<2\}$?

If x>2, what happens to x^2?

Solution · Answer

If $x\gt 2$, then $x^2\gt 4\gt 2$, so such an $x$ cannot belong to $S$. Therefore every member of $S$ lies at most $2$, so $2$ is an upper bound.

Checkpoint

Why does the density of Q not imply completeness of Q?

Answer in one careful sentence.

Solution · Answer

Density only says there are rationals between nearby rationals, while completeness requires least upper bounds for all bounded nonempty subsets; the set $S=\{x\in Q\mid x^2\lt 2\}$ has arbitrarily good rational approximations to its boundary but no rational supremum.

Exercises

Checkpoint

Explain why every finite totally ordered set is complete.

Use the fact that nonempty finite subsets always attain their largest and smallest elements.

Solution · Guided solution

Take any nonempty subset $Y$ of a finite totally ordered set $X$. Because $Y$ is finite, one can compare its elements and find a largest and smallest element. Those are respectively the maximum and minimum of $Y$, so they also serve as supremum and infimum. Hence every finite totally ordered set is complete.

Checkpoint

Suppose Q were complete. What would that force for the set $S=\{x\in Q:x^2<2\}$, and why is that impossible?

Connect the definition of completeness directly to the counterexample.

Solution · Guided solution

If $Q$ were complete, then every nonempty subset of $Q$ with an upper bound would have a supremum in $Q$. The set $S$ is nonempty and bounded above, so it would have some $s\in Q$ with $s=\sup(S)$. But the three-case argument shows no rational $s$ can do that. So completeness of $Q$ is impossible.

Related notes

Read this after 3.5 Gaps in Q and sqrt(2) and 4.2 Upper bounds, supremum, and infimum. Then continue to 4.4 Axioms for the reals and first approximations.