4.3 Completeness and gaps in Q

Completeness is an existence property

An ordered field can support arithmetic and comparison while a nonempty subset bounded above has no least upper bound. Completeness rules out precisely this failure. The rational set defined by $x^2\lt2$ will show that $Q$ does not satisfy that stronger requirement.

The definition

Definition 4.3.1

(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 4.3.2

(Finite totally 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 4.3.3

(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 requires a supremum for every nonempty subset bounded above in the ambient order, and an infimum for every nonempty subset bounded below.

Worked example 4.3.4

(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 $\sup\limits_X(Y)=3$; the only lower bound is $-2$, so $\inf\limits_X(Y)=-2$. 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 4.3.5

(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 $\inf\limits_Q(Q_{>0})=0$, even though the minimum does not exist. It is the same membership distinction that separates a bound in the ambient set from an element of the subset.

Density is not the same as completeness

For rational $a<b$, the midpoint $(a+b)/2$ is rational and lies strictly between them. Repeating this gives more rationals in every rational interval. This density property compares points already in $Q$; completeness asks whether every nonempty set bounded above has a least upper bound in $Q$. The set $S$ satisfies the boundedness hypotheses but has no such rational bound.

Common mistake

(Approximation does not establish a boundary object)

Rational brackets for the equation $x^2=2$ can be made progressively narrower. That process alone neither supplies a rational supremum of $S$ nor defines a new number. A construction must specify an object and prove that it has the required order properties.

Reading the perturbation proof as a completeness test

The two perturbations attack different clauses of the supremum definition: $s+h$ disproves the upper-bound claim, while the smaller upper bound $s-h$ disproves leastness. Both comparisons require positive candidates, since squaring is not increasing on all of $Q$. Neither test names an already existing irrational boundary.

Worked example 4.3.6

(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 $\sup\limits_Q(T)=1$. 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 $\inf\limits_Q(T)=0$, and $T$ has no minimum. This example is deliberately close to the set $S$: the same language of bounds is used, but the endpoint $1$ is rational and present, so no completeness failure occurs.

For $T$, the bound $1$ exists in $Q$; for $S$, no rational candidate survives. Thus the existence of a supremum for one set does not imply completeness of the ambient order. The axiom quantifies over every nonempty subset bounded above, with the dual requirement for subsets bounded below.

After a complete ordered field containing $Q$ has been constructed, $S$ will have a supremum there. The notations $\sup\limits_Q(S)$ and $\sup\limits_R(S)$ make different existence claims: the first fails, while the second will follow from the construction. Keeping the ambient set visible distinguishes a missing bound from a failure of density between existing rational points.

Quick checks

Quick check

(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.

Quick check

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

Answer in one careful sentence.

Solution · Answer

Density says that between any two distinct rationals there is another rational. Completeness requires a least upper bound for every nonempty subset bounded above in the ambient order; the set $S=\{x\in Q\mid x^2\lt 2\}$ has arbitrarily good rational approximations to its boundary but no rational supremum.

Exercises

Quick check

(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.

Quick check

(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.