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

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.

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

The rational counterexample

The standard counterexample is

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

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.

Why no rational number can be sup(S)

The proof splits into three cases according to the square of a rational candidate s.

Case 1: $s^2<2$

Then s is still too small. In fact, one can move a little to the right and remain inside $S$.

Case 2: $s^2>2$

Now s really is an upper bound, but it is too large to be the least upper bound. We can move slightly left while staying above $S$.

Case 3: $s^2=2$

This is impossible in $Q$, because the previous chapter proved that $\sqrt{2}$ is irrational.

Putting the three cases together:

• $s^2<2$ means s is not even an upper bound;
• $s^2>2$ means s is an upper bound, but not the least one;
• $s^2=2$ cannot happen in $Q$.

So $S$ has no supremum in $Q$.

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.

Quick checks

Exercises

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.