4.5 Dedekind cuts and the embedding of Q

Why cuts are the next step

The previous note ended with a picture:

• a real number should determine everything in $Q$ that lies to its left;
• it should also determine everything in $Q$ that lies to its right;
• the real number itself should behave like the boundary between those two camps.

The next step is to turn that picture into a formal definition. The important move is this: instead of assuming the real number already exists and then asking what lies below it, we define the real number by the left-hand part of $Q$ itself.

That is the basic idea of a Dedekind cut.

Two equivalent definitions

In words, $A$ is the entire left side of the boundary, and $B$ is the entire right side. There is no overlap, nothing is missing, and the left side does not contain its own last point.

The second form is often easier to use. Once the left set $A$ is known, the right set is automatically $Q\setminus A$. So the cut is completely determined by which rationals count as "already to the left of the boundary".

What each condition is doing

The definition is short, but every clause has a job.

• $A\ne\varnothing$ and $A\ne Q$ prevent a fake cut with no left side or no right side.
• Downward closure says that if a rational is already on the left, then every smaller rational must also be on the left.
• The "no maximum" condition says the boundary itself is not stored as the last element of $A$; there is always room to move a little farther right while staying on the left.

This last point is the subtle one. It is exactly what prevents one rational number from being represented twice.

Rational cuts and the embedding of $Q$

Call a cut (A,B) rational if $B$ has a minimum element. In that case the boundary is already achieved by a rational number.

If $\min(B)=q$, then necessarily

$$A=\{x\in Q\mid x<q\}.$$

That motivates the notation

$$q_R=\{x\in Q\mid x<q\}.$$

So each rational number q gives a Dedekind cut $q_R$, and the map

$$i:Q\to R,\qquad i(q)=q_R$$

embeds $Q$ into the cut model of the reals.

This matters conceptually. The cut construction is not trying to throw away the rationals and start from scratch. It is enlarging $Q$ by adding new boundaries that were missing before.

Order and the first operations on cuts

Defining the set $R$ of all cuts is not enough. To match the target from the previous note, we must also define order and arithmetic on $R$.

For order, the natural rule is:

$$A\le A' \quad\text{if and only if}\quad A\subseteq A'.$$

This is exactly the right comparison for left sets. If every rational already on the left of $A$ is also on the left of A', then the boundary represented by $A$ cannot lie to the right of the boundary represented by A'.

For addition, define

$$A+A'=\{a+a' \mid a\in A,\ a'\in A'\}.$$

The idea is that the left side of a sum should consist of rational numbers that can already be reached by adding something strictly left of the first boundary to something strictly left of the second boundary.

For multiplication, first handle the case of nonnegative cuts and then extend the definition by sign rules. When $0_R\le A$ and $0_R\le A'$, they use

$$A\cdot A' = \{aa' \mid a\in A,\ a'\in A',\ a\ge 0,\ a'\ge 0\}\cup\{x\in Q\mid x<0\}.$$

This is more technical than addition, but the guiding idea is the same: build the left side of the product from rational data that already lies to the left.

The proofs are not short, but the overall message is clean: Dedekind cuts do exactly what chapter 4 asked a model of the real numbers to do.

Seeing the boundary on the number line

Figure. A cut stores every rational to the left of a boundary. When the boundary is $\sqrt{2}$, the right side has no smallest rational element.

Compare rational and irrational cuts interactively

The explorer below places the rational boundary 3/2 next to the irrational boundary $\sqrt{2}$. The key structural question is whether the right-hand side starts with a least rational element.

Common mistakes

Quick checks

Exercises

Related notes

Read this after 4.4 Axioms for the reals and first approximations. Then continue with 4.6 Decimal expansions and irrational numbers, which reconnects cuts with familiar decimal notation and introduces $\sqrt2$ as an irrational cut.