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. The construction has four tasks: identify the boundary data, recover the rational numbers inside it, define compatible order and arithmetic, and prove that every nonempty bounded family has a least upper bound. The last task is what fills the gaps in $Q$.

Two equivalent definitions

Definition

Dedekind cut as a pair

A Dedekind cut is a pair $(A,B)$ of nonempty subsets of $Q$ such that:

• $Q=A\cup B$;
• for every $a\in A$ and $b\in B$, we have $a\lt b$;
• $A$ has no maximum element.

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.

Definition

Dedekind cut as a single subset

Equivalently, a subset $A\subset Q$ is called a Dedekind cut if:

• $A\ne\varnothing$ and $A\ne Q$;
• whenever $x\in A$ and $y\in Q$ satisfy $y\lt x$, then $y\in A$;
• for every $x\in A$, there exists $y\in A$ with $x\lt y$.

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".

Why the two definitions agree

For a pair $(A,B)$, the strict separation forces disjointness, so $B=Q\setminus A$. If $x\in A$ and $y\lt x$, then $y$ cannot belong to $B$, because separation would require $x\lt y$; hence $y\in A$. Nonemptiness, properness, and no greatest member follow from the pair axioms. Conversely, start with a lower cut $A$ and put $B=Q\setminus A$. Both sides are nonempty. If $a\in A$, $b\in B$, then $b\le a$ is impossible: equality contradicts membership, and $b\lt a$ contradicts downward closure. Thus $a\lt b$, and all pair axioms hold.

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.

Worked example

The cut for $q=3/2$

Let

$$A=\{x\in Q\mid x\lt 3/2\}, \qquad B=\{x\in Q\mid x\ge 3/2\}.$$

Then $A$ and $B$ are both nonempty, every element of $A$ is smaller than every element of $B$, and $A$ has no maximum.

To see the last point, start with any $x\lt 3/2$ and define

$$y=\frac{x+3/2}{2}.$$

Then $x\lt y\lt 3/2$, so $y\in A$ and $x$ was not maximal. Thus $(A,B)$ is a Dedekind cut. This cut is the rational number $3/2$ viewed inside the real number system built from cuts.

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

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

That motivates the notation

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

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.

Theorem

The rational numbers sit inside the cut model

Rational Dedekind cuts are in bijection with rational numbers. After this embedding is established, it is standard to identify $q$ with its cut $q_R$. In particular, $0_R$ and $1_R$ play the roles of $0$ and $1$ inside $R$.

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.

Common mistake

Using $x\le q$ instead of $x\lt q$

The set $\{x\in Q\mid x\le q\}$ is nonempty, proper, and downward closed, but it fails the "no maximum" condition because $q$ itself is the largest element. If we allowed that set as well as $q_R$, then the same rational number would be represented in two different ways. The strict inequality is not cosmetic; it removes that ambiguity.

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 $A'$.

Theorem

Cut inclusion is a total order and preserves rational order

Inclusion is reflexive, antisymmetric, and transitive. If cuts $A,D$ satisfy $A\nsubseteq D$, choose $a\in A\setminus D$. Every $d\in D$ has $d\lt a$: otherwise downward closure of $D$ would force $a\in D$. Thus $D\subseteq A$, so cuts are totally comparable.

For rationals $p,q$,

$$p\le q\quad\Longleftrightarrow\quad p_R\subseteq q_R.$$

The forward implication follows from transitivity. For the converse, if $q\lt p$, the midpoint $(p+q)/2$ belongs to $p_R$ but not to $q_R$, contradicting the inclusion. In particular, equal rational cuts have equal rational boundaries, so the embedding is injective.

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.

Addition produces a cut and has the expected laws

Choose $a\in A$, $a'\in A'$ to witness nonemptiness. Choose $b\notin A$, $b'\notin A'$; every sum $a+a'$ is strictly below $b+b'$, so the sum is proper. If $x=a+a'$ and $y\lt x$, then $y-a'\lt a$ lies in $A$, hence $y=(y-a')+a'$ lies in the sum. To improve $x$, choose $c\in A$ with $a\lt c$; then $x\lt c+a'$ lies in the sum. These verify all four lower-cut conditions.

Associativity follows because both $(A+A')+D$ and $A+(A'+D)$ consist exactly of the rational sums $a+a'+d$; commutativity follows likewise. Moreover $A+0_R=A$: adding a negative rational moves downward inside $A$; conversely, for $x\in A$ choose $a\in A$ with $x\lt a$, and write $x=a+(x-a)$.

Worked example

Why $1_R+1_R=2_R$

Write

$$1_R=\{x\in Q\mid x\lt 1\}.$$

If $a\lt 1$ and $a'\lt 1$, then $a+a'\lt 2$, so every element of $1_R+1_R$ lies in $2_R$.

Conversely, if $x\lt 2$, then $x/2\lt 1$, and

$$x=\frac{x}{2}+\frac{x}{2}.$$

So every rational in $2_R$ lies in $1_R+1_R$. Hence

$$1_R+1_R=2_R.$$

This example shows that the cut definition really extends the ordinary rational operations rather than inventing new arithmetic.

Additive inverses require a strict gap

Define

$$-A=\{q\in Q:\exists r\in Q\setminus A,\ q\lt -r\}.$$

The strict inequality matters. Simply reflecting the complement would put $-p$ into the proposed inverse of $p_R$, creating a greatest element.

Existence and uniqueness of the additive inverse

The displayed set is nonempty: if $r\notin A$, then $-r-1$ belongs to it. It is proper: for $a\in A$, every such $q$ satisfies $q\lt -r\lt -a$, so $-a$ is excluded. Downward closure is immediate, and the midpoint of $q$ and its witness $-r$ gives a larger member. Thus $-A$ is a cut.

We need a rational gap lemma. For any rational $\delta\gt 0$, choose $a_0\in A$ and $b\notin A$. The Archimedean property of $Q$ gives a positive integer $n$ with $a_0+n\delta\gt b$. Among the positive integers $k$ for which $a_0+k\delta\notin A$, take the least. Then $a=a_0+(k-1)\delta\in A$ and $r=a+\delta\notin A$.

If $a\in A$ and $q\in-A$ with witness $r\notin A$, then $a\lt r$ and $a+q\lt 0$, proving $A+(-A)\subseteq0_R$. Conversely, for any rational $x\lt 0$, apply the lemma with $\delta=-x/2$. It gives $a\in A$ and $r=a+\delta\notin A$; then $q=x-a\lt -r$, so $q\in-A$ and $x=a+q$. Therefore $A+(-A)=0_R$.

If another cut $D$ satisfies $A+D=0_R$, associativity and the zero identity give $D=D+(A+(-A))=(D+A)+(-A)=-A$. This proves uniqueness. For $A=p_R$, the definition gives $-A=(-p)_R$, with witness $r=p$ for every $q\lt -p$. If $A$ is irrational, its complement has no least member; therefore the strict gap definition also equals $Q\setminus\{-a:a\in A\}$ in that case.

Multiplication on nonnegative cuts

Addition and additive inverses are now established. For multiplication, first define the product for nonnegative cuts, where multiplying rational members respects the direction of the boundaries; sign rules will then handle the other cases. For $A,B\ge0_R$, define $A\cdot B=P(A,B)$ by

$$P(A,B)=\{ab:a\in A,\ b\in B,\ a\ge0,\ b\ge0\}\cup\{q\in Q:q<0\}.$$

The negative rationals are included deliberately. They provide the whole negative part of the new lower set, while the products of nonnegative members locate its nonnegative part.

If $A=0_R$ or $B=0_R$, the corresponding nonnegative member set is empty, so $P(A,B)=0_R$. Suppose now that $A,B>0_R$. Each cut contains a positive rational: if a nonnegative member is available, use the no-greatest-member property; if not, the cut would be $0_R$. Choose positive $a\in A$ and $b\in B$. This proves nonemptiness of $P(A,B)$.

Choose positive rationals $r\notin A$ and $s\notin B$. Every nonnegative $a\in A,b\in B$ satisfies $a<r$ and $b<s$, hence $ab<rs$. Thus $rs$ is not in $P(A,B)$, proving properness. Downward closure is immediate for negative $y$. If $0\le y<x=ab$ with $a,b$ nonnegative, then $a,b>0$ and $y=a(y/a)$ with $0\le y/a<b$; downward closure of $B$ puts $y/a$ in $B$, so $y\in P(A,B)$.

Finally, a negative member can be increased slightly while staying negative. A zero member can be increased to a positive product because both cuts are positive. If $x=ab>0$, choose $a'>a$ in $A$; then $a'b>x$, and $a'b$ is still in $P(A,B)$. These cases prove that $P(A,B)$ has no greatest member. Therefore the nonnegative product is itself a Dedekind cut.

Product laws and the rational restriction

Commutativity follows by interchanging the two rational factors. For the identity, let $A\ge0_R$. If $0\le b<1$, then $ab\le a$ for every nonnegative $a\in A$, so downward closure gives $A\cdot1_R\subseteq A$. Conversely, if $0\le x\in A$, choose $a\in A$ with $x<a$; then $a>0$ and $x=a(x/a)$ with $0\le x/a<1$, proving $x\in A\cdot1_R$. Negative rationals are present on both sides. The zero case is already covered by $0_R\cdot B=0_R$.

For positive rationals $p,q$, every nonnegative product of members of $p_R$ and $q_R$ is below $pq$, so $p_Rq_R\subseteq(pq)_R$. Conversely, if $0\le x<pq$, choose rational $a$ with

$$\max(0,x/q)<a<p$$

and put $b=x/a$. Then $0\le b<q$, so $a\in p_R$, $b\in q_R$, and $x=ab$. The zero cases and sign rules give $p_Rq_R=(pq)_R$ for all rational signs.

For positive cuts $A,B,C$, a nonnegative member of $(A B) C$ has the form $(ab)c$, where $a,b,c$ are nonnegative members of the three cuts. The same description, regrouped as $a(bc)$, gives a member of $A(B C)$, and the reverse inclusion is identical. Negative rationals occur in both products. If one cut is zero, both bracketings are zero. This proves associativity directly from rational associativity and the cut definitions; no real-number associativity has been assumed.

Distributivity: the shared-factor argument

We first record a representation fact. If $U,V>0_R$ and $x\ge0$ lies in $U+V$, then $x=u+v$ for nonnegative $u\in U,v\in V$. Indeed, start with any representation $x=u_0+v_0$. If one summand is negative, replace it by $0$ and replace the other by $x$; the replacement is smaller than the old positive summand and therefore remains in its cut. This lemma is intentionally restricted to strictly positive cuts: $0_R$ contains no nonnegative rational, so it would be false in a zero case, for example with $U=0_R,V=1_R,x=1/2$.

For positive cuts $A,B,C$, take a nonnegative $z\in A(B+C)$. Write $z=a(b+c)=ab+ac$ with nonnegative $a\in A,b\in B,c\in C$. The representation fact puts $z$ in $AB+AC$. Conversely, apply the representation fact to the sum $AB+AC$ and write its nonnegative member as $z=u+v$, with nonnegative $u\in AB,v\in AC$. By the product definition, write $u=ab$ and $v=a'c$ with nonnegative factors in the corresponding cuts. Choose $\alpha\in A$ strictly above $a$ and $a'$; it is positive. Then

$$z=\alpha\left[\left(\frac a\alpha\right)b+ \left(\frac {a'}\alpha\right)c\right].$$

Both coefficients lie in $[0,1)$, so the bracket belongs to $B+C$ and hence $z\in A(B+C)$. Negative rationals occur on both sides. If $A=0_R$, then $0_R(B+C)=0_R=0_RB+0_RC$; the cases $B=0_R$ or $C=0_R$ are the same direct calculation using the zero product and addition laws. This proves distributivity for nonnegative cuts without limits or an unconstructed real root.

Signs and order compatibility

Extend the product by

$$A B=-((-A)B)\quad(A<0_R\le B),\qquad A B=(-A)(-B)\quad(A<0_R,\ B<0_R),$$

For the other mixed-sign order, if $A>0_R>B$, set $AB=BA=-(A(-B))$. Use the zero product when either factor is $0_R$. A product of two nonzero positive cuts is positive, so these rules give the expected sign of every product. Associativity for arbitrary signs reduces to the already proved positive-magnitude associativity together with additive inversion.

The mixed-sign distributive step is also explicit. For nonnegative $A,B,C$ with $B\ge C$, the cut difference $B-C$ is nonnegative and $B=(B-C)+C$. Nonnegative distributivity gives

$$A(B-C)=AB-AC.$$

If $B<C$, interchange them and negate the resulting equality. This identity handles a mixed-sign second or third summand; a negative first factor is then handled by the sign rule and additive inversion. Thus distributivity holds for all signs.

The order is compatible with translation because $A\subseteq B$ implies $A+D\subseteq B+D$; the converse follows by adding $-D$ to both sides. Products of nonnegative cuts are nonnegative by definition. If $B>A$ and $D>0_R$, then $B-A>0_R$, so $(B-A)D>0_R$; distributivity gives $BD-AD>0_R$, hence $AD<BD$. This proves the ordered-field compatibility of the cut operations.

General multiplicative inverses

Let $A>0_R$. Choose a fixed positive rational $\alpha\in A$. Every rational $r\notin A$ then satisfies $r>\alpha>0$. Define

$$I=\{q\in Q:\exists r\notin A\text{ with }q<1/r\}.$$

This definition works for rational and irrational cuts alike. It is nonempty, and contains a positive element such as $1/(2r)$ for any chosen outside $r$. Every member is below $1/\alpha$, so $I$ is proper; downward closure is immediate, and the midpoint between $q$ and its witness $1/r$ gives a larger member. Thus $I$ is a cut.

If $a\in A,b\in I$ are nonnegative and $b<1/r$ witnesses membership in $I$, then $a<r$, so $ab<1$. Hence $AI\subseteq1_R$. For the reverse inclusion, let $0\le z<1$ be rational and choose rational

$$0<\delta<\alpha(1-z).$$

The rational stepping lemma gives $a\in A$ and $r=a+\delta\notin A$, with $a\ge\alpha$. Then $a/r=1-\delta/r>z$. Put $b=z/a$. We have $0\le b<1/r$, so $b\in I$ and $z=ab\in AI$. Negative rationals are automatic, therefore $AI=1_R$.

For $A<0_R$, define $A^{-1}=-((-A)^{-1})$; zero is excluded because no product with $0_R$ can equal $1_R$. For positive rational $p$, the same definition reduces to $(1/p)_R$, and signs give the negative rational case. If $AX=AY=1_R$, then associativity and the identity yield

$$X=X(AY)=(XA)Y=Y,$$

so the inverse is unique.

Completeness in the cut model

The cut construction makes the least-upper-bound property visible. Let $C$ be a nonempty family of cuts bounded above by a cut $U$. Define

$$L=\bigcup_{A\in C}A.$$

The set $L$ is nonempty because $C$ is nonempty. It is proper because every $A\in C$ lies below $U$, so $L\subseteq U$, and $U\ne Q$. Downward closure passes to a union: if $x\in L$, then $x\in A$ for some $A$, and every rational below $x$ is in $A$ and hence in $L$. If $x\in L$, choose $y\in A$ with $x\lt y$ using the no-greatest-element axiom for $A$; then $y\in L$. Therefore $L$ is a cut. It is an upper bound of $C$, and any other upper bound contains each $A$, hence contains their union. Thus $L$ is the supremum of $C$.

Theorem

The union construction supplies a supremum

For nonempty bounded families of Dedekind cuts, the union of the left sets is the least upper bound. The proof uses exactly nonemptiness, properness, downward closure, and no greatest element; none can be silently dropped.

General nonnegative square roots

Now that the ordered-field laws and union completeness have been established, we can use the supremum property inside the complete ordered field $R$. Let $A\ge0_R$ and define

$$S=\{x\in R:x\ge0_R\text{ and }x^2\le A\}.$$

The set is nonempty because $0_R\in S$. It is bounded above by $\max(1_R,A)$: if $x$ is larger than both $1_R$ and $A$, then $x^2>x>A$, using positivity and multiplication of positive elements. Let $s=\sup S$, so $s\ge0_R$.

The perturbations $h$ below are elements of the complete field $R$; no rationality of this $h$ is being claimed. The earlier rational gap lemma used for cut construction is a separate statement with rational $\delta$.

Suppose first that $s^2<A$. Choose

$$h=\frac12\min\left(1_R,\frac{A-s^2}{2s+1_R}\right)>0_R.$$

Then $(s+h)^2=s^2+h(2s+h)<s^2+h(2s+1_R)<A$, so $s+h\in S$, contradicting that $s$ is an upper bound. Suppose instead that $s^2>A$. Then $s>0_R$. Choose

$$h=\frac12\min\left(s,\frac{s^2-A}{2s}\right)>0_R, \qquad r=s-h>0_R.$$

The estimate $r^2=s^2-2sh+h^2>A$ holds. Every $x\in S$ satisfies $x<r$, because $x\ge r>0_R$ would imply $x^2\ge r^2>A$. Thus $r<s$ is also an upper bound for $S$, contradicting the leastness of $s$. Therefore $s^2=A$.

If $0_R\le s<t$, then $t^2-s^2=(t-s)(t+s)>0_R$, so two nonnegative square roots cannot differ. We have proved existence and uniqueness of the nonnegative square root of every nonnegative cut, including the boundary case $A=0_R$, whose root is $0_R$.

The multiplicative Archimedean corollary

For positive cuts $A,B$, the additive Archimedean argument applied to $B/A$ gives an integer $N$ with $N_R>B/A$. Multiplying this strict inequality by the positive cut $A$ preserves its direction, and therefore

$$N_R\,A>B.$$

This corollary uses the inverse and order-compatibility laws already proved. The earlier rational gap lemma depended only on the arithmetic of $Q$.

Worked classifications of cuts

A formula defining a subset of $Q$ must pass all the cut axioms. Odd powers give a useful family of examples; even powers and arbitrary set operations show why each axiom needs its own check.

Odd powers give lower cuts

For an odd positive integer $k$ and positive rational $t$, the set

$$D_{k,t}=\{x\in Q:x^k<t\}$$

is a cut. Odd powers are strictly increasing on $Q$: for nonnegative $x<y$, factor $y^k-x^k=(y-x)(y^{k-1}+y^{k-2}x+\cdots+x^{k-1})>0$; for negative $x<y$, apply the same positive argument to $-y<-x$ and use that $k$ is odd; if $x<0<y$, then $x^k<0<y^k$. If $x\in D_{k,t}$, choose a rational $h\in Q$ with

$$0<h<\min\left(1,\frac{t-x^k}{k(|x|+1)^{k-1}}\right).$$

The factorisation of $(x+h)^k-x^k$ has $k$ terms, each bounded in absolute value by $(|x|+1)^{k-1}$, so $(x+h)^k<t$; this proves there is no greatest member. Downward closure follows from monotonicity: if $y<x$, then $y^k<x^k<t$. An integer $M>\max(1,t)$ supplies nonemptiness and properness.

Worked example

Polynomial conditions: locate the failed axiom

Both $\{x\in Q:x^3<17\}$ and $\{x\in Q:x^3<5\}$ are cuts by the odd-power argument. In contrast, $\{x\in Q:x^4<100\}$ is not a cut: $0$ belongs but $-4$ does not, so downward closure fails. The set $\{x\in Q:x^2>-3\}=Q$ fails properness.

The set $\{x\in Q:x^5\le2\}$ is a cut despite its weak inequality. No rational fifth power is $2$: in a lowest terms fraction $a/b$, the equation $a^5=2b^5$ makes both $a$ and $b$ even. The weak inequality therefore defines the same cut as $x^5<2$.

Worked example

Set operations that preserve or destroy cuts

Let $A$ and $B$ be cuts. The arbitrary product set $\{ab:a\in A,b\in B\}$ is not necessarily a cut: for $A=B=0_R$ it is $Q_{>0}$, which contains $1$ but not $0$, violating downward closure. The difference set $\{a-b:a\in A,b\in B\}$ is always all of $Q$: given $q$, choose $a_0\in A,b_0\in B$ and a rational $b<\min(b_0,a_0-q)$; then $b\in B$, $a=q+b\in A$, and $q=a-b$. It is therefore not proper. The set $Q\setminus\{-a:a\in A\}$ is not necessarily a cut: for $A=0_R$ it is $Q_{\le0}$, whose greatest element is $0$. For an irrational cut it agrees with the strict-gap inverse; the zero cut shows why the formula does not work uniformly for all cuts. The intersection $A\cap B$ is always a cut because total comparability makes it the smaller of the two cuts; the no-greatest witness is the smaller of two members chosen above a given element.

Worked example

The zero cut and its additive inverse

The embedded zero is $0_R=\{q\in Q:q\lt 0\}$. The cut representing its additive inverse is itself, because adding two negative rational left parts produces a left part below zero and every rational below zero can be split as $q/2+q/2$. This illustrates why the inverse is an equality of cuts, not merely an informal reflection of a picture.

Worked example

Adding embedded rationals

For $p_R$ and $q_R$, the cut addition is $p_R+q_R=\{a+b:a\lt p,b\lt q\}$. Every such sum is below $p+q$; conversely, if $x\lt p+q$, put $a=p-(p+q-x)/2\lt p$ and $b=q-(p+q-x)/2\lt q$, so $a+b=x$. Hence the sum equals $(p+q)_R$, so the embedding respects addition.

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 boundaries

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

Common mistake

The pair version and the subset version are not two different theories

They describe the same object from two angles. The pair $(A,B)$ records both sides of the boundary explicitly, while the subset form keeps only the left side and recovers the right side as $Q\setminus A$.

Common mistake

Order on cuts is not a comparison between every element of two sets

The statement $A\le A'$ means $A\subseteq A'$. It does not mean that every element of $A$ is less than every element of $A'$. The two left sets usually overlap heavily, especially when one boundary lies to the left of the other.

Quick checks

Checkpoint

What is the cut $0_R$?

Write it directly from the definition $q_R=\{x\in Q\mid x\lt q\}$.

Solution · Answer

$$0_R=\{x\in Q\mid x\lt 0\},$$

so it is the set of all negative rational numbers.

Checkpoint

If $A\subseteq A'$, which boundary lies to the left?

Answer in terms of the cut order.

Solution · Answer

If $A\subseteq A'$, then $A\le A'$. The boundary represented by $A$ lies at or to the left of the boundary represented by $A'$.

Exercises

Checkpoint

Show that $q_R=\{x\in Q\mid x\lt q\}$ is a Dedekind cut for every rational number $q$.

Check the three conditions in the subset definition.

Solution · Guided solution

The set $q_R$ is nonempty because $q-1\lt q$, so $q-1\in q_R$. It is not all of $Q$ because $q\notin q_R$.

It is downward closed: if $x\lt q$ and $y\lt x$, then certainly $y\lt q$, so $y\in q_R$.

It has no maximum: if $x\lt q$, then

$$\frac{x+q}{2}$$

is rational and satisfies $x\lt (x+q)/2\lt q$, so $(x+q)/2\in q_R$. Therefore every element of $q_R$ can be improved by a larger element still in $q_R$.

Checkpoint

Does the intersection of every nonempty family of Dedekind cuts have to be a cut? Test the family $A_n=(1/n)_R$, where $n\ge1$.

Compare finite intersections with the intersection of this decreasing family.

Solution · Model solution

Every nonpositive rational belongs to every $A_n$. If $q>0$, choose an integer $n>1/q$; then $1/n<q$, so $q\notin A_n$. Hence

$$\bigcap_{n\ge1}A_n=\{q\in Q:q\le0\}.$$

This set has greatest element $0$ and is not a cut. Finite intersections are cuts because a finite family has a smallest member under inclusion. That argument cannot be extended to an arbitrary family without checking that such a smallest member exists.

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.