# Gracious Living

The Construction of the Reals: Metric Completion and Dedekind Cuts
March 6, 2011, 21:31
Filed under: Algebra, Math, Set Theory, Topology | Tags: , , , ,

It looks like I’m getting views now, which is surprising.  I’ve been pretty busy with schoolwork, but I really want to get this blog up to speed, particularly because I’d like to start discussing things as I’m learning about them.  I’d also like to make more non-mathematical posts, but maybe these are best left to a separate blog?  Thoughts?

Our first example of a field was the field of rationals, $\mathbb{Q}$.  Recall that this was the field of fractions of the integers, which were in turn the free abelian group on one generator with their natural multiplication.  But now it appears that we’re stuck.  While we intuitively know what $\mathbb{R}$ should be — it’s a line, for crying out loud — there seems to be no algebraic way of “deriving” it from $\mathbb{Q}$.  A first guess might be to add in solutions of polynomials, like $\sqrt{2}$ as the solution of $f(x)=x^2-2$, but not only does this include some complex numbers, it also misses some real numbers like $e$ and $\pi$.  (We call such numbers — those that aren’t solutions of polynomials with rational coefficients — transcendental.  It’s actually quite difficult to prove that transcendental numbers even exist.)

Instead, we turn to topology.  Below, I give two ways of canonically defining $\mathbb{R}$, one using the metric properties of $\mathbb{Q}$, one using its order properties.  I found this really interesting when I first saw it, but I can’t see it interesting everyone, so be warned if you’re not a fan of set theory or canonical constructions.  One of the topological techniques we’ll see will be useful later, but at that point it’ll be treated in its own right.

First, note that $\mathbb{Q}$ is naturally ordered.  This is easiest to see starting in $\mathbb{Z}$, which can be generated by either $1$ or $-1$.  We call elements that are the sum of copies of $1$ “positive,” and those that are the sum of copies of $-1$ “negative.”  Since $\mathbb{Z}$ is free, only one of the two properties can hold, and then we can define $a>b$ if $a-b$ is positive.  Likewise, we can define an order among fractions in the expected way: the quotient of two numbers of the same sign is positive, and the quotient of two numbers with opposite signs is negative.  This makes the map $a\mapsto a/1$ of $\mathbb{Z}$ into $\mathbb{Q}$ order-preserving.  In both cases, the order is also total, since the difference between two numbers is either positive, negative, or zero.

Starting with a number $a\in \mathbb{Q}$, we see that $a$ cuts the set into two parts: the set $S_a=\{x\in\mathbb{Q}:x\le a\}$, and its complement.  We will call this partition (or equivalently, just $S_a$ itself) the cut at $a$.  Cuts have some very nice properties, in terms of order.  If $a\ne b$, then $S_a\ne S_b$, and in fact $S_a\subseteq S_b$ iff $a\le b$.  Thus, the set of cuts is totally ordered as well, and in order-preserving bijection with $\mathbb{Q}$.  The set $S_a$ is a lower set for any $a$: if it contains $b$, then it contains $c$ for all $c\le b$.  Its complement, meanwhile, is an upper set, with a definition I’m sure you can fill in.

In the late nineteenth century, a number theorist and set theorist named Richard Dedekind had an important realization.  “The sets $S_a$,” cried he from a feverish sweat, earning angry looks from the other theatregoers, “are not the only ones of their kind!”  And indeed, he was right.  There are many other ways to partition $\mathbb{Q}$ into an upper set and a lower set.  An a posteriori example is a set we can call “$S_{\sqrt{2}}$“, consisting of all the rational numbers below $\sqrt{2}$ in the ordering on the reals.  If we don’t know what the reals are, then we can’t really talk about this set, but we can still build it by noticing that the sequence of rationals $1,1.4,1.41,1.414,\dotsc$ has no limit in $\mathbb{Q}$, and the set of rational numbers below at least one number in this sequence is a lower set.

So we just define $\mathbb{R}$ to be the set of cuts (which I’ll hereafter write as lower sets) of $\mathbb{Q}$.  As each $a\in\mathbb{Q}$ gives a unique $S_a$, there is an embedding of $\mathbb{Q}$ into $\mathbb{R}$.  Given two lower sets $X$ and $Y$, we define their sum to be $X+Y=\{x+y:x\in X,y\in Y\}$, giving a group structure with $-X=\{-x:x\not\in X\}$ and $0=S_0$.  It is clear that cuts are totally ordered, so we can define “positive” and “negative,” and then define, for positive $X,Y$, the product $XY$ to be the complement of $\{xy:x\not\in X,y\not\in Y\}$.  This is extended to the general case by defining $(-X)Y=-(XY)=X(-Y)$.  As you can check, this makes $\mathbb{R}$ into a field, and the field structure is compatible with the ordering in that $X>Y$ implies $X+Z>Y+Z$ for all $Z$, and $XZ>YZ$ for all $Z>0$.  We’ll call this an ordered field.  On top of that, when we restrict to the embedded $\mathbb{Q}$, we recover its ordering and field structure.  Unlike $\mathbb{Q}$, $\mathbb{R}$ has the least upper bound property — any subset of it with an upper bound has a least upper bound.

This method of adjoining cuts can be done to any ordered set, even a partially ordered one, and will return a set with the least upper bound property.  But unfortunately, it’s difficult to see where the field structure comes from and how it’s related to that of $\mathbb{Q}$.  To answer this question, we turn to a more topological point of view.

$\mathbb{Q}$ is a metric space under the metric $d(x,y)=|x-y|$.  (All right, technically, we don’t know what a metric space is yet because we defined a metric as a map to $\mathbb{R}$.  But note that in this case, $d(x,y)\in\mathbb{Q}$, so we can consider $\mathbb{Q}$ a “$\mathbb{Q}$-metric space.”)  We define a Cauchy sequence $(x_i)$ to be one where for all $\epsilon>0$, there is a positive integer $N$ for which $|x_n-x_m|<\epsilon$ for every $n,m>N$.

This looks pretty similar to the definition of convergence… but we make no reference to any limit.  Instead, the points in the sequence become arbitrarily close to each other.  We can characterize this as a sequence that “should converge,” and indeed, in $\mathbb{Q}$, sequences like $1.4,1.41,1.414,1.4142,\dotsc$ are examples of non-convergent Cauchy sequences.  A metric space is complete if all Cauchy sequences converge.  We will construct a space, called the completion of $\mathbb{Q}$, which is the smallest complete space containing the rationals.

We define an equivalence relation on the set of Cauchy sequences: $(x_i)$ is equivalent to $(y_i)$ if $(x_i-y_i)$ is a sequence converging to $0$.  (More properly, since we can’t subtract points in a general metric space, if $(d(x_i-y_i))$ is a sequence in $\mathbb{R}$ [or in this case, $\mathbb{Q}$] converging to $0$.)  We define the completion of $\mathbb{Q}$ to be the set of Cauchy sequences, mod this equivalence relation.  We can add and multiply Cauchy sequences by adding and multiplying their individual terms, and I leave it to you to show that this is well-defined at the level of equivalence classes (and produces a field structure).  Ordering is a little less obvious to come by.  If $(x_i-y_i)$ doesn’t converge to $0$, then for some positive $\epsilon$, $x_i-y_i>\epsilon$ for all $i$.  But the sequences are Cauchy, so after a certain point in each sequence, their terms are less than $\epsilon/2$ apart from each other.  Then all of one sequence’s terms are rational numbers greater than all of the other sequence’s terms, and we say that the equivalence class of that sequence is greater than the equivalence class of the other sequence.  Again, you should check that this is well-defined and works well with addition and multiplication, as described above.  We now give our space the order topology, and note that $\mathbb{Q}$ embeds in it as a field and topological space, via $x\mapsto$ the equivalence class of $(x,x,x,\dotsc)$.

To me, this feels more natural than the construction by Dedekind cuts, but really they’re just useful in different ways.  Completion works on a metric space and gives you the smallest metric space containing it in which all Cauchy sequences converge.  Dedekind cuts work on an ordered set and give you the smallest ordered set containing it in which every bounded subset has a least upper bound.  (It’s intuitively clear why the results are the “smallest”: if they were any smaller, they would generate a different topology or order on the embedded original set.  We can make this clearer by talking about “universal properties,” which I’ll probably do in the context of category theory.)  But how do we know that they do actually behave the same way on $\mathbb{Q}$?  How do we know there’s just one $\mathbb{R}$?

We answer this by describing one more property, ensuring that $\mathbb{R}$ is not “too large.”  We say that an ordered abelian group is Archimedean if for all $x,y>0$, there is an integer $n$ such that $nx>y$.  An example of a non-Archimedean group would be the “hyperreals,” which include $\mathbb{R}$ but also an element $\infty$ such that $nx<\infty$ for all real $x$.  Now we are ready for our theorem:

• Theorem.  There is a unique ordered Archimedean field with the least upper bound property.

Proof.  Let $F$ be a field satisfying the above.  In fields, $1\ne 0$.  By changing the direction of the ordering if necessary, we can have $1>0$.  Then $1+1>1,1+1+1>1+1$, and so on; this proves that $F$ contains an embedded $\mathbb{Z}$ (as opposed to finite fields like $\mathbb{Z}/2\mathbb{Z}$).  But since it’s a field, it must contain the field of fractions of $\mathbb{Z}$, which is $\mathbb{Q}$; since $1$ is positive, the sign on $F$ agrees with the sign on $\mathbb{Q}$, and thus the orderings are the same.

Since $\mathbb{R}$, by the Dedekind cut definition, is the smallest ordered field containing $\mathbb{Q}$ with the least upper bound property, it follows that $F\supseteq\mathbb{R}$.  If they aren’t equal, then there is some $x\in F-\mathbb{R}$, which we can take to be positive.  Consider the set $A$ of real numbers less than $x$.  This is nonempty (it includes $0$ and all the negatives).  If it is bounded above by a real number, then it has a real least upper bound $y$; but then $x implies $x\in A$, and $x>y$ implies $y\in A$, both of which are contradictions.  If it is not bounded above by a real number, then $x$ is greater than every real number.  But then $F$ is not Archimedean, since $n\cdot 1 for all $n$.  $\square$

As a corollary, the two definitions of $\mathbb{R}$ given are the same.  We can rewrite the least upper bounds of sets of rationals as limits of Cauchy sequences, so the completion of $\mathbb{Q}$ also has the least upper bound property.  Since $\mathbb{Q}$ is Archimedean, so is its completion: given two positive Cauchy sequences $(x_i,y_i)$, we can (without loss of generality) take their terms to be positive, and then find a nonzero lower bound $x\in\mathbb{Q}$ for the $x_i$ and an upper bound $y\in\mathbb{Q}$ for the $y_i$.  Then if $nx>y$, we of course have $nx_i>y_i$ for each $i$, so $(nx_i)=n(x_i)>(y_i)$ as desired.

So now we see that we were justified in talking about $\mathbb{R}$ nonrigorously and using its nice properties in different places.  Of course, the story doesn’t end here.  We’ll want to talk about the complex numbers sometime soon, hopefully in the context of algebraic closure and field extensions, and then as an aside the quaternions and octonions, hopefully in the context of Hopf fibrations.  I’d also like to start discussion what I’m learning in and outside of class, so I hope to speed through the point-set topology to get right to it.  See you next time!

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