# Completeness, the Archimedean Property, and other miscellany

## Completeness of The Reals

As we discussed in class on Friday (9.9.2016), the Completeness Axiom for $\mathbb{R}$ figures largely in the all-important fact that

Every Cauchy sequence (of real numbers) converges.

We can prove this in a sort of step-by-step procedure, building up relevant theorems that “link together.”  In particular, we have

• Completeness Axiom.  Every non-empty subset $S \subset \mathbb{R}$ that is bounded above has a supremum.
• Theorem A.  Every Cauchy sequence (of real numbers) is bounded.
• Theorem B. (Bolzano-Weierstrass)  Every infinite, bounded subset $S \subset \mathbb{R}$ has at least one accumulation point.
• Theorem C.  If a sequence of real numbers $\left\{a_n\right\}$ is Cauchy and has an accumulation point, then it has only one accumulation point and $\left \{a_n\right\}$ converges to this accumulation point.

One of your previous homework questions asks you to prove Theorem A, and another question asks you to analyze the proof of the Bolzano-Weierstrass Theorem to determine exactly how

Completeness Axiom $\Rightarrow$ Theorem B

Note that Theorem C is stated in our book as Theorem 1.7: Every Cauchy sequence of real numbers converges.

In this blog post I want to summarize some of our other important theorems about sequences (and/or real numbers).  In particular we have the following chain/table of implications (that I’ve drawn up in class multiple times):

$\begin{array}{cccc} \text{Completeness of } \mathbb{R} &\Rightarrow \text{ Nested Interval Property } & \Rightarrow \text{Bolzano Weierstrass } & \Rightarrow \text{ Cauchy Criterion} \\ \Downarrow & & \Downarrow & \\ \text{ Monotone Convergence Theorem } & & \text{Bolzano Weierstrass for seq's} & \end{array}$

All of these implications can be reversed, but the work required to do so can vary.  In addition, more implications can be introduced between these statements.  For example it is not very difficult to argue that

$\displaystyle \text{Monotone Convergence Thm} \Rightarrow \text{Nested Interval Property}$.

Here is a proof sketch: Suppose $I_n = \left[a_n, b_n\right]$ is a sequence of nested, non-empty closed intervals of real numbers.  We can note that, by these assumptions, the sequence of left-endpoints $\{a_n\}$ forms a bounded, monotone (increasing) sequence.  (The boundedness comes from observing that $b_1 \geq a_n \geq a_1$ for all $n \in \mathbb{N}$.)  This sequence therefore converges to some number, call it $a \in \mathbb{R}$, and one can prove by contradiction that $a \in \bigcap I_n$.

As I mentioned in class, the Cauchy Criterion — i.e. that every Cauchy sequence converges — provides mathematicians a way to do analysis on arbitrary metric spaces $(X, d)$.  If one has Cauchy sequences in $X$ that do not converge, then a larger, better-behaved metric space may be used instead, one called the metric completion of $X$.

It is typically denoted by $\overline{X}$ and it carries with it an inherited metric (which we continue to call $d$).  As a set $\overline{X}$ has a rather strange definition:

$\displaystyle \overline{X} = \left\{ \text{equivalence classes of Cauchy sequences of points } a_n \in X \right \}$

where $\left\{a_n\right\} \sim \left\{b_n\right\}$ means that

$\displaystyle \lim_{n\to\infty} d\left(a_n, b_n\right) = 0.$

Here is a link to the wiki page on completing a metric space, but the basic idea to keep in mind is that whereas $X$ may feature many “holes” or “gaps” (points where some Cauchy sequences are piling up but failing to converge), the bigger space $\overline{X}$ “fills them in” by simply declaring the missing points to be there.  Quite literally, the completed metric space gathers up every possible sequence that “wanted” to converge to one of these gaps and then declares the collection (or equivalence class) of these sequences to be the missing point.  This is one way to construct the ordered fields of real numbers from the ordered field of rationals, i.e.

$\displaystyle \mathbb{R} = \overline{\mathbb{Q}}$.

Making this precise requires a fair amount of work, including redefining notions of convergence by including the adjective “rational” in front of every $\varepsilon$ symbol.  Moreover, one has to define addition, subtraction, multiplication, and division of equivalence classes of Cauchy sequences of rationals, as well as a notion of “greater than or equal to.”  Lastly, one needs to confirm all 12 axioms for this construction!

Back to our chain of equivalent completeness statements, though…

Exactly how difficult or tricky it is to prove any of the statements in the above collection starting from any of the others is an interesting question.  Some are rather immediate and cute, but others highlight some easy-to-overlook assumptions about the real numbers that even we have not yet discussed. Here are some relevant forum discussions: discussion 1 and discussion 2 (please do not use or reference these when taking an exam).

In the second link the idea of using the Nested Interval Property to prove the Axiom of Completeness is discussed, and a helpful mathematician commented that “Actually Nested Interval Theorem implies Completeness Axiom, only if you assume that Archimedean Property holds…

Completeness Axiom implies both Archimedean Property and Nested Interval Theorem!!!! Though in Non-Archimedean Ordered Fields (like the field of Rational Polynomials) the Nested Interval Property ALONE does not entails Completeness Axiom

So in a general setting, Completeness Axiom is equivalent to Nested Interval Property+Archimedean Property.”

As a class we have not read about or used this so-called “Archimedean Property” (its not even mentioned in our textbook), so what is it?  How is it useful?  This leads naturally to…

## The Archimedean Property

The Archimedean Property claims nothing all that fancy-sounding.  In fact, its downright boring:

The Archimedean Property: The set of natural numbers $\mathbb{N} \subset \mathbb{R}$ is unbounded (in particular, it is not bounded above).

Said differently: for every $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ so that $n \geq x$.

How on earth can something so obvious be so important?  Moreover, if it is so important, why did we omit it from our class discussions? Let’s tackle this second question first.

Once we decided to axiomatize the set $\mathbb{R}$ (as opposed to constructing it rigorously from, say, the rationals), we asserted various properties to be true.  We did this based on our intuition of what $\mathbb{R}$ is or ought to be — a solid line of points, arranged in an order.  The natural numbers $\mathbb{N}$ are easily seen or assumed to be a subset of this line, one that clearly goes on without any upper bounds.  Since we began this class with this intuitive approach, our textbook and I decided to keep the naturals intuitive, too — at least for a while.

Once you axiomatize the set $\mathbb{R}$ so that other familiar sets set inside of it in obvious ways, the need to carefully state how to see $\mathbb{N}$ as a subset vanishes.  However, if you formally construct the set $\mathbb{R}$ as, say, a set of equivalence classes of Cauchy sequences or, as we will do in January, as a set of Dedekind cuts, then it is less obvious that $\mathbb{N}$ is a bounded set.  “Perhaps one of these weird, artificial “real number” guys we built some how winds up being larger than every natural number?” a cautious student may suggest.  And given how abstract these real numbers are / will be, this is something we’d need to actually check!

The first question, though  — how on earth can this property be so important? — is not too difficult to answer, either.  Every time we write down a convergence argument (or even simply state the definition of convergence) for a sequence of real numbers, we are using the Archimeddean Property.

For instance, when you claim that the sequence $1/n$ converges to the real number $0$, you are obligated to find or describe an index $N \in \mathbb{N}$ so that

$\displaystyle n \geq N \Rightarrow d\left(\frac{1}{n}, 0\right) = \frac{1}{n} < \varepsilon$.

As we’ve been practicing, the choice of $n \geq N$ will depend on the positive number $\varepsilon$.  In the instance of this famous $1/n$ sequence, we need to choose something like $N > 1/\varepsilon$, and here comes the potential problem: how do we even know there is a natural number $N$ that is larger than the real number $1/\varepsilon$?

Every time we’ve said “$choose N \geq$ blah blah blah” we’ve been using this Archimedean Property.  Every time we’ve used such an idea in a definition — be it for a convergent sequence, a Cauchy sequence, etc. — we’ve needed it.

Fortunately, the property is as easy to prove is it to forget to mention!  Indeed, it follows rather instantly from our Completeness Axiom.  Here is a sketch of a proof by contradiction.

Suppose $\mathbb{N} \subset \mathbb{R}$ is bounded above.  By the completeness axiom there exists $x \in \mathbb{R}$ that serves as the supremum for $\mathbb{N}$.  This means that, as an upper bound,

$n \leq x \text{ for all } n \in \mathbb{N}$.

Because $n \in \mathbb{N} \Rightarrow n+1 \in \mathbb{N}$, we also have that

$n + 1 \leq x \text{ for all } n \in \mathbb{N} \, \iff \, n \leq x-1 \text{ for all } n \in \mathbb{N}$

which implies that $x-1$ is an upper bound, too.  This, of course, contradicts the definition of supremum.

Going back to the aforementioned forum comment, there are, evidently, ordered fields $\left(\mathbb{F}, +, \cdot, \leq\right)$ that satisfy a Nested Interval Property but do not satisfy the Completeness Axiom.  The problem is that these strange fields also do not satisfy the Archimedean Property, and so notions of convergence are “thrown off.”  An interesting discussion of these sorts of constructions and counterexamples can be found in this lovely paper by James Propp.

## The Axiom of Choice

Since we’re on the subject of “secret things we’ve been saying without actually saying,” now seems like an ideal time to mention this Axiom of Choice thingy that comes up every so often.

There are many ways to formulate this Axiom, but the basic English version goes something like this:

Axiom of Choice: Given an infinite family of sets, $S_i$ (where $i \in \mathcal{I}$ is an element of some infinite index set), one can form a new set $S$ by choosing exactly one element from each $S_i$.

Said slightly differently:

Axiom of Choice: Let $S_i$ be an infinite family of sets.  Then the infinite Cartesian product of all these sets is non-empty, e.g.

$\displaystyle \prod_{i\in\mathcal{I}} S_i \neq \varnothing.$

Some mathematicians distinguish between the countable axiom of choice (wherein $i \in \mathcal{I} = \mathbb{N}$) and a more general, potentially uncountable axiom (wherein $\left|\mathcal{I}\right| \geq \left|\mathbb{N}\right|$).

Most mathematicians accept this axiom, but its “controversial” status results from some of the surprising Theorems one can prove with it.  There is a lot of history here, and we have free access to a pretty good book that details much of it.  Don’t believe me?  Then click right here.

We’ve used this axiom when constructing sequences and subsequences.  Most recently, we used this axiom when we proved that

$\displaystyle \lim_{x\to x_0} f(x) \text{ exists } \iff \text{ for every sequence } x_n \to x_0 \, \text{ the sequence } f(x_n) \text{ converges }.$

Technical note: In the above result we also requested that $x_n \in D\smallsetminus\{x_0\}$ for all $n$.

Using this axiom — either implicitly or explicitly — is absolutely acceptable mathematical practice, so don’t worry about it too much.  (Although, it can be interesting to identify where it is secretly used and to wander how or if a proof might work without it.)