Completion of rational numbers via Cauchy sequences

Can anyone recommend a good self-contained reference for completion of rationals to get reals using Cauchy sequences?

Solution 1:

Update 1

This should be the final edit. Lots of typos have been corrected and the presentation in the section on existence has been greatly improved. It should be very close to a form that can be used as a basis for a project for students.

Update 2

Even more typos corrected. Added example of non-archimedean ordered field. Fleshed out a few bits more.

Update 3

Fixed up the proof sketch in the last exercise. Also I keep finding these small annoying errors... I'm sorry that I keep up bumping this thread, I just hope that you don't mind too much.

Introduction

I'll do one better: I'll write you one (mostly because I actually have already written one - consider this sort of a Cliff's notes version with most of the details left as exercises). Recall that a first course in analysis typically asserts that there exists a field with an order and the least upper bound property, and you then develop the theory of sequences and series from there. Thus our goal should be to prove existence and uniqueness (up to some isomorphism notion) of such an object.

Uniqueness

Definition 1. An ordered field is a field, $K$ along with a total order $\leq$ on $K$ such that

1. $\forall x,y,z \in K : x \leq y \implies x+z \leq y+z$ (Transitivity).
2. $\forall x,y \in K : 0 \leq x \land 0 \leq y \implies 0 \leq xy$ (Positivity of product).

Moreover, define the absolute value function $\vert \cdot \vert: K \to K$ by $$\vert x \vert = \begin{cases} x & \text{if } 0 \leq x \\\\ -x & \text{otherwise.} \end{cases} $$

Exercise 1. There is an alternate definition of an ordered field: $K$ is said to be ordered if there is a subset $P$ of $K$ - called the positive elements of $K$ - such that

1. $K$ is the disjoint union of $P$, $-P$ and $\lbrace 0 \rbrace$.
2. $x,y \in P \implies x+y \in P$ and $xy \in P$.

Make formal the colloquial statement "The two definitions of ordered field are equivalent" and prove said equivalence.

Note that we could just as well require that the inequality in Definition 1 be strict since $x \lt y \lor x = y \iff x \leq y$ allows us to pass between strict and non-strict orderings as we please.

Definition 2. Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq y \implies f(x)\leq f(y)$) ringhomomorphism is called an embedding.

Note that any embedding must be injective since the kernel of a ringhomomorphism is and ideal and $0 \lt 1$.

Proposition 3. If $K$ is an ordered field, then $K$ has a subfield isomorphic to $\mathbb{Q}$.

Exercise 2. Prove Proposition 3 by showing that $\varphi: \mathbb{N} \to K$ defined by $$ \varphi(n) = \underbrace{1+\dotsb+1}_{n \text{ } 1\text{s}} $$ extends to an embedding of $\mathbb{Q}$ in $K$.

Definition 4. An ordered field $K$ is called an archimedean field if for all nonzero $x$ in $K$ there exists a natural number $n$ such that $n|x| \gt 1$. Or equivalently: there exists a natural number $n$ such that $n \gt |x|$.

Proposition 5. Let $K$ be an ordered field. If $K$ has the least upper bound property, then $K$ is archimedean.

Exercise 3. Show the contrapositive of Proposition 5 by considering the set $$ I = \lbrace x \in K \;\vert\; 0 \lt x \land \forall n \in \mathbb{N} : nx < 1 \rbrace. $$ Show that $I$ is nonempty, $1$ is an upper bound for $I$ and $I$ has no least upper bound [Hint: assume that $y$ is the least upper bound and split in to cases (i) for $y \in I$ look at $2y$, and (ii) for $y \notin I$ look at $y/2$].

In a bit of an abuse of language, the elements of $I$ can be said to be "infinitely small". For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of rational functions in one variable over $\mathbb{Q}$ (see [Wikipedia] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.

Now it's time to consider the order topology on ordered and archimedean fields, i.e. the topology generated by the open intervals $(a,b)$ defined in the obvious way.

Exercise 4. Show that the open intervals in an ordered field $K$ are a basis for a topology on $K$.

Proposition 6. $\mathbb{Q}$ is dense in any archimedean field $K$.

Sketch of proof. Since the open intervals are a basis, we just need to verify that any open interval contains a rational. Define the ceiling function $\lceil \cdot \rceil : K \to \mathbb{Z}$ by $$ \lceil x \rceil = \min \lbrace n \in \mathbb{Z} \;\vert\; n \geq x\rbrace, $$ which is well-defined since $K$ is archimedean.

Now let $x \in K$ and a $K \owns \varepsilon \gt 0$ be given. Since $K$ is archimedean we can find $n \in \mathbb{N}$ such that $n\varepsilon > 1$. Now we have: $$ 0 \leq \left\lvert \frac{\lceil nx \rceil}{n} - x \right\rvert \leq \left\lvert \frac{\lceil nx \rceil - nx}{n} \right\rvert \leq \frac{1}{n} \lt \varepsilon $$ Hence $\frac{\lceil nx \rceil}{n} \in (x-\varepsilon,x+\varepsilon)$ from which that result follows.

Corollary 7. $\mathbb{Q}$ is densely ordered in any archimedean field, i.e. if $K$ is archimedean, $x,y \in K$, and $x \lt y$, then there exists a $q \in \mathbb{Q}$ such that $x \lt q \lt y$.

Proposition 8. Any archimedean field is 2nd-countable.

Exercise 5. Prove Proposition 8 [Hint: Consider the set of intervals with rational endpoints. Show that it is a basis for a topology. Argue that it is coarser than the standard topology, and then show that any basis set for the standard topology (i.e. an interval) contains an interval with rational endpoints]

Proposition 9. Any archimedean field $K$ is a topological field, i.e. $K \times K \owns (x,y) \mapsto x+y \in K$, $K^\times \times K^\times \owns (x,y) \mapsto xy \in K^\times$ and $K^\times \owns x \mapsto x^{-1} \in K^\times$ are continuous.

Exercise 6. Prove Proposition 9 [Hint: Consider sequences $x_n$ and $y_n$ that converge to $x$ and $y$ respectively, then use that convergent sequences are bounded].

Note that proposition 9 in fact holds for ordered fields too. You just have to consider nets instead.

Definition 10. Let $G$ be a topological group. A sequence $(x_n)$ in $G$ is said to be a Cauchy sequence if for every neighbourhood $U$ of $e$ exists a $N$ such that $m,n \geq N \implies x_nx_m^{-1} \in U$. We shall denote the set of Cauchy sequences in $G$ by $G^\mathbb{N}_c$.

Note that this definition too can be used to define Cauchy nets in topological groups, but we shall not need that. Observe that the intervals with rational endpoints gives us a countable basis at $0$ in any archimedean field, so we see that for archimedean fields, the definition boils down to this: $$ \forall \varepsilon \in \mathbb{Q}^+ \exists N : n,m \geq N \implies |x_n - x_m| < \varepsilon. $$

Proposition 11. Let $K$ be an archimedean field and $(x_n)$ a sequence in $K$. The following hold:

1. If $(x_n)$ is convergent, then $(x_n)$ is a Cauchy sequence, and any subsequence of $(x_n)$ converges to the same limit.
2. If $(x_n)$ is a Cauchy sequence, then it is bounded.
3. If $(x_n)$ is bounded, then $(x_n)$ has a Cauchy subsequence ("pre-Bolzano-Weierstraß")
4. If $(x_n)$ is a Cauchy sequence then any subsequence is a Cauchy sequence. Moreover if $(x_n)$ has a convergent subsequence, then $(x_n)$ converges to the same limit.
5. If $(x_n)$ is a Cauchy sequence and $(x_{k_n})$ is a subsequence, then $\lim_{n \to \infty} |x_n - x_{k_n}| = 0$.

Exercise 7. Prove Proposition 11.

Proposition 12. Let $K$ be an archimedean field. TFAE:

1. $K$ has the least upper bound property.
2. Any Cauchy sequence in $K$ is convergent ("$K$ is complete").

Exercise 8. Flesh out the following sketch of a proof of Proposition 12:
$1 \implies 2$: Let $(x_n)$ be a Cauchy sequence in $K$. If $(x_n)$ is eventually constant, then it is obviously convergent, so assume that it is not so, and consider the set $$ X = \lbrace x \in K \;\vert\; \exists N \in \mathbb{N} : n \geq N \implies x \lt x_n \rbrace. $$

$X$ is nonempty and has an upper bound (why?) and thus $\sup X$ exists, and we claim that $x_n \to \sup X$. Given an $\varepsilon > 0$ find $N$ such that $|x_n - x_m| < \frac{\varepsilon}{2}$, and consider the interval $(x_N-\frac{\varepsilon}{2}, x_N + \frac{\varepsilon}{2})$. There are only finitely many elements of the sequence outside of this interval, so $x_N - \frac{\varepsilon}{2} \in X$. Moreover, $x + \frac{\varepsilon}{2}$ is an upper bound for $X$. The triangle inequality (note: prove the triangle inequality for archimedean fields) now gives $$ |x_n - \sup X| \leq |x_n - x_N| + |x_N - \sup X| \lt \varepsilon. $$ $2 \implies 1$: Let $A \subset K$ be nonempty and bounded above by $y_0 \in K$. Let $x_0$ be some non-upper bound for $A$ e.g. $a-1$ for some $a \in A$, and recursively define a pair of sequences $(x_n)$ and $(y_n)$ over $\mathbb{N}_0$ by $$ \begin{align} y_{n+1} & = \begin{cases} \tfrac{y_n + x_n}{2} & \text{if } \tfrac{y_n+x_n}{2} \text{ is an upper bound for } A, \\\\ y_n & \text{otherwise} \end{cases} \\\\ x_{n+1} & = \begin{cases} \tfrac{y_n+x_n}{2} & \text{if } \tfrac{y_n+x_n}{2} \text{ is a non-upper bound for } A, \\\\ x_n & \text{otherwise.} \end{cases} \end{align} $$ By induction we have that $(x_n)$ is increasing, $(y_n)$ is decreasing, and for all $i,j$: $x_i \leq y_j$. Likewise by induction we get that for all $i$: $0 \leq y_i - x_i \leq 2^{-i}(y_0 - x_0)$. It is now easy to show that $(y_n)$ is a Cauchy sequence. Assume that $n \geq m$ and we have: $$ \begin{align} y_m-y_n & = y_m - y_{m+1} + y_{m+1} - \dotsb - y_{n-1} + y_{n-1} - y_n \\\\ & \leq (2^{-m-1} + 2^{-m-2} + 2^{-n})(y_0 - x_0) \\\\ & \leq 2^{-m}(y_0 - x_0) \end{align} $$ from which it is easy to conclude that $(y_n)$ is Cauchy. Similarly $(x_n)$ is shown to be Cauchy. Thus the two sequences are convergent, and their limits must be the same since $y_n - x_n \leq 2^{-n}(y_0 - x_0)$ -- denote the limit by $s$. Finally observe that for all $a \in A$, $a \leq y_n$, so $a \leq s$ (why?) i.e. $s$ is an upper bound for $A$. Moreover if $u$ is an upper bound for $A$ then $x_n \leq u$ for all $n$, so $s \leq u$ (why?). Hence $s$ is the least upper bound for $A$. QED.

Proposition 12. Let $K$ and $L$ be archimedean fields. If $f: K \to L$ and $g: L \to K$ are embeddings, then $K$ and $L$ are isomorphic.

Proof. Let $g \circ f = \phi: K \to K$. By definition $\phi$ is an embedding. Assume that $\phi$ is not the identity. Then there exists an $x$ for which $\phi(x) \neq x$. Find a $q \in \mathbb{Q}$ between $x$ and $\phi(x)$ and observe that $\phi$ has to be the identity on $\mathbb{Q}$, but it can't preserve the ordering of $x$ and $q$. This gives a contradiction, and thus $\phi$ must be the identity. The proof that $f \circ g$ is the identity is similar. QED.

Lemma 13. Let $G$ and $H$ be topological groups and let $G'$ be a dense subgroup of $G$. If $\phi: G \to H$ is continuous and $\phi|_{G'}$ is a homomorphism, then $\phi$ is a homomorphism.

Proof. Let $x,y \in G$ and assume that $\phi(xy) \neq \phi(x)\phi(y)$. Since $G'$ is dense there exists nets $(x_\alpha)$ and $(y_\alpha)$ in $G'$ converging to $x$ and $y$ respectively. Since $\phi$ is a homomorphism on $G'$ we have for all $\alpha$ that $\phi(x_\alpha y_\alpha) = \phi(x_\alpha)\phi(y_\alpha)$, and since $\phi$ is continuous and $\cdot$ is continuous we have that $\phi(xy) = \phi(x)\phi(y)$.

Remark. This is not a vacuous use of reductio ad absurdum, since a direct proof would have to work with arbitrary nets converging to $x$ and $y$. Of course this lemma immediately generalizes to also apply to rings in an obvious manner.

Theorem 14. (Embedding theorem for archimedean fields). Let $K$ and $L$ be archimedean fields where $L$ is complete. Then there exists an embedding $\phi: K \to L$.

Proof. Define $\phi: K \to L$ by $\phi(x) = \sup \lbrace q \in \mathbb{Q} \;\vert\; q \lt x \rbrace$. It is easy to show that $\phi$ is order-preserving, so the only hard part is showing that it's a homomorphism. We wish to use the preceding lemma, so first we note that it's obvious that $\phi$'s restriction to $\mathbb{Q}$ is the identity, so the only hard part is verifying that it's continuous. Consider a $k \in K$ and let $V$ be a neighbourhood of $\phi(k)$. Find rationals $p$ and $q$ such that $\phi(k) \in (p,q) \subset V$. We claim that $\phi((p,q)) \subset (p,q)$. Let $k' \in (p,q) \subset K$ and find $q' \in \mathbb{Q}$ such that $p \lt k' \lt q' \lt q$. It's easy to see that $p \lt \phi(k') \leq q' \lt q$, so $\phi(k') \in (p,q)$, and we conclude that $\phi$ is continuous. QED.

Theorem 15. (Uniqueness of complete ordered field) There exists up to order-preserving isomorphism one and only one complete ordered field.

Proof. Let $K$ and $L$ be complete ordered fields. Theorem 14 gives embeddings $\phi: K \to L$ and $\varphi: L \to K$. Proposition 12 gives that $K$ and $L$ are isomorphic. QED.

Existence

Much more has been written on this subject than uniqueness, so I'll write a lot less here. Recall that $\mathbb{Q}$ under addition is a topological group, so we use the previously established notation for its set of Cauchy sequences: $\mathbb{Q}^\mathbb{N}_c$.

Proposition 16. $\mathbb{Q}^\mathbb{N}_c$ is a commutative unital ring under coordinatewise addition and multiplication.

Exercise 8. Prove Proposition 16.

Proposition 17. The set of rational Cauchy sequences that converge to zero, i.e. $$ \mathfrak{z} = \left\lbrace (z_n) \in \mathbb{Q}^\mathbb{N}_c \;\bigg\vert\; z_n \to 0 \right\rbrace,$$ is a maximal ideal in $\mathbb{Q}^\mathbb{N}_c$.

Exercise 9. Prove Proposition 17. [Hint: Consider an ideal $\mathfrak{a} \supset \mathfrak{z}$ with a $(q_n) \in \mathfrak{a}\setminus\mathfrak{z}$. Show that there exists an $N$ such that $n \geq N \implies q_n \neq 0$ and consider the sequence $(z_n) = (1-q_1,\dotsc,1-q_N,0,0,\dotsc)$. Argue that $(q_n + z_n) \in \mathfrak{a}$ and conclude that $\mathfrak{a} = \mathbb{Q}^\mathbb{N}_c$.]

Thus $\mathfrak{z}$ is a maximal ideal and hence the quotient, which we shall denote by $R$, must be a field. From here on out the rest is sketchy as there are a lot of details to check and it's really rather boring.

We adopt the convention that we write sequences with bold, e.g. $(x_n) = {\bf x}$ and use the notation $[{\bf x}]$ for the equivalence class containing ${\bf x}$.

Define $\lt$ on $R$ by $[{\bf p}] \lt [{\bf q}] \iff \exists (p_n) \in [{\bf p}], (q_n) \in [{\bf q}], \alpha \in \mathbb{Q}^+ : p_n \lt q_n + \alpha$ eventually. In particular note that the existence of the positive $\alpha$ cannot be dropped; e.g. look at $(1/n)$, $(0)$ and $(-1/n)$. Without the requirement that such a positive $\alpha$ exists, you'd thus have $0 \lt 0$, which is clearly rubbish.

Exercise 10. Verify that if $(p_n) \in [{\bf p}]$ and $(q_n) \in [{\bf q}]$ are witnesses that $[{\bf p}] \lt [{\bf q}]$, then any pair of representatives will do just as well, i.e. if $(p'_n) \in [{\bf p}]$ and $(q'_n) \in [{\bf q}]$, then $(p'_n)$ and $(q'_n)$ are witnesses as well.

Exercise 11. Show that $\lt$ is a strict total order, i.e. $\lt$ is irreflexive, transitive and total [Hint (for totality): You need to show that exactly one of $[{\bf x}] \lt [{\bf y}]$, $[{\bf x}] = [{\bf y}]$ and $[{\bf y}] \lt [{\bf x}]$ holds, so first show that at least one holds by assuming that none of them holds and derive a contradiction, and then show that at most one holds likewise by contradiction and using that $[{\bf x}] \nless [{\bf y}] \iff [{\bf y}] \leq [{\bf x}]$. You'll probably need to pass to "nicer" subsequences many times, so Proposition 11(5) will come in handy].

And sit back and reap the reward with the following easy

Exercise 12. Show that $R$ is an ordered field.

Exercise 13. Show that $R$ is archimedean.

Exercise 14. Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent:
Proof. Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i,j$ with $j \geq i$ that $\left\lvert[{\bf q}]^i - [{\bf q}]^j]\right\rvert \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for some representatives, so we are free to pick nice ones. Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$. Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$. Convince yourself that this is the same as this: $$\forall \varepsilon \in \mathbb{Q}^+ \exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$ Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon > 0$. So let $\varepsilon \gt 0$ be given. We use the standard "trick" of inserting a clever element into the inequality: $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that "$4^{-m}$"-property to conclude that if we choose $N,K$ such that $N=K$ and $2^{-N} \lt \varepsilon$ then RHS is eventually smaller than $\varepsilon$.

CONGRATULATIONS! You're done and you now have a fully certified driver's license for the real numbers!

Solution 2:

These are the handout articles, given to me by our Professor, when i took Real Analysis class. I hope you find it useful. (TeXing it up was tough work. Phew!!)

Often i have omitted the proof, since i felt that one can do it with little bit of thinking. But if you find that you need a proof, of theorem or a lemma, please inform, i shall be happy to edit it.

Definition 1. Let $\mathbb{Q}$ be the set of all rational numbers. A sequence $(x_{n}), x \in \mathbb{Q}$ is said to be Cauchy if for every $\epsilon \in \mathbb{Q}$, there exists a positive integer $n_{0}$ such that $|x_{n}-x_{m}|<\epsilon$ for all $n,m \geq n_{0}$.

Definition 2. A sequence $(x_{n})$ in $\mathbb{Q}$ is said to be convergent in $\mathbb{Q}$, to a rational number $a$, if for every $\epsilon^{+} \in \mathbb{Q}$, there exists a $n_{0} \in \mathbb{N}$, such that $|x_{n}-a| < \epsilon$ for all $n \geq n_{0}$. We denote this by $\lim{x_{n}}=a$.

Notations: Let $\mathcal{C}$ be the set of all Cauchy sequences in $\mathbb{Q}$ and let $\mathcal{N}$ be the set of all $(x_{n}) \in \mathcal{C}$ such that $\lim{x_{n}}=0$. Elements of $\mathcal{N}$ are called null sequences. We define addition and multiplication of two Cauchy sequences $(x_{n})$ and $(y_{n})$ in $\mathcal{C}$ as follows: $(x_{n}) + (y_{n})= (x_{n}+y_{n})$ and $(x_{n}) \cdot (y_{n}) = (x_{n}y_{n})$.

Lemma 1. $\mathcal{C}$, is a commutative ring under the addition and multiplication defined above.

Definition 3. If $R$ is a ring then a non empty subset $I \subseteq R$ is said to be an ideal of $R$, if for all $x,y \in I$ and $r \in R$, $x+y \in I$ and $rx \in I$.

Lemma 2. $\mathcal{N}$ is an ideal of $\mathcal{C}$.

Definition 4. We define a relation $\sim$ on $\mathcal{C}$ as follows: For $(x_{n}), (y_{n}) \in \mathcal{C}$, we say that $(x_{n}) \sim (y_{n})$ iff $(x_{n})-(y_{n})= (x_{n}-y_{n}) \in \mathcal{N}$. This $\sim$ is an equivalence relation. We define the equivalence class $\mathcal{C}/\mathcal{N}= \{(x_{n}) + \mathcal{N} \ | \ (x_{n}) \in \mathcal{C}\}$ as the set of real numbers denoted by $\mathbb{R}$.

We now make $\mathcal{C}/\mathcal{N}$ into a ring. This is a very common result in algebra if $R$ is a ring and $I$ is an ideal of $R$, then $R/I$ can be made into a ring. We define addition and multiplication in $\mathcal{C}/\mathcal{N}$ as follows:

• For $(x_{n}) + \mathcal{N}, (y_{n}) +\mathcal{N} \in \mathcal{C}/\mathcal{N}$

• $(x_{n})+\mathcal{N} + (y_{n})+\mathcal{N} = (x_{n}+y_{n})+\mathcal{N}$, and $(x_{n}+\mathcal{N})\cdot ((y_{n})+\mathcal{N}) = (x_{n}y_{n})+\mathcal{N}$.

Lemma 3. Let $R$ be a commutative ring and $I$ be an ideal in $R$. Let $R/I$ the quotient of $R$, w.r.t the equivalence relation $\sim$ on $R$ as: $x \sim y$ iff $x-y \in I$. If $x \sim x_{1}$ and $y \sim y_{1}$, then $x+y+I=x_{1}+y_{1}+I$ and $xy+I=x_{1}y_{1}+I$.

Lemma 4. Let $(x_{n}) \in \mathcal{C}/\mathcal{N}$. There exists $\epsilon>0$ and $n_{0} \in \mathbb{N}$, such that $|x_{n}|>\epsilon$, for all $n \geq n_{0}$. In fact, there exists $\epsilon >0$ and $n_{0} \in \mathbb{N}$ such that only one of the following is true:

• Either $x_{n} \geq \epsilon$, for all $n \geq n_{0}$, or

• $x_{n} \leq - \epsilon$, for all $n \geq n_{0}$.

Theorem. $\mathbb{R} = \mathcal{C}/\mathcal{N}$ is a field.

Proof. It is easy to show that $\mathbb{R}$ is a commutative ring witht he zero element $\mathcal{N}$ and the identity element $1+\mathcal{N}$. We need to check that if $x+\mathcal{N} \in \mathcal{C}/\mathcal{N}$ and $x \notin \mathcal{N}$, then it is invertible. That is, there exists $y+\mathcal{N}$ such that $(x+\mathcal{N})\cdot(y+\mathcal{N}=1+\mathcal{N}$.

Let $x+\mathcal{N} \in \mathcal{C}/\mathcal{N}$ and $x \notin \mathcal{N}$. By Lemma 4 there exists $\epsilon >0$, and $N \in \mathbb{N}$ such that $x_{n}>\epsilon$ for all $n \geq N$. Define $y=(y_{1},y_{2},\cdots,y_{N},0,...)$ such that $x_{i}+y_{i} \neq 0$, for $1 \leq i \leq N$. Note that $x+\mathcal{N}=(x+y)+\mathcal{N}$. Define $$(x+y)^{-1}=\biggl( \frac{1}{x_{1}+y_{1}},\cdots,\frac{1}{x_{N}+y_{N}},\cdots\biggr)$$ We claim that $(x+y)^{-1} \in\mathcal{C}$. Let $\delta \in\mathbb{Q}^{+}$ be given. For all $m,n \geq N$, $$\biggl|\frac{1}{x_{n}}-\frac{1}{x_{m}}\biggr|=\frac{|x_{m}-x_{n}|}{|x_{n}\cdot |x_{m}|}<\frac{|x_{n}-x_{m}|}{\epsilon^{2}}$$

Since $(x_{n})\in\mathcal{C}$, for the above $\delta$ there exists $n_{1} \in \mathbb{N}$, such that $|x_{m}-x_{n}| < \delta \epsilon^{2}$, for all $n,m \geq n_{1}$. Choose $n_{0}=\max (N,n_{1})$, and conclude the result.

Definition 5. An ideal $I$ of a ring $R$ is said to be a maximal ideal if $J$ is an ideal containing $I$ properly, then $J=R$.

Remark 1. In fact we have proved that $\mathcal{N}$ is a maximal ideal of $\mathcal{C}$.

Definition 6. A cauchy sequence $(x_{n}) \in \mathbb{Q}$ is said to be positive if there exists $\epsilon \in \mathbb{Q}^{+}$, and $N \in \mathbb{N}$ such that $x_{n} > \epsilon$, for all $n \geq N$.

Definition 7. A real number $\alpha \in \mathbb{R}$ is said to be positive, if $(x_{n}) \in \alpha$, then $(x_{n})$ is a positive sequence in $\mathbb{Q}$.

Theorem. If $(x_{n})$ is a positive sequence in $\mathcal{C}$, and $(z_{n}) \in \mathcal{N}$, then $(x_{n}+z_{n})$ is a positive sequence in $\mathcal{C}$. If $(x_{n})$ and $(y_{n})$ are positive sequences in $\mathcal{C}$, then $(x_{n}+y_{n})$ and $(x_{n}y_{n})$ are positive sequences in $\mathcal{C}$

We denote the set of all positive sequences in $\mathcal{C}/\mathcal{N}$ by $\mathbb{R}^{+}$.

Definition 8. Let $\mathbb{F}$ be a field. By an order on $\mathbb{F}$, we mean a subset $\mathbb{F}^{+}$ of $\mathbb{F}$, with the following properties:

• Any $x \in \mathbb{F}$, lies in exactly one of the sets $\mathbb{F}^{+},\{0\},$ and $\mathbb{F}^{-}=-\mathbb{F}^{+}$.

• For any $x,y \in \mathbb{F}^{+}$ their sum $x+y$ and the product $xy$ again lie in $\mathbb{F}^{+}$.

Theorem. $\mathbb{R}$ is an ordered field with an order $\mathbb{R}^{+}$.

Definition 9. Let $\overline{x},\overline{y} \in \mathbb{R}$. We say that $\overline{x}>\overline{y}$ if $\overline{x}-\overline{y} \in \mathbb{R}^{+}$.

Theorem. $\mathbb{R}$ has the Archimedean property, that is, if $\overline{x} \in \mathbb{R}^{+}$ and $\overline{y}\in\mathbb{R}$, then there exists $n \in \mathbb{N}$ such that $n\overline{x}>\overline{y}$.

Corollary. $\mathbb{N}$ is not bounded in $\mathbb{R}$.

Theorem. $\mathbb{R}=\mathcal{C}/\mathcal{N}$, has the l.u.b property, that is if $S$ is a non empty subset of $\mathbb{R}$, which is bounded above, then there exists a real number which is the least upper bound for $S$.

Proof. Let $S \subseteq \mathbb{R}$, be non empty and bounded above. Let $M \in \mathbb{R}$, be such that $x \leq M$ for all $x \in S$. Without loss of generality we can assume that $M \in \mathbb{Z}$. Fix $x \in S$. We claim that there exists $m \in \mathbb{Z}$, such that $m \leq x$.

For otherwise, $m>x$, for all $m \in \mathbb{Z}$, which implies that $-m<-x$, for all $m \in \mathbb{Z}$. This implies that $\mathbb{N}$ is bounded, which is a contradiction. Hence, $m \leq x \leq M$ for some $m, M \in \mathbb{Z}$. Since $S$ is bounded above by $M$, if at all the l.u.b exists, it has to lie in $[m,M]$. For each $n \in \mathbb{N}$, consider the set $$B_{n} = \Bigl\{ \frac{c}{2^{n}} \ | \ m \leq \frac{c}{2^{n}} \leq M, \ c \in \mathbb{Z}\Bigr\}$$ Note that, since $M= \frac{M.2^{n}}{2^{n}}$, $M \in B_{n}$. Hence $B_{n}$ is not empty.Also, if $\frac{c}{2^{m}}$, then $\frac{2c}{2^{n+1}} \in B_{n+1}$. Hence $B_{n} \subseteq B_{n+1}$ for all $n \in \mathbb{N}$. Since there are only finitely many integers between $m2^{n}$ and $M2^{n}$, $B_{n}$ is finite. Hence there are only finitely many upper bounds for $S$ in $B_{n}$. Let $a_{n}$ be the smallest such upper bound for $S$ in $B_{n}$, Since for $n \geq m$, $B_{m} \subseteq B_{n}$, it follows that $a_{m} \in B_{n}$ and hence $a_{n} \leq a_{m}$ for all $n \geq m$.

Now we claim that for each $n \in \mathbb{N}$, $a_{n} - \frac{1}{2^{n}}$ is not an upper bound for $S$. If $m \leq a_{n} - \frac{1}{2^{n}}$, then we are through, since $a_{n}$ is the least upper bound for $S$ in $B_{n}$, $a_{n}-\frac{1}{2^{n}}<a_{n}$ cannot be an upper bound.

Suppose $m > a_{n}-\frac{1}{2^{n}}$. Since $m \leq x$, for $x \in S$, $a_{n}-\frac{1}{2^{n}}< m\leq x$. Hence $a_{n}-\frac{1}{2^{n}}$ cannote be an upper bound for $S$. Since $a_{m}-\frac{1}{2^{m}} \leq x \leq a_{n}$, it follows that $a_{m}-a_{n}<\frac{1}{2^{m}}$. Hence $(a_{n})$ is cauchy sequence. Define $\alpha = (a_{n}) + \mathcal{N}$. First of all we claim that $a_{n}-\frac{1}{2^{n}} < \alpha < a_{n}$, where $a_{n}=(a_{n},a_{n},a_{n}, \cdots)$ and $a_{n}-\frac{1}{2^{n}} = (a_{n}-\frac{1}{2^{n}},a_{n}-\frac{1}{2^{n}},\cdots)$. Since $a_{k}$ is a decreasing sequence $a_{n} \geq a_{k}$ for all $k \geq n$. Hence $a_{n} - a_{k} >0,$ for all $k \geq n$. This shows that $(a_{n}- \alpha$ is a positive sequence in $\mathcal{C}$ and hence $\alpha < a_{n}$. Next $\alpha-\Bigl(a_{n}-\frac{1}{2^{n}}\Bigr)=(a_{1}-a_{n}+\frac{1}{2^{n}},\cdots)$ . Since $a_{k}-\frac{1}{2^{k}}$ is not an upper bound for $S$, $a_{k}>(a_{n}-\frac{1}{2^{n}})$ for all $k$, $a_{k}-(a_{n}-\frac{1}{2^{n}})>0$, for all $k$. Hence $\alpha-(a_{n}-\frac{1}{2^{n}}$ is a positive sequence and hence $\alpha > a_{n}-\frac{1}{2^{n}}$

Finally we claim that $\alpha$ is the least upper bound for S. First of all we have to show that $\alpha$ is an upper bound for S. Suppose not. Then there exists $x \in S$, such that $x > \alpha$, hence $x - \alpha >0$. Using the Archimedean property there exists $N\in\mathbb{N}$ such that $N(x-\alpha)>1$. Hence $x-\alpha>\frac{1}{N}>\frac{1}{2^{N}}$. Thus we se that $x>\alpha+\frac{1}{2^{N}}>a_{N}-\frac{1}{2^{N}}+\frac{1}{2^{N}}=a_{N}$. This is a contradiction, since $a_{N}$ is the least upper bound for $S$ in $B_{n}$.

Next, to show that $\alpha$ is the least upper bound. Suppose not. Then there exists $b \in \mathbb{R}$, such that $b$ is an upper bound for $S$ and $b < \alpha$. By the archimedean property there exists $N \in \mathbb{N}$, such that $N(\alpha - b)>1$. Hence $\alpha -b > \frac{1}{N}>\frac{1}{2^{N}}$. This implies that $\alpha -b>\frac{1}{2^{N}}$ and hence $\alpha -\frac{1}{2^{N}}>b$, or $b < \alpha-\frac{1}{2^{N}}< a_{N}-\frac{1}{2^{N}}$ as $\alpha < a_{n}$, for all $n$.But $a_{N}-\frac{1}{2^{N}}$ is not an upper bound for $S$. Hence anything less than this cannot be an upper bound for $S$. In particular, $b$ is not a upper bound for $S$, which is a contradiction. Hence $\alpha$ is the least upper bound for $S$.

Solution 3:

Here is a link that came up in a Google search. You can also search for "construction of real numbers". Since this is standard material, you can find it in many books on analysis.

http://www.math.ucsd.edu/~tkemp/140A/Construction.of.R.pdf

Solution 4:

Pages 142-154 of Stoll's Set theory and logic looks like a good self-contained treatment. The chapter starts on page 130 and is called "The extension of the natural numbers to the real numbers."