From Naturals to Reals

From Naturals to Reals#

Johannes Siedersleben, December 2025

Introduction#

Hamlet has been interpreted countless times, and yet, whenever you read or watch it, you are likely to get new impressions, new insights. Over and over again, its depth and beauty overwhelm you.

The same is true of the path from naturals to reals, which lies at the very heart of mathematics. It has been taught, presented, published countless times, and yet, every time I go over it (I can only speak for myself), I find new insights, and I am amazed by the clarity and elegance of this intellectual edifice.

What lies ahead is a fast track from naturals to reals. Starting with the naturals, I introduce the integers as the smallest ring containing the naturals, the rationals as the smallest field containing the integers, and the reals as the smallest complete set containing the rationals. So, the reals are there, available and ready to be used as a starting point for further expeditions, such as the calculus. Along the way, we encounter the concepts of countable and uncountable infinity, and we end with an outlook on the continuum hypothesis.

The naturals are a model of what can be counted, such as coins, fingers, or pebbles. If you cut a pie into, say, eight pieces, you have done some rational arithmetic. Naturals and rationals are an abstraction of some aspects of the world we encounter or perceive. Negative numbers are a bit more complicated: A debt, a negative temperature, is an abstraction itself. You can physically count the coins in your pocket but not the ones you owe your friend. The reals, the continuum, are quite different. The continuum is an abstraction of straight lines in space or time. However, quantum physics tells us that straight lines break down into discrete quanta when observed closely. And yet, the continuum is one of the most important and most useful abstractions ever conceived.

Much of the material presented can be found in [Heuser, 2009], [Forster, 2016], [Stillwell, 2010], to name but a few prominent examples. Why did I write this paper? The starting point was the idea of a fast track from naturals to reals, with all difficulties included, but the boring parts sketched or omitted. The result is, hopefully, a roadmap you see before your eyes, easy to remember, you can talk about to a friend over a pint or two. It is, at best, a distilled version of what you find in the books cited.

The first section, Lewis Carroll’s paradox, is a reflection about what mathematical reasoning really is, meant as a nudge to always take a step back and ask: What am I doing here? How do I know I am right? Never stop asking!

What the Tortoise Said to Achilles (Lewis Carroll)#

This paradox sounds Greek, but it is in fact due to Lewis Carroll, the author of Alice in Wonderland. It questions what mathematicians really do. Imagine the tortoise explaining a mathematical argument to Achilles. This is what it says, in the words of Lewis Carroll:

What the Tortoise Said to Achilles

Listen:

(A) Things that are equal to the same are equal to each other.
(B) The two sides of this Triangle are things that are equal to the same.
(C) If A and B are true, Z must be true.
(Z) The two sides of this Triangle are equal to each other.

“You should call it D, not Z,” said Achilles. “It comes next to the other three. If you accept A and B and C, you must accept Z.” “And why must I?” “Because it follows logically from them. If A and B and C are true, Z must be true. You don’t dispute that, I imagine?”

“If A and B and C are true, Z must be true,” the Tortoise thoughtfully repeated. “That’s another Hypothetical, isn’t it? And, if I failed to see its truth, I might accept A and B and C, and still not accept Z, mightn’t I?” “You might,” the candid hero admitted, “though such obtuseness would certainly be phenomenal. Still, the event is possible. So I might ask you to grant one more Hypothetical!” “Very good. I’m quite willing to grant it, as soon as you’ve written it down. We will call it

(D) If A and B and C are true, Z must be true.

“Have you entered that in your notebook?” “I have!” Achilles joyfully exclaimed, as he ran the pencil into its sheath. “And at last we’ve got to the end of this ideal racecourse! Now that you accept A and B and C and D, of course you accept Z.” “Do I?” said the Tortoise innocently. “Let’s make that quite clear. I accept A and B and C and D. Suppose I still refused to accept Z?” “Then Logic would take you by the throat and force you to do it!” Achilles triumphantly replied. “Logic would tell you, ‘You can’t help yourself. Now that you’ve accepted A and B and C and D, you must accept Z!’ So you’ve no choice, you see.” “Whatever Logic is good enough to tell me is worth writing down,” said the Tortoise. “So enter it in your book, please. We will call it

(E) If A and B and C and D are true, Z must be true.”

“Until I’ve granted that, of course I needn’t grant Z. So it’s quite a necessary step, you see?” “I see,” said Achilles; and there was a touch of sadness in his tone.

Here the narrator, having pressing business at the Bank, was obliged to leave the happy pair, and did not again pass the spot until some months afterwards. When he did so, Achilles was still seated on the back of the much-enduring Tortoise and was writing in his notebook, which appeared to be nearly full. The Tortoise was saying, “Have you got that last step written down? Unless I’ve lost count, that makes a thousand and one. There are several million more to come. And would you mind, as a personal favour, considering what a lot of instructions this colloquy of ours will provide for the Logicians of the Nineteenth Century—would you mind adopting a pun that my cousin the Mock-Turtle will then make, and allowing yourself to be re- named Taught-Us?” “As you please!” replied the weary warrior, in the hollow tones of despair, as he buried his face in his hands. “Provided that you, for your part, will adopt a pun the Mock-Turtle never made, and allow yourself to be re-named A Kill-Ease!”

So much from Lewis Carroll. With a touch of sadness, I must admit that this paradox puts all mathematical reasoning at risk. How can we be sure that a proof is correct? What rules do we apply, and who granted us the right to apply them? Is there a formal, indisputable way of proving theorems, or is mathematics subject to some kind of majority decision? Here, formal logic – the Gödel-Tarski stuff – comes to our rescue, but that is a different story.

Naturals#

Where do natural numbers come from? Who invented them? Did they preexist in some kind of Platonian heaven? I don’t know, and, honestly, I don’t care too much. What seems clear to me is that there must have been first humans, probably at different times and different locations, to grasp the concept of a number. Today, many millennia later, the Peano axioms perfectly describe the set \(\mathbb{N}\) of natural numbers for all mathematical purposes.

Definition 12 (Peano Axioms)

\(\mathbb{N}\) is not empty: \(0 \in \mathbb{N}\)
Every natural number \(n\) has exactly one successor \(S(n)\), and every natural number \(n\) except \(0\) has exactly one predecessor \(P(n)\). The function \(S\)

\[\begin{split} S: \left\{ \begin{array}{lr} \mathbb{N} \to \mathbb{N} - \{0\} \\ n \mapsto S(n) \end{array} \right .\end{split}\]

is a bijection, and \(P = S^{-1}\) is its inverse. We normally write, of course, \(n + 1\) and \(n - 1\) to indicate successor and predecessor.

Induction: If \(\phi\) is a unary predicate such that \(\phi(0)\) is true, and for every \(n \in \mathbb{N}\), \(\phi(n)\) being true implies that \(\phi(S(n))\) is true, then \(\phi(n)\) is true for every \(n \in \mathbb{N}\).

From these axioms, order, addition, and multiplication can be defined recursively:

Definition 13 (Addition, Multiplication, Ordering)

Let \(n, m \in \mathbb{N}\).

a) Addition

\[\begin{split}&n + 0 = n \\ &n + S(m) = S(n + m)\end{split}\]

b) Multiplication

\[\begin{split}&n \cdot 0 = 0 \\ &n \cdot S(m) = n · m + n\end{split}\]

c) Ordering

\[m < n \Longleftrightarrow \exists k \in \mathbb{N}: k + m = n\]

With this ordering, \(\mathbb{N}\) is a totally ordered set with lower bound \(0\).

The Peano axioms characterize \(\mathbb{N}\) uniquely up to isomorphism: any two models satisfying these axioms are structurally identical. These axioms not only allow the definition of all basic arithmetical operations, but are the foundation of number theory and open the door to infinity: We can start at \(0\) and construct as many natural numbers as we like by applying the successor function. The following is a formal definition of infinity, with no handwaving, no dots, and no “and so on.”

Definition 14 (Infinity)

Let \(A\) be any set.

a)\(A\) is finite iff there is an \(n \in \mathbb{N}\) and an injection \(\phi: A \to \{0, 1, \ldots, n\}\).

b)\(A\) is countable iff there is an injection \(\phi: A \to \mathbb{N}\). This injection is often called an enumeration, a way of arranging all elements of \(A\) in a sequential order.

c)\(A\) is infinite iff there is an a proper subset \(B\) of \(A\) and an injection \(\phi: A \to B\).

The set \(\mathbb{N}\) is infinite because the successor function \(S\) is exactly such an injection, with \(B = \mathbb{N} - \{0\}\). That’s a formal version of Hilbert’s Hotel, which can always accommodate an extra guest by moving everyone else up one room. Another injection would be \(\phi(n) = 10^n\), with \(B = \{1, 10, 100, \ldots\}\). \(B\) is a set with huge gaps, and yet, it’s still infinite: you never run out of natural numbers.

The equivalence of not finite and infinite in Definition 14 is not obvious. To prove it, we need a weak form of the axiom of choice.

Definition 15 (Axiom of Countable Chioce)

Every not finite set contains a countably infinite subset

Theorem 17 (Countable and Uncountable Sets)

(a) If \(A\) is infinite iff it is not finite.

(b) If \(A\) is countably infinite iff there is an bijection \(\phi: A \to \mathbb{N}\).

(c) A countable union of countable sets is countable.

(d) A finite crossproduct of countable sets is countable

(e) Let \(A\) be any set. It is never possible to define an injection from the power set \(\mathcal{P}(A)\) to \(A\). Therefore, the power set of a countable set is always uncountable.

Let \(A\) be not finite, and \(\{a_0, a_1, \ldots \}\) be a countably infinite subset as granted by Definition 15. The function \(\phi\) defined by:

\[\begin{split}\phi: \left\{ \begin{array}{l} A &\to &A - \{a_0\} \\ a_k &\mapsto &a_{k+1} \text{ for } a_k \in \{a_0, a_1, \ldots \} \\ x &\mapsto &x \text{ for } x \notin \{a_0, a_1, \ldots \} \end{array} \right .\end{split}\]

is the desired injection.

(b) If there is a bijection \(\phi : A \to \mathbb{N}\), then \(A\) is countable and infinite, hence countably infinite.

Now, let \(A\) be countably infinite and \(\phi : A \to \mathbb{N}\) an injection. We set:

\[\begin{split}&n_0 = \min \phi(A) \\ &a_0 = \phi^{-1}(n_0) \\ \\ &n_k = \min \phi(A - \{a_0, a_1, \ldots, a_{k-1} \}) \\ &a_k = \phi^{-1}(n_k) \end{split}\]

The function \(\psi\) defined by:

\[\psi(a_k) = k\]

is the desired bijection: it is defined for all \(a_k\), it is an injection by construction, and it is a surjection because of \(\psi(A) = \mathbb{N}\).

(c, d)

The shortest way to prove assertions (c) and (d) relies on two facts from elementary arithmetic:

Any natural number \(a > 1\) can be uniquely decomposed into a product of primes \(a = \prod_{i=1}^n p_i^{\alpha_i}\) where \(p_i\) are prime numbers and \(\alpha_i\) are naturals (see Theorem 11).
There are infinitely many primes \(p_1, p_2, \ldots\) (see Theorem 12).

Let \(\{A_i\}\) be a family of sets, countable for assertion (c), finite for (d), and let

\[\phi_i : A_i \to \mathbb{N}\]

be the injection corresponding to \(A_i\).

(c) Countable Union: The function

\[\begin{split}\phi: \left\{ \begin{array}{lr} \cup_i A_i \to \mathbb{N} \\ a \mapsto p_i^{\phi_i(a)} \text{ if } a \in A_i \end{array} \right .\end{split}\]

is an injection: The set \(A_i\) is mapped to the powers of \(p_i\) (we are using Theorem 12).

(d) Finite Crossproduct: The function

\[\begin{split}\phi: \left\{ \begin{array}{lr} \prod_{i=1}^n A_i \to \mathbb{N} \\ (a_1, a_2, \ldots, a_n) \mapsto \prod_{i=1}^n p_i^{\phi_i(a_i)} \end{array} \right .\end{split}\]

is an injection: Each \((a_1, a_2, \ldots, a_n)\) is mapped to the unique natural number whose decomposition is \((\phi_1(a_1), \phi_2(a_2), \ldots, \phi_n(a_n))\) (we are using Theorem 11).

e) Assume there is an injection

\[\phi: \mathcal{P}(A) \to A\]

We investigate the set

\[K = \{\phi(U) \mid U \subseteq A, U \ne \emptyset, \phi(U) \notin U\}\]

If \(K\) were empty, we would have

\[\begin{split}&\forall U \subseteq A, U \ne \emptyset: \phi(U) \in U \\ \Rightarrow &\forall a \in A: \phi(\{a\}) \in \{a\} \\ \Rightarrow &\forall a \in A: \phi(\{a\}) = a\end{split}\]

But in this case, for any subset \(U\) of \(A\) with two or more elements, there would be no room left for \(\phi(U)\), since \(\phi\) is injective. Therefore, \(K\) is not empty, and we can ask if \(\phi(K)\) belongs to \(K\):

(37)#\[\begin{split}&\phi(K) \in K \Rightarrow \phi(K) \notin K \\ &\phi(K) \notin K \Rightarrow \phi(K) \in K\end{split}\]

This contradiction proves that there can be no injection from a power set of a set to the set itself. It is the formal variant of the barber who shaves all men who do not shave themselves. Note that, with \(K\) empty, contradiction would be meaningless: ex falso quodlibet.

Integers#

The set \(\mathbb{N}\) is a semigroup with respect to addition and a semigroup with respect to multiplication. But \(\mathbb{N}\) lacks the negative numbers: the equation \(a + x = b\) is solvable in \(\mathbb{N}\) only if \(a \le b\).

We introduce the set \(\mathbb{Z}\) of integers as the smallest ring containing \(\mathbb{N}\). While this definition is concise, the question remains whether such a ring exists at all. The answer is yes, it does, and it can be easily constructed. We introduce a new element \(-1\) as the unique solution of \(1 + x = 0\) and define:

\[ \begin{align}\begin{aligned}(-1) (-1) &= 1\\-a &= (-1) a \text{ for } a \in \mathbb{N}\\-\!\mathbb{N} &= \{-a \mid a \in \mathbb{N} \}\\\mathbb{Z} &= -\!\mathbb{N} \cup \mathbb{N} \end{aligned}\end{align} \]

Starting from \((-1) (-1) = 1\), the basic arithmetic operations are easily constructed. As an example, we explain how \(a - b\) can be given a meaning if \(a < b\):

\[a - b = a + (-1)b = (-1)((-1)a + b) = (-1)(b - a) = - (b-a)\]

Continuing in this way, the definition of addition, subtraction, and multiplication on \(\mathbb{Z}\) is a straightforward exercise. With these operations in place, the set \(\mathbb{Z}\) is a ring. As any ring containing \(\mathbb{N}\) must also contain the negative numbers \(-\!\mathbb{N}\), the set \(\mathbb{Z}\) is indeed the smallest ring containing \(\mathbb{N}\). \(\mathbb{Z}\) is countable, since it is the union of two countable sets.

Definition 16 (Ordering of Integers)

We extend the order on \(\mathbb{N}\) to \(\mathbb{Z}\) by setting, for \(m, n \in \mathbb{N}\):

\[\begin{split}&-n < 0 \\ &-n < -m \Leftrightarrow m < n\end{split}\]

With this ordering, \(\mathbb{Z}\) is a totally ordered set without lower or upper bound.

From now on, we take \(\mathbb{Z}\) as given, with its standard operations and ordering, and focus on how to extend it to \(\mathbb{Q}\), and \(\mathbb{R}\). Our construction proceeds from here with \(\mathbb{Z}\) as the base case, showing how successively richer number systems emerge through a consistent pattern of equivalence class constructions.

Rationals#

The set \(\mathbb{Z}\) is a ring but not a field. \(\mathbb{Z}\) lacks the fractions: the equation \(ax = 1\) is solvable in \(\mathbb{Z}\) only for \(a = 1\).

To this end, we introduce the set \(\mathbb{Q}\) of rationals as the smallest field containing \(\mathbb{Z}\). Again, the question remains whether such a field exists at all. The answer is yes, it does, and it can be easily constructed. We introduce an equivalence relation on the set \(\mathbb{Z} \times (\mathbb{Z}-\{0\})\) as

\[(a, b) \sim (c, d) \Longleftrightarrow ad = bc\]

and set

\[\mathbb{Q} = (\mathbb{Z} \times (\mathbb{Z}-\{0\}))/\sim\]

This immediately gives the cancellation rule:

\[(a, b) \sim (c, d) \Longleftrightarrow (na, nb) \sim (c, d)\]

Setting

\[a^{-1} = \frac{1}{a} = (1, a)\]

we can easily define the well-known rules of rational arithmetic, such as:

\[\begin{split}&\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \\ &\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}\end{split}\]

With these operations in place, the set \(\mathbb{Q}\) is a field. As any field containing \(\mathbb{Z}\) must also contain the rationals, the set \(\mathbb{Q}\) is indeed the smallest field containing \(\mathbb{Z}\). \(\mathbb{Q}\) is countable, since it is the cross-product of two countable sets.

The normal form of a rational number \(a/b\) requires \(b > 0\) and \(a, b\) coprime.

Definition 17 (Ordering of Rationals)

We extend the order on \(\mathbb{Z}\) to \(\mathbb{Q}\) by setting, for \(a/b, c/d \in \mathbb{Q}\) in normal form:

\[\frac{a}{b} < \frac{c}{d} \Longleftrightarrow ad < bc\]

With this ordering, \(\mathbb{Q}\) is a totally ordered set without lower or upper bound.

Theorem 18 (Archimedean Property)

For any \(q \in \mathbb{Q}\), there exists \(n \in \mathbb{N}\) such that \(n > q\).

Proof. Write \(q = a/b\) in normal form, so \(b > 0\). The assertion follows from \(a/b < a\).

Reals#

What worked fine for integers and rationals proves to be more tricky for reals. The set \(\mathbb{Q}\) is a field with many holes: the equation \(x^2 = a\) is solvable in \(\mathbb{Q}\) only for square numbers, but not for, say, \(a=2\). \(\mathbb{Q}\) famously does not contain \(\sqrt 2, \pi, e\), and many other numbers. It is tempting to think of the set \(\mathbb{R}\) of real numbers as the smallest set containing \(\mathbb{Q}\) with no holes. While this idea captures the essence of the continuum, it lamentably lacks rigour. The rest of this section is about curing this defect. The key concept is the Cauchy sequence, a formalization of the idea of a hole. The following definitions are, for now, restricted to rationals (even the epsilons are rational!) because that’s all we have. But the definitions remain valid, of course, in \(\mathbb{R}\) and in any metric space.

Definition 18 (Convergence of Rational Sequences)

(a) We call a sequence \(\{x_n\}\) convergent to \(x\), written as

\[\lim_{n \to \infty} x_n = x\]

iff \(\left | x_n - x \right |\) becomes arbitrariliy small:

\[\forall \epsilon > 0 : \exists n_0(\epsilon) \in \mathbb{N} : \forall n > n_0(\epsilon) : \left | x_n - x \right | < \epsilon\]

(b) We call a sequence \(\{x_n\}\) a Cauchy sequence iff \(\left | x_m - x_n \right |\) becomes arbitrariliy small:

\[\forall \epsilon > 0 : \exists n_0(\epsilon) \in \mathbb{N} : \forall m, n > n_0(\epsilon) : \left | x_m - x_n \right | < \epsilon\]

(c) We call two Cauchy sequences \(\{x_n\}, \{y_n\}\) equivalent iff \(\left | x_n - y_n \right |\) becomes arbitrariliy small:

\[\forall \epsilon > 0 : \exists n_0(\epsilon) \in \mathbb{N} : \forall n > n_0(\epsilon) : \left | x_n - y_n \right | < \epsilon\]

This is written as:

\[\{x_n\} \sim \{y_n\}\]

(d) We call a set \(A\) complete iff every Cauchy sequence converges.

Theorem 19 (Arithmetic on Sequences)

a) Sum, difference, product, and quotient of convergent sequences are convergent and converge to the sum, difference, product, and quotient of their limits. For the denominator \(\{y_n\}\) of a quotient we require that \(\lim_{n \to \infty} y_n \ne 0\).

b) Sum, difference, product, and quotient of Cauchy sequences are Cauchy sequences. For the denominator \(\{y_n\}\) of a quotient we require an \(\epsilon > 0\) and an \(n_0 \in \mathbb{N}\) such that \(|y_n| > \epsilon\) for all \(n > n_0\).

Proof. We only prove that the quotient of two Cauchy sequences \(\{x_n\}\), \(\{y_n\}\) is again a Cauchy sequence.

As \(\{|y_n|\}\) is positive and bounded, we can choose \(\epsilon_0, n_0, M\) such that \(\epsilon_0 \le |y_n| \le M\) for all \(n > n_0\). Then:

\[\left | \frac{x_n}{y_n} - \frac{x_m}{y_m} \right | = \left | \frac{x_n y_m - x_m y_n}{y_m y_n} \right | \le \frac{1}{\epsilon_0^2} \left | x_n y_m - x_m y_n \right | \le \frac{M}{\epsilon_0^2} \left | x_n - x_m \right |\]

This proves the assertion.

With these definitions, we can repeat the approach that worked for integers and rationals. We introduce the set \(\mathbb{R}\) of reals as the smallest complete field containing \(\mathbb{Q}\). As before, the question remains whether such a set exists at all. The answer is yes, it does, and goes back to Cantor. He defined

\[ \begin{align}\begin{aligned}&\mathbb{S} = \text{ set of all Cauchy sequences in } \mathbb{Q}\\&\mathbb{R} = \mathbb{S}/\sim\end{aligned}\end{align} \]

This reads as follows: \(\mathbb{R}\) is the set of the equivalence classes of all Cauchy sequences, or, more informally: It is the set of whatever numbers that can be approximated by rationals. This gives us the following theorem for free:

Theorem 20 (Rationals and Reals)

Every real number can be approximated by rationals to any degree of precision.

Example: The sequence \(\{x_n\}\) defined by

\[\begin{split}x_0 &= 2 \\ x_{n+1} &= x_n - \frac{2-x_n^2}{2x_n}\end{split}\]

is a Cauchy sequence in \(\mathbb{Q}\) and converges to \(\sqrt 2\) in \(\mathbb{R}\). Cantor’s construction identifies the sequence \(\{x_n\}\) (and all sequences \(\{y_n\}\) with \(\{y_n\} \sim \{x_n\}\)) with its limit \(\sqrt 2\). The notation

\[\{x_n\} = \sqrt 2\]

is a shorthand representation of this fact.

All basic arithmetic are extended to \(\mathbb{R}\) in a natural way. Example: For \(x = \{x_n\}\), \(y = \{y_n\}\) we define:

\[x + y = \{x_n\} + \{y_n\}\]

Thanks to Theorem 19 we know the meaning of \(\{x_n\} + \{y_n\}\) and of all other arithmetic operations. Therefore, the set \(\mathbb{R}\) is a field.

\(\mathbb{R}\) not only inherits the basic arithmetics but also the topology: We must define the distance and convergence of reals in terms of the distance and convergence of Cauchy sequences of rationals, an undertaking that involves a considerable number of epsilons.

Definition 19 (Convergence of Real Sequences)

Let \(r, s\) be real numbers and \(\{r_n\}_n\) a sequence of reals numbers. Here is how they are represented by their defining Cauchy series of rationals:

\[ \begin{align}\begin{aligned}r &= \{y_k\}_k\\s &= \{z_k\}_k\\r_n &= \{x_{n_k}\}_k\\\{r_n\}_n &= \{\{x_{n_k}\}\}_n\end{aligned}\end{align} \]

Each \(r_n\) corresponds to (or: is represented by, can be identified with) a Cauchy sequence \(\{x_{n_k}\}_k\), and the sequence \(\{r_n\}_n\) is itself a sequence of Cauchy sequences. This can be represented as a matrix, with a row for each \(r_n\):

\[ \begin{align}\begin{aligned}r_1 & = \{x_{1_k}\} = \{x_{1_1}, x_{1_2}, \ldots, x_{1_k}, \ldots \} \\r_2 & = \{x_{2_k}\} = \{x_{2_1}, x_{2_2}, \ldots, x_{2_k}, \ldots \} \\&\vdots\\r_n & = \{x_{n_k}\} = \{x_{n_1}, x_{n_2}, \ldots, x_{n_k}, \ldots \}\\&\vdots\end{aligned}\end{align} \]

a) Distance of reals
We define what it means for a real number to be close to zero or to another real number:

(38)#\[\begin{split}\left | r \right | < \epsilon &\Leftrightarrow \exists k_0(\epsilon): \forall k > k_0(\epsilon): \left | y_k \right | < \epsilon \\ \left | r - s \right | < \epsilon &\Leftrightarrow \exists k_0(\epsilon): \forall k > k_0(\epsilon): \left | y_k - z_k \right | < \epsilon\end{split}\]

Note that the equivalence of rational Cauchy sequences is the same as the equality of their corresponding reals:

\[r = s \Leftrightarrow \{y_k\}_k \sim \{z_k\}_k\]

b) Convergence of reals
We define what it means for a sequence of real numbers to converge to zero or to another real number:

(39)#\[\begin{split}\lim_{n \to \infty} r_n = 0 &\Leftrightarrow \forall \epsilon > 0: \exists n_0(\epsilon): \forall n > n_0(\epsilon) \left | r_n \right | < \epsilon \\ \lim_{n \to \infty} r_n = r &\Leftrightarrow \forall \epsilon > 0: \exists n_0(\epsilon): \forall n > n_0(\epsilon) \left | r_n - r \right | < \epsilon\end{split}\]

Expanding statement (39) with (38), gives us the long version:

\[ \begin{align}\begin{aligned}\lim_{n \to \infty} r_n = 0 &\Leftrightarrow \forall \epsilon > 0: \exists n_0(\epsilon): \forall n > n_0(\epsilon): \exists k_0(\epsilon, n): \forall k > k_0(\epsilon, n): \left | x_{n_k} \right | < \epsilon\\\lim_{n \to \infty} r_n = r &\Leftrightarrow \forall \epsilon > 0: \exists n_0(\epsilon): \forall n > n_0(\epsilon): \exists k_0(\epsilon, n): \forall k > k_0(\epsilon, n): \left | x_{n_k} - y_k \right | < \epsilon\end{aligned}\end{align} \]

c) Cauchy convergence of reals
We define what it means for a sequence of real numbers to be a Cauchy sequence: The sequence \(\{r_n\}_n = \{\{x_{n_k}\}\}_n\) is a Cauchy sequence iff:

(40)#\[\forall \epsilon > 0: \exists n_0(\epsilon): \forall m, n > n_0(\epsilon): \left | r_n - r_m \right | < \epsilon\]

Expanding statement (40) with (38), gives us the long version:

\[\forall \epsilon > 0: \exists n_0(\epsilon): \forall m, n > n_0(\epsilon): \exists k_0(\epsilon, m, n): \forall k > k_0(\epsilon, m, n): \left | x_{n_k} - x_{m_k} \right | < \epsilon\]

We take as an example the sequences

\[ \begin{align}\begin{aligned}x_{n_0} &= 2 + \frac{1}{n} \\x_{n_{k+1}} &= x_{n_k} - \frac{2 + \frac{1}{n} -x_{n_k}^2}{(2 + \frac{1}{n})x_{n_k}}\end{aligned}\end{align} \]

For each \(n \in \mathbb{N}\), \(\{x_{n_k}\}_k\) is a Cauchy sequence in \(\mathbb{Q}\) that converges in \(\mathbb{R}\) to \(\sqrt{2 + \frac{1}{n}}\) in \(\mathbb{R}\).

Theorem 21 (\(\mathbb{R}\) is Complete)

Every Cauchy sequence \(\{r_n\}\) in \(\mathbb{R}\) converges.

Proof. Let \(\{r_n\}\) be a Cauchy sequence in \(\mathbb{R}\). We proceed in two steps: (1) We show that the diagonal sequence \(r\) is a Cauchy sequence. (2) We show that \(\{r_n\}\) converges to \(r\).

Every number \(r_n\) in that sequence is represented by a Cauchy sequence \(\{x_{n_k}\}\) of rationals:

\[r_n = \{x_{n_k}\}_k\]

Each \(r_n\) being a Cauchy sequence, we get:

(41)#\[\forall n \in \mathbb{N}, \epsilon > 0: \exists k_0(n, \epsilon): \forall k, j > k_0(n, \epsilon) : \left | x_{n_k} - x_{n_j} \right | < \epsilon\]

The sequence \(\{r_n\}_n = \{\{x_{n_k}\}\}_n\) being a Cauchy sequence itself, we get:

(42)#\[\forall \epsilon > 0: \exists n_0(\epsilon): \forall m, n > n_0(\epsilon): \exists k_1(\epsilon, m, n): \forall k > k_1(\epsilon, m, n): \left | x_{n_k} - x_{m_k} \right | < \epsilon\]

We conclude that the diagonal sequence \(\{x_{n_n}\}\) is also Cauchy sequence. To this end, we choose \(\epsilon >0\), \(n, m > n_0(\epsilon)\), and \(k > k_1(\epsilon, m, n)\) and, using inequality (41) twice and (42) once, get the inequality:

\[\left | x_{n_n} - x_{m_m} \right | \le \left | x_{n_n} - x_{n_k} \right | + \left | x_{n_k} - x_{m_k} \right | + \left | x_{m_k} - x_{m_m} \right | < 3 \epsilon\]

So, the sequence \(\{x_{n_n}\}\) is a Cauchy sequence and represents a real number \(r\):

\[r = \{x_{n_n}\}\]

It remains to show that

\[\lim_{n \to \infty}r_n = r\]

We observe that in statement (42) we can choose \(m\) as we like. Setting \(m = k\) we get (compare Definition 19):

\[\forall \epsilon > 0: \exists n_0(\epsilon): \forall n > n_0(\epsilon): \exists k_1(\epsilon, n): \forall k > k_1(\epsilon, n): \left | x_{n_k} - x_{k_k} \right | < \epsilon\]

which is what we want.

This proof, however dry or abstract it might appear, is nothing but the obstinate application of known facts: We apply the triangular inequality to show that the diagonal sequence is a Cauchy sequence, and, in statement (42), we replace the general \(m\) with a specific choice — this is how “forall” is meant to be employed.

Definition 20 (Ordering of Reals)

We extend the order on \(\mathbb{Q}\) to \(\mathbb{R}\) by setting, for \(x = \{x_n\}, y = \{y_n\} \in \mathbb{R}\):

\[a < b \Longleftrightarrow \exists n_0 \in \mathbb{N}: \forall n > n_0: x_n < y_n\]

With this ordering, \(\mathbb{R}\) is a totally ordered set without lower or upper bound.

We close this chapter with a note about the uncountability of the real numbers and a glimpse at the continuum hypothesis.

Theorem 22 (\(\mathbb{R}\) is uncountable)

(a)

Let \(B \ge 2\) be a natural number, the basis. Every rational number \(x = p/q \in [0, 1)\) can be represented by a series

\[\frac{p}{q} = \sum_{k=1}^\infty \frac{a_k}{B^k}\]

where \(a_k \in \{0,1, \ldots, B-1\}\). For \(B = 10\), these are the decimal fractions.

(b) Every open interval of \(\mathbb{R}\) is uncountable.

Proof. (a) We are going to generalize the decimal expansion as it is taught at secondary schools. Look at the sequences \(\{a_k\}, \{r_k\}\) defined by:

\[\begin{split}&a_1 = pB // q \\ &r_1 = pB \, \% \, q \\ \\ &a_{n+1} = r_n // q \\ &r_{n+1} = r_n \, \% \, q\end{split}\]

The definition of % and // implies:

\[\begin{split}pB &= (pB // q)q + pB \, \% \, q \\ &= qa_1 + r_1\end{split}\]

\[\begin{split}r_n B &= (r_nB // q)q + r_nB \, \% \, q \\ &= qa_{n+1} + r_{n+1}\end{split}\]

Dividing by \(qB\) or \(qB^{n+1}\) resp. gives:

\[\frac{p}{q} = \frac{a_1}{B} + \frac{r_1}{qB} \]

\[\frac{r_n}{qB^n} = \frac{a_{n+1}}{B^{n+1}} + \frac{r_{n+1}}{qB^{n+1}}\]

from which follows for any \(n\):

(43)#\[\frac{p}{q} = \sum_{k=1}^n \frac{a_k}{B^k} + \frac{r_{n}}{qB^n}\]

(b) The proof uses the famous diagonalization method. We prove in a first step that it is impossible to enumerate the rationals in \([0,1)\). Assume that \(\{x_n\}\) is such an enumeration and

\[x_n = \sum_{k=0}^\infty \frac{a_{n_k}}{2^k}\]

is the binary representation of \(x_n\) as presented in part (a). Then

\[y = \sum_{k=0}^\infty \frac{1- a_{k_k}}{2^k}\]

cannot be one of the \(x_n\) because \(y = x_{n^*}\) for some \(n^*\) would imply:

\[1 - a_{{n^*}_{n^*}} = a_{{n^*}_{n^*}}\]

But this is impossible, as \(a_{{n^*}_{n^*}}\) is a natural number.

To prove that any open interval \((a, b)\) is uncountable we use the injection:

\[\begin{split}\phi: \left\{ \begin{array}{lr} (0, 1) \to (a, b) \\ x \mapsto a + x(b-a) \end{array} \right .\end{split}\]

As any complete field containing \(\mathbb{Q}\) necessarily contains the set \(\mathbb{R}\), it is indeed the smallest complete field containing \(\mathbb{Q}\). From now on, \(\mathbb{R}\) will be our home. Whatever happens, happens in \(\mathbb{R}\).

Remark 7 (Continuum Hypothesis)

Two sets are said to be equinumerous, or have the same cardinality, if there is a bijection between them. Theorem 17 (b) tells us that all countably infinite sets share the same cardinality, denoted as \(\aleph_0\). Theorem 22 tells us that \(\mathbb{R}\) and \(\mathcal{P}(\mathbb{N})\) have the same cardinality, denoted as \(\mathfrak{c}\). Theorem 17 (e) tells us that, for any set \(A\), the cardinality of \(\mathcal{P}(A)\) is always strictly higher than that of \(A\). And if two sets have the same cardinality, then their power sets necessarily do too. This leads to an infinite sequence of cardinalities \(\{\aleph_0, \aleph_1, \ldots \}\) with \(\aleph_{n+1} = 2^{\aleph_n}\). Are there any others?

The continuum hypothesis (abbreviated as CH) claims that there aren’t. If this is true, there is no cardinality between countable infinity \(\aleph_0\) and the continuum \(\mathfrak{c} = \aleph_1 = 2^{\aleph_0}\). CH is independent of the ZFC-Axioms: Both “ZFC + CH” and “ZFC + \(\neg\) CH” are consistent, assuming ZFC is. But this is a different story, to be told in another paper.