There is no greatest cardinal

We have seen in this article several sets that are infinite and countable. Namely, the set of natural numbers \(\mathbb N\), the set of integers \(\mathbb Z\), the set of rational numbers \(\mathbb Q\). We have seen that the product \(\mathbb N \times \mathbb N\) is also countable. The question then arises, “Are all infinite sets countable?”.

The answer is negative and we prove that \(\mathbb R\) is not countable.

The set of all reals \(\mathbb R\) is not countable

Suppose for the sake of contradiction that \(\mathbb R\) is countable and that \(\mathbb R = \{x_n \ | \ n \in \mathbb N\}\), the \(x_n\) being all distinct. We build by induction two sequences \((a_n)_{n \in \mathbb N}\) increasing and \((b_n)_{n \in \mathbb N}\) decreasing such that:

  • \(a_n < b_n\) for all \(n \in \mathbb N\),
  • \(\vert a_n – b_n \vert \) converges to zero,
  • and \(x_n \notin [a_{n+1},b_{n+1}]\).

Continue reading There is no greatest cardinal

Counterexamples around cardinality (part 2)

I follow on here from the initial article on cardinality. In that article, we have seen that \(\mathbb N\) contains an infinite countable subset whose complement is also infinite. Namely the subsets of even and odd integers.

Let’s now consider the set \(\mathbb N \times \mathbb N\). \(\mathbb N \times \mathbb N\) contains a countable infinite number of copies of \(\mathbb N\), as for all \(i \in \mathbb N\), \(\{i\} \times \mathbb N \subset \mathbb N \times \mathbb N\). Surprisingly, it is however possible to define a bijection between \(\mathbb N \times \mathbb N\) and \(\mathbb N\).

\(\mathbb N \times \mathbb N\) is equinumerous to \(\mathbb N\)

Consider the map \[\begin{array}{l|rcl}
\varphi : & \mathbb N \times \mathbb N & \longrightarrow & \mathbb N\\
& (n,m) & \longmapsto & 2^{n-1} (2m-1) \end{array}\] \(\varphi\) is one-to-one. Suppose that \(2^{n_1-1} (2m_1-1) = 2^{n_2-1} (2m_2-1)\). If \(n_1=1\), we must also have \(n_2=1\) as \(2 m_1 – 1\) is odd. That implies \((n_1,m_1)=(n_2,m_2)\). Let’s now suppose \(n_1 >1\). According to Euclid’s lemma, \(2^{n_1-1}\) divides \(2^{n_2-1}\) as \(2^{n_1-1}\) is even and \(2m_2 – 1\) odd. By symmetry \(2^{n_2-1}\) also divides \(2^{n_1-1}\). Finally, \(2^{n_1 – 1} = 2^{n_2 – 1}\), \(n_1 = n_2\) and consequently \(m_1 = m_2\). Continue reading Counterexamples around cardinality (part 2)

A non Archimedean ordered field

Let’s recall that an ordered field \(K\) is said to be Archimedean if for any \(a,b \in K\) such that \(0 \lt a \lt b\) it exists a natural number \(n\) such that \(na > b\).

The ordered fields \(\mathbb Q\) or \(\mathbb R\) are Archimedean. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions
\[\mathbb R(x) = \left\{\frac{S(x)}{T(x)} \ | \ S, T \in \mathbb R[x] \right\}\] For \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) we can suppose that the polynomials have a constant polynomial greatest common divisor.

Now we define \(P\) as the set of elements \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) in which the leading coefficients of \(S\) and \(T\) have the same sign.

One can verify that the subset \(P \subset \mathbb R(x)\) satisfies following two conditions:

Given \(f(x) \in \mathbb R(x)\), we have either \(f(x) \in P\), or \(f(x)=0\), or \(-f(x) \in P\), and these three possibilities are mutually exclusive. In other words, \(\mathbb R(x)\) is the disjoint union of \(P\), \(\{0\}\) and \(-P\).
For \(f(x),g(x) \in P\), \(f(x)+g(x)\) and \(f(x)g(x)\) belong to \(P\).

This means that \(P\) is a positive cone of \(\mathbb R(x)\). Hence, \(\mathbb R(x)\) is ordered by the relation
\[f(x) > 0 \Leftrightarrow f(x) \in P.\]

Now let’s consider the rational fraction \(h(x)=\frac{x}{1} \in \mathbb R(x)\). \(h(x)\) is a positive element, i.e. belongs to \(P\) as \(h-1 = \frac{x-1}{1}\). For any \(n \in \mathbb N\), we have
\[h – n 1=\frac{x-n}{1} \in P\] as the leading coefficients of \(x-n\) and \(1\) are both equal to \(1\). Therefore, we have \(h \gt n 1\) for all \(n \in \mathbb N\), proving that \(\mathbb R(x)\) is not Archimedean.

A trigonometric series that is not a Fourier series (Riemann-integration)

We’re looking here at convergent trigonometric series like \[f(x) = a_0 + \sum_{k=1}^\infty (a_n \cos nx + b_n \sin nx)\] which are convergent but are not Fourier series. Which means that the terms \(a_n\) and \(b_n\) cannot be written\[
a_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \cos nt \, dt & (n= 0, 1, \dots) \\
b_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \sin nt \, dt & (n= 1, 2, \dots)
\end{array}\] where \(g\) is any integrable function.

This raises the question of the type of integral used. We cover here an example based on Riemann integral. I’ll cover a Lebesgue integral example later on.

We prove here that the function \[
f(x)= \sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}\] is a convergent trigonometric series but is not a Fourier series. Continue reading A trigonometric series that is not a Fourier series (Riemann-integration)

Counterexamples around cardinality (part 1)

In mathematics, the cardinality of a set is a measure of the “number of elements of the set”. We denote by \(|A|\) the cardinality of the set \(A\). We say that \(A\) and \(B\) have the same cardinality if there exists a bijection from \(A\) onto \(B\). Such sets said to be equinumerous.

The finite sets satisfy the property that if \(A \subset B\) and \(A\) and \(B\) have the same cardinality then \(A=B\). For example take \(A=\{2,5,42\}\). If \(A \subset B\) and \(B\) has more elements than \(A\), a bijection \(f\) from \(A\) to \(B\) cannot exist as an element in \(B \setminus \{f(2),f(5),f(42)\}\) won’t have an inverse image under \(f\).

This becomes false for infinite sets. Let’s give some examples.

Proper subsets of \(\mathbb N\) equinumerous to \(\mathbb N\)

The set \(A\) of the positive even integers is strictly included in \(\mathbb N\): there exist odd positive integers! However, \(A\) is equinumerous to \(\mathbb N\). The map \(f : n \to 2n\) is a bijection from \(A\) onto \(\mathbb N\).

For our second example, consider the set \(\mathbb P\) of prime numbers. We recall that \(\mathbb P\) is infinite. The proof is quite simple by contradiction. Suppose that there is only a finite number \(\{p_1, \dots, p_n\}\) of prime numbers and consider the number \(N = p_1 p_2 \cdots p_n + 1\). As a consequence of the fundamental theorem of arithmetic, a prime \(q\) divides \(N\). However \(q \notin \{p_1, \dots, p_n\}\) as this would imply \(q\) divides \(1\). From there, we can define by induction a bijection from \(\mathbb N\) onto \(\mathbb P\) by defining \[f(n)=\begin{cases}
2 & \text{for } n=1\\
\min (\mathbb P \setminus \{p_1, \dots, p_n\}) & \text{else }
\end{cases}\] Continue reading Counterexamples around cardinality (part 1)