# An uncountable chain of subsets of the natural numbers

Consider the set $$\mathcal P(\mathbb N)$$ of the subsets of the natural integers $$\mathbb N$$. $$\mathcal P(\mathbb N)$$ is endowed with the strict order $$\subset$$. Let’s have a look to the chains of $$(\mathcal P(\mathbb N),\subset)$$, i.e. to the totally ordered subsets $$S \subset \mathcal P(\mathbb N)$$.

### Some finite chains

It is easy to produce some finite chains like $$\{\{1\}, \{1,2\},\{1,2,3\}\}$$ or one with a length of size $$n$$ where $$n$$ is any natural number like $\{\{1\}, \{1,2\}, \dots, \{1,2, \dots, n\}\}$ or $\{\{1\}, \{1,2^2\}, \dots, \{1,2^2, \dots, n^2\}\}$

### Some infinite countable chains

It’s not much complicated to produce some countable infinite chains like $\{\{1 \},\{1,2 \},\{1,2,3\},…,\mathbb{N}\}$ or $\{\{5 \},\{5,6 \},\{5,6,7\},…,\mathbb N \setminus \{1,2,3,4\} \}$

Let’s go further and define a one-to-one map from the real interval $$[0,1)$$ into the set of countable chains of $$(\mathcal P(\mathbb N),\subset)$$. For $$x \in [0,1)$$ let $$\displaystyle x = \sum_{i=1}^\infty x_i 2^{-i}$$ be its binary representation. For $$n \in \mathbb N$$ we define $$S_n(x) = \{k \in \mathbb N \ ; \ k \le n \text{ and } x_k = 1\}$$. It is easy to verify that $$\left(S_n(x))_{n \in \mathbb N}\right)$$ is a countable chain of $$(\mathcal P(\mathbb N),\subset)$$ and that $$\left(S_n(x))\right) \neq \left(S_n(x^\prime))\right)$$ for $$x \neq x^\prime$$.

What about defining an uncountable chain? Continue reading An uncountable chain of subsets of the natural numbers

# Counterexamples on real sequences (part 3)

Let $$(u_n)$$ be a sequence of real numbers.

### If $$u_{2n}-u_n \le \frac{1}{n}$$ then $$(u_n)$$ converges?

This is wrong. The sequence
$u_n=\begin{cases} 0 & \text{for } n \notin \{2^k \ ; \ k \in \mathbb N\}\\ 1- 2^{-k} & \text{for } n= 2^k\end{cases}$
is a counterexample. For $$n \gt 2$$ and $$n \notin \{2^k \ ; \ k \in \mathbb N\}$$ we also have $$2n \notin \{2^k \ ; \ k \in \mathbb N\}$$, hence $$u_{2n}-u_n=0$$. For $$n = 2^k$$ $0 \le u_{2^{k+1}}-u_{2^k}=2^{-k}-2^{-k-1} \le 2^{-k} = \frac{1}{n}$ and $$\lim\limits_{k \to \infty} u_{2^k} = 1$$. $$(u_n)$$ does not converge as $$0$$ and $$1$$ are limit points.

### If $$\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$$ then $$(u_n)$$ has a finite or infinite limit?

This is not true. Let’s consider the sequence
$u_n=2+\sin(\ln n)$ Using the inequality $$\vert \sin p – \sin q \vert \le \vert p – q \vert$$
which is a consequence of the mean value theorem, we get $\vert u_{n+1} – u_n \vert = \vert \sin(\ln (n+1)) – \sin(\ln n) \vert \le \vert \ln(n+1) – \ln(n) \vert$ Therefore $$\lim\limits_n \left(u_{n+1}-u_n \right) =0$$ as $$\lim\limits_n \left(\ln(n+1) – \ln(n)\right) = 0$$. And $$\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$$ because $$u_n \ge 1$$ for all $$n \in \mathbb N$$.

I now assert that the interval $$[1,3]$$ is the set of limit points of $$(u_n)$$. For the proof, it is sufficient to prove that $$[-1,1]$$ is the set of limit points of the sequence $$v_n=\sin(\ln n)$$. For $$y \in [-1,1]$$, we can pickup $$x \in \mathbb R$$ such that $$\sin x =y$$. Let $$\epsilon > 0$$ and $$M \in \mathbb N$$ , we can find an integer $$N \ge M$$ such that $$0 < \ln(n+1) - \ln(n) \lt \epsilon$$ for $$n \ge N$$. Select $$k \in \mathbb N$$ with $$x +2k\pi \gt \ln N$$ and $$N_\epsilon$$ with $$\ln N_\epsilon \in (x +2k\pi, x +2k\pi + \epsilon)$$. This is possible as $$(\ln n)_{n \in \mathbb N}$$ is an increasing sequence and the length of the interval $$(x +2k\pi, x +2k\pi + \epsilon)$$ is equal to $$\epsilon$$. We finally get $\vert u_{N_\epsilon} - y \vert = \vert \sin \left(\ln N_\epsilon \right) - \sin \left(x + 2k \pi \right) \vert \le \left(\ln N_\epsilon - (x +2k\pi)\right) \le \epsilon$ proving that $$y$$ is a limit point of $$(u_n)$$.

# A Commutative Ring with Infinitely Many Units

In a ring $$R$$ a unit is any element $$u$$ that has a multiplicative inverse $$v$$, i.e. an element $$v$$ such that $uv=vu=1,$ where $$1$$ is the multiplicative identity.

The only units of the commutative ring $$\mathbb Z$$ are $$-1$$ and $$1$$. For a field $$\mathbb F$$ the units of the ring $$\mathrm M_n(\mathbb F)$$ of the square matrices of dimension $$n \times n$$ is the general linear group $$\mathrm{GL}_n(\mathbb F)$$ of the invertible matrices. The group $$\mathrm{GL}_n(\mathbb F)$$ is infinite if $$\mathbb F$$ is infinite, but the ring $$\mathrm M_n(\mathbb F)$$ is not commutative for $$n \ge 2$$.

The commutative ring $$\mathbb Z[\sqrt{2}] = \{a + b\sqrt{2} \ ; \ (a,b) \in \mathbb Z^2\}$$ is not a field. However it has infinitely many units.

### $$a + b\sqrt{2}$$ is a unit if and only if $$a^2-2b^2 = \pm 1$$

For $$u = a + b\sqrt{2} \in \mathbb Z[\sqrt{2}]$$ we denote $$\mathrm N(u) = a^2- 2b^2 \in \mathbb Z$$. For any $$u,v \in \mathbb Z[\sqrt{2}]$$ we have $$\mathrm N(uv) = \mathrm N(u) \mathrm N(v)$$. Therefore for a unit $$u \in \mathbb Z[\sqrt{2}]$$ with $$v$$ as multiplicative inverse, we have $$\mathrm N(u) \mathrm N(v) = 1$$ and $$\mathrm N(u) =a^2-2b^2 \in \{-1,1\}$$.

### The elements $$(1+\sqrt{2})^n$$ for $$n \in \mathbb N$$ are unit elements

The proof is simple as for $$n \in \mathbb N$$ $(1+\sqrt{2})^n (-1 + \sqrt{2})^n = \left((1+\sqrt{2})(-1 + \sqrt{2})\right)^n=1$

One can prove (by induction on $$b$$) that the elements $$(1+\sqrt{2})^n$$ are the only units $$u \in \mathbb Z[\sqrt{2}]$$ for $$u \gt 1$$.

# A strictly increasing continuous function that is differentiable at no point of a null set

We build in this article a strictly increasing continuous function $$f$$ that is differentiable at no point of a null set $$E$$. The null set $$E$$ can be chosen arbitrarily. In particular it can have the cardinality of the continuum like the Cantor null set.

### A set of strictly increasing continuous functions

For $$p \lt q$$ two real numbers, consider the function $f_{p,q}(x)=(q-p) \left[\frac{\pi}{2} + \arctan{\left(\frac{2x-p-q}{q-p}\right)}\right]$ $$f_{p,q}$$ is positive and its derivative is $f_{p,q}^\prime(x) = \frac{2}{1+\left(\frac{2x-p-q}{q-p}\right)^2}$ which is always strictly positive. Hence $$f_{p,q}$$ is strictly increasing. We also have $\lim\limits_{x \to -\infty} f_{p,q}(x) = 0 \text{ and } \lim\limits_{x \to \infty} f_{p,q}(x) = \pi(q-p).$ One can notice that for $$x \in (p,q)$$, $$f_{p,q}^\prime(x) \gt 1$$. Therefore for $$x, y \in (p,q)$$ distinct we have according to the mean value theorem $$\frac{f_{p,q}(y)-f_{p,q}(x)}{y-x} \ge 1$$.

### Covering $$E$$ with an appropriate set of open intervals

As $$E$$ is a null set, for each $$n \in \mathbb N$$ one can find an open set $$O_n$$ containing $$E$$ and measuring less than $$2^{-n}$$. $$O_n$$ can be written as a countable union of disjoint open intervals as any open subset of the reals. Then $$I=\bigcup_{m \in \mathbb N} O_m$$ is also a countable union of open intervals $$I_n$$ with $$n \in \mathbb N$$. The sum of the lengths of the $$I_n$$ is less than $$1$$. Continue reading A strictly increasing continuous function that is differentiable at no point of a null set