# 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

# A monotonic function whose points of discontinuity form a dense set

Consider a compact interval $$[a,b] \subset \mathbb R$$ with $$a \lt b$$. Let’s build an increasing function $$f : [a,b] \to \mathbb R$$ whose points of discontinuity is an arbitrary dense subset $$D = \{d_n \ ; \ n \in \mathbb N\}$$ of $$[a,b]$$, for example $$D = \mathbb Q \cap [a,b]$$.

Let $$\sum p_n$$ be a convergent series of positive numbers whose sum is equal to $$p$$ and define $$\displaystyle f(x) = \sum_{d_n \le x} p_n$$.

### $$f$$ is strictly increasing

For $$a \le x \lt y \le b$$ we have $f(y) – f(x) = \sum_{x \lt d_n \le y} p_n \gt 0$ as the $$p_n$$ are positive and dense so it exists $$p_m \in (x, y]$$.

### $$f$$ is right-continuous on $$[a,b]$$

We pick-up $$x \in [a,b]$$. For any $$\epsilon \gt 0$$ is exists $$N \in \mathbb N$$ such that $$0 \lt \sum_{n \gt N} p_n \lt \epsilon$$. Let $$\delta > 0$$ be so small that the interval $$(x,x+\delta)$$ doesn’t contain any point in the finite set $$\{p_1, \dots, p_N\}$$. Then $0 \lt f(y) – f(x) \le \sum_{n \gt N} p_n \lt \epsilon,$ for any $$y \in (x,x+\delta)$$ proving the right-continuity of $$f$$ at $$x$$. Continue reading A monotonic function whose points of discontinuity form a dense set

# A normal extension of a normal extension may not be normal

An algebraic field extension $$K \subset L$$ is said to be normal if every irreducible polynomial, either has no root in $$L$$ or splits into linear factors in $$L$$.

One can prove that if $$L$$ is a normal extension of $$K$$ and if $$E$$ is an intermediate extension (i.e., $$K \subset E \subset L$$), then $$L$$ is a normal extension of $$E$$.

However a normal extension of a normal extension may not be normal and the extensions $$\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension $$k \subset K$$ , i.e. an extension of degree two is normal. Suppose that $$P$$ is an irreducible polynomial of $$k[x]$$ with a root $$a \in K$$. If $$a \in k$$ then the degree of $$P$$ is equal to $$1$$ and we’re done. Otherwise $$(1, a)$$ is a basis of $$K$$ over $$k$$ and there exist $$\lambda, \mu \in k$$ such that $$a^2 = \lambda a +\mu$$. As $$a \notin k$$, $$Q(x)= x^2 – \lambda x -\mu$$ is the minimal polynomial of $$a$$ over $$k$$. As $$P$$ is supposed to be irreducible, we get $$Q = P$$. And we can conclude as $Q(x) = (x-a)(x- \lambda +a).$

The entensions $$\mathbb Q \subset \mathbb Q(\sqrt{2})$$ and $$\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ are quadratic, hence normal according to previous lemma and $$\sqrt[4]{2}$$ is a root of the polynomial $$P(x)= x^4-2$$ of $$\mathbb Q[x]$$. According to Eisenstein’s criterion $$P$$ is irreducible over $$\mathbb Q$$. However $$\mathbb Q(\sqrt[4]{2}) \subset \mathbb R$$ while the roots of $$P$$ are $$\pm \sqrt[4]{2}, \pm i \sqrt[4]{2}$$ and therefore not all real. We can conclude that $$\mathbb Q \subset \mathbb Q(\sqrt[4]{2})$$ is not normal.

# The image of an ideal may not be an ideal

If $$\phi : A \to B$$ is a ring homomorphism then the image of a subring $$S \subset A$$ is a subring $$\phi(A) \subset B$$. Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take $$A=\mathbb Z$$ the ring of the integers and for $$B$$ the ring of the polynomials with integer coefficients $$\mathbb Z[x]$$. The inclusion $$\phi : \mathbb Z \to \mathbb Z[x]$$ is a ring homorphism. The subset $$2 \mathbb Z \subset \mathbb Z$$ of even integers is an ideal. However $$2 \mathbb Z$$ is not an ideal of $$\mathbb Z[x]$$ as for example $$2x \notin 2\mathbb Z$$.

# A function whose Maclaurin series converges only at zero

Let’s describe a real function $$f$$ whose Maclaurin series converges only at zero. For $$n \ge 0$$ we denote $$f_n(x)= e^{-n} \cos n^2x$$ and $f(x) = \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty e^{-n} \cos n^2 x.$ For $$k \ge 0$$, the $$k$$th-derivative of $$f_n$$ is $f_n^{(k)}(x) = e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right)$ and $\left\vert f_n^{(k)}(x) \right\vert \le e^{-n} n^{2k}$ for all $$x \in \mathbb R$$. Therefore $$\displaystyle \sum_{n=0}^\infty f_n^{(k)}(x)$$ is normally convergent and $$f$$ is an indefinitely differentiable function with $f^{(k)}(x) = \sum_{n=0}^\infty e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right).$ Its Maclaurin series has only terms of even degree and the absolute value of the term of degree $$2k$$ is $\left(\sum_{n=0}^\infty e^{-n} n^{4k}\right)\frac{x^{2k}}{(2k)!} > e^{-2k} (2k)^{4k}\frac{x^{2k}}{(2k)!} > \left(\frac{2kx}{e}\right)^{2k}.$ The right hand side of this inequality is greater than $$1$$ for $$k \ge \frac{e}{2x}$$. This means that for any nonzero $$x$$ the Maclaurin series for $$f$$ diverges.

# A group that is not a semi-direct product

Given a group $$G$$ with identity element $$e$$, a subgroup $$H$$, and a normal subgroup $$N \trianglelefteq G$$; then we say that $$G$$ is the semi-direct product of $$N$$ and $$H$$ (written $$G=N \rtimes H$$) if $$G$$ is the product of subgroups, $$G = NH$$ where the subgroups have trivial intersection $$N \cap H= \{e\}$$.

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group $$D_{2n}$$ with $$2n$$ elements is isomorphic to a semidirect product of the cyclic groups $$\mathbb Z_n$$ and $$\mathbb Z_2$$. $$D_{2n}$$ is the group of isometries preserving a regular polygon $$X$$ with $$n$$ edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

### The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group $$\mathbb H_8$$ is the group consisting of the symbols $$\pm 1, \pm i, \pm j, \pm k$$ where$-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.$ One can prove that $$\mathbb H_8$$ endowed with the product operation above is indeed a group having $$8$$ elements where $$1$$ is the identity element.

$$\mathbb H_8$$ is not abelian as $$ij = k \neq -k = ji$$.

Let’s prove that $$\mathbb H_8$$ is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup $$N$$ and a subgroup $$H$$ such that $$G=N \rtimes H$$.

• If $$\vert N \vert = 4$$ then $$H = \{1,h\}$$ where $$h$$ is an element of order $$2$$ in $$\mathbb H_8$$. Therefore $$h=-1$$ which is the only element of order $$2$$. But $$-1 \in N$$ as $$-1$$ is the square of all elements in $$\mathbb H_8 \setminus \{\pm 1\}$$. We get the contradiction $$N \cap H \neq \{1\}$$.
• If $$\vert N \vert = 2$$ then $$\vert H \vert = 4$$ and $$H$$ is also normal in $$G$$. Noting $$N=\{1,n\}$$ we have for $$h \in H$$ $$h^{-1}nh=n$$ and therefore $$nh=hn$$. This proves that the product $$G=NH$$ is direct. Also $$N$$ is abelian as a cyclic group of order $$2$$. $$H$$ is also cyclic as all groups of order $$p^2$$ with $$p$$ prime are abelian. Finally $$G$$ would be abelian, again a contradiction.

We can conclude that $$G$$ is not a semi-direct product.

Can you paint a surface with infinite area with a finite quantity of paint? For sure… let’s do it!

Consider the 3D surface given in cylindrical coordinates as $S(\rho,\varphi):\begin{cases} x &= \rho \cos \varphi\\ y &= \rho \sin \varphi\\ z &= \frac{1}{\rho}\end{cases}$ for $$(\rho,\varphi) \in [1,\infty) \times [0, 2 \pi)$$. The surface is named Gabriel’s horn.

### Volume of Gabriel’s horn

The volume of Gabriel’s horn is $V = \pi \int_1^\infty \left( \frac{1}{\rho^2} \right) \ d\rho = \pi$ which is finite.

### Area of Gabriel’s horn

The area of Gabriel’s horn for $$(\rho,\varphi) \in [1,a) \times [0, 2 \pi)$$ with $$a > 1$$ is: $A = 2 \pi \int_1^a \frac{1}{\rho} \sqrt{1+\left( -\frac{1}{\rho^2} \right)^2} \ d\rho \ge 2 \pi \int_1^a \frac{d \rho}{\rho} = 2 \pi \log a.$ As the right hand side of inequality above diverges to $$\infty$$ as $$a \to \infty$$, we can conclude that the area of Gabriel’s horn is infinite.

### Conclusion

Gabriel’s horn could be filled with a finite quantity of paint… therefore painting a surface with infinite area. Unfortunately the thickness of the paint coat is converging to $$0$$ as $$z$$ goes to $$\infty$$, leading to a paint which won’t be too visible!

# A normal subgroup that is not a characteristic

Let’s $$G$$ be a group. A characteristic subgroup is a subgroup $$H \subseteq G$$ that is mapped to itself by every automorphism of $$G$$.

An inner automorphism is an automorphism $$\varphi \in \mathrm{Aut}(G)$$ defined by a formula $$\varphi : x \mapsto a^{-1}xa$$ where $$a$$ is an element of $$G$$. An automorphism of a group which is not inner is called an outer automorphism. And a subgroup $$H \subseteq G$$ that is mapped to itself by every inner automorphism of $$G$$ is called a normal subgroup.

Obviously a characteristic subgroup is a normal subgroup. The converse is not true as we’ll see below.

### Example of a direct product

Let $$K$$ be a nontrivial group. Then consider the group $$G = K \times K$$. The subgroups $$K_1=\{e\} \times K$$ and $$K_2=K \times \{e\}$$ are both normal in $$G$$ as for $$(e, k) \in K_1$$ and $$(a,b) \in G$$ we have
$(a,b)^{-1} (e,x) (a,b) = (a^{-1},b^{-1}) (e,x) (a,b) = (e,b^{-1}xb) \in K_1$ and $$b^{-1}K_1 b = K_1$$. Similar relations hold for $$K_2$$. As $$K$$ is supposed to be nontrivial, we have $$K_1 \neq K_2$$.

The exchange automorphism $$\psi : (x,y) \mapsto (y,x)$$ exchanges the subgroup $$K_1$$ and $$K_2$$. Thus, neither $$K_1$$ nor $$K_2$$ is invariant under all the automorphisms, so neither is characteristic. Therefore, $$K_1$$ and $$K_2$$ are both normal subgroups of $$G$$ that are not characteristic.

When $$K = \mathbb Z_2$$ is the cyclic group of order two, $$G = \mathbb Z_2 \times \mathbb Z_2$$ is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group.

### Example on the additive group $$\mathbb Q$$

Consider the additive group $$(\mathbb Q,+)$$ of rational numbers. The map $$\varphi : x \mapsto x/2$$ is an automorphism. As $$(\mathbb Q,+)$$ is abelian, all subgroups are normal. However, the subgroup $$\mathbb Z$$ is not sent into itself by $$\varphi$$ as $$\varphi(1) = 1/ 2 \notin \mathbb Z$$. Hence $$\mathbb Z$$ is not a characteristic subgroup.