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.

# A non complete normed vector space

Consider a real normed vector space $$V$$. $$V$$ is called complete if every Cauchy sequence in $$V$$ converges in $$V$$. A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like $$(\ell_p, \Vert \cdot \Vert_p)$$ the space of real sequences endowed with the norm $$\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}$$ for $$p \ge 1$$, the space $$(C(X), \Vert \cdot \Vert)$$ of real continuous functions on a compact Hausdorff space $$X$$ endowed with the norm $$\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert$$ or the Lebesgue space $$(L^1(\mathbb R), \Vert \cdot \Vert_1)$$ of Lebesgue real integrable functions endowed with the norm $$\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx$$.

Let’s give an example of a non complete normed vector space. Let $$(P, \Vert \cdot \Vert_\infty)$$ be the normed vector space of real polynomials endowed with the norm $$\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert$$. Consider the sequence of polynomials $$(p_n)$$ defined by
$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.$ For $$m < n$$ and $$x \in [0,1]$$, we have $\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}$ which proves that $$(p_n)$$ is a Cauchy sequence. Also for $$x \in [0,1]$$ $\lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.$ As uniform converge implies pointwise convergence, if $$(p_n)$$ was convergent in $$P$$, it would be towards $$p$$. But $$p$$ is not a polynomial function as none of its $$n$$th-derivative always vanishes. Hence $$(p_n)$$ is a Cauchy sequence that doesn't converge in $$(P, \Vert \cdot \Vert_\infty)$$, proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

# Uniform continuous function but not Lipschitz continuous

Consider the function $\begin{array}{l|rcl} f : & [0,1] & \longrightarrow & [0,1] \\ & x & \longmapsto & \sqrt{x} \end{array}$

$$f$$ is continuous on the compact interval $$[0,1]$$. Hence $$f$$ is uniform continuous on that interval according to Heine-Cantor theorem. For a direct proof, one can verify that for $$\epsilon > 0$$, one have $$\vert \sqrt{x} – \sqrt{y} \vert \le \epsilon$$ for $$\vert x – y \vert \le \epsilon^2$$.

However $$f$$ is not Lipschitz continuous. If $$f$$ was Lipschitz continuous for a Lipschitz constant $$K > 0$$, we would have $$\vert \sqrt{x} – \sqrt{y} \vert \le K \vert x – y \vert$$ for all $$x,y \in [0,1]$$. But we get a contradiction taking $$x=0$$ and $$y=\frac{1}{4 K^2}$$ as $\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert$