# Cantor set: a null set having the cardinality of the continuum

## Definition of the Cantor set

The Cantor ternary set (named Cantor set below) $$K$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$K_0=I$$
• If $$K_n$$ is a finite disjoint union of segments $$K_n=\cup_k \left[a_k,b_k\right]$$, $K_{n+1}=\bigcup_k \left(\left[a_k,a_k+\frac{b_k-a_k}{3}\right] \cup \left[a_k+2\frac{b_k-a_k}{3},b_k\right]\right)$

And finally $$K=\displaystyle \bigcap_{n \in \mathbb{N}} K_n$$. The Cantor set is created by repeatedly deleting the open middle third of a set of line segments starting with the segment $$I$$.

The Cantor set is a closed set as it is an intersection of closed sets. Continue reading Cantor set: a null set having the cardinality of the continuum

# A module without a basis

Let’s start by recalling some background about modules.

Suppose that $$R$$ is a ring and $$1_R$$ is its multiplicative identity. A left $$R$$-module $$M$$ consists of an abelian group $$(M, +)$$ and an operation $$R \times M \rightarrow M$$ such that for all $$r, s \in R$$ and $$x, y \in M$$, we have:

1. $$r \cdot (x+y)= r \cdot x + r \cdot y$$ ($$\cdot$$ is left-distributive over $$+$$)
2. $$(r +s) \cdot x= r \cdot x + s \cdot x$$ ($$\cdot$$ is right-distributive over $$+$$)
3. $$(rs) \cdot x= r \cdot (s \cdot x)$$
4. $$1_R \cdot x= x$$

$$+$$ is the symbol for addition in both $$R$$ and $$M$$.
If $$K$$ is a field, $$M$$ is $$K$$-vector space. It is well known that a vector space $$V$$ is having a basis, i.e. a subset of linearly independent vectors that spans $$V$$.
Unlike for a vector space, a module doesn’t always have a basis. Continue reading A module without a basis

# An unbounded convex not containing a ray

We consider a normed vector space $$E$$ over the field of the reals $$\mathbb{R}$$ and a convex subset $$C \subset E$$.

We suppose that $$0 \in C$$ and that $$C$$ is unbounded, i.e. there exists points in $$C$$ at distance as big as we wish from $$0$$.

The following question arises: “does $$C$$ contains a ray?”. It turns out that the answer depends on the dimension of the space $$E$$. If $$E$$ is of finite dimension, then $$C$$ always contains a ray, while if $$E$$ is of infinite dimension $$C$$ may not contain a ray. Continue reading An unbounded convex not containing a ray

# A compact whose convex hull is not compact

We consider a topological vector space $$E$$ over the field of the reals $$\mathbb{R}$$. The convex hull of a subset $$X \subset E$$ is the smallest convex set that contains $$X$$.

The convex hull may also be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X.

The convex hull of $$X$$ is written as $$\mbox{Conv}(X)$$. Continue reading A compact whose convex hull is not compact

# A vector space not isomorphic to its double dual

In this page $$\mathbb{F}$$ refers to a field. Given any vector space $$V$$ over $$\mathbb{F}$$, the dual space $$V^*$$ is defined as the set of all linear functionals $$f: V \mapsto \mathbb{F}$$. The dual space $$V^*$$ itself becomes a vector space over $$\mathbb{F}$$ when equipped with the following addition and scalar multiplication:
$\left\{ \begin{array}{lll}(\varphi + \psi)(x) & = & \varphi(x) + \psi(x) \\ (a \varphi)(x) & = & a (\varphi(x)) \end{array} \right.$ for all $$\phi, \psi \in V^*$$, $$x \in V$$, and $$a \in \mathbb{F}$$.
There is a natural homomorphism $$\Phi$$ from $$V$$ into the double dual $$V^{**}$$, defined by $$(\Phi(v))(\phi) = \phi(v)$$ for all $$v \in V$$, $$\phi \in V^*$$. This map $$\Phi$$ is always injective. Continue reading A vector space not isomorphic to its double dual

# A field that can be ordered in two distinct ways

For a short reminder about ordered fields you can have a look to following post. We prove there that $$\mathbb{Q}$$ can be ordered in only one way.

That is also the case of $$\mathbb{R}$$ as $$\mathbb{R}$$ is a real-closed field. And one can prove that the only possible positive cone of a real-closed field is the subset of squares.

However $$\mathbb{Q}(\sqrt{2})$$ is a subfield of $$\mathbb{R}$$ that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways

# An infinite field that cannot be ordered

## Introduction to ordered fields

Let $$K$$ be a field. An ordering of $$K$$ is a subset $$P$$ of $$K$$ having the following properties:

ORD 1
Given $$x \in K$$, we have either $$x \in P$$, or $$x=0$$, or $$-x \in P$$, and these three possibilities are mutually exclusive. In other words, $$K$$ is the disjoint union of $$P$$, $$\{0\}$$, and $$-P$$.
ORD 2
If $$x, y \in P$$, then $$x+y$$ and $$xy \in P$$.

We shall also say that $$K$$ is ordered by $$P$$, and we call $$P$$ the set of positive elements. Continue reading An infinite field that cannot be ordered

# A function that is everywhere continuous and nowhere differentiable

Let $$f_1(x) = |x|$$ for $$| x | \le \frac{1}{2}$$, and let $$f_1$$ be defined for other values of $$x$$ by periodic continuation with period $$1$$. $$f_1$$ graph looks like following picture:

$$f_1$$ is continuous everywhere and differentiable on $$\mathbb{R} \setminus \mathbb{Z}$$. Continue reading A function that is everywhere continuous and nowhere differentiable

# A compact convex set whose extreme points set is not close

Let’s remind that an extreme point $$c$$ of a convex set $$C$$ in a real vector space $$E$$ is a point in $$C$$ which does not lie in any open line segment joining two points of $$C$$.

## The specific case of dimension $$2$$

Proposition: when $$C$$ is closed and its dimension is equal to $$2$$, the set $$\hat{C}$$ of its extreme points is closed.
Continue reading A compact convex set whose extreme points set is not close