We aim at defining a continuous function \(\varphi : [0,1] \rightarrow [0,1]^2\). At first sight this looks quite strange.

Indeed, \(\varphi\) cannot be a bijection. If \(\varphi\) would be bijective, it would also be an homeomorphism as a continuous bijective function from a compact space to a Haussdorff space is an homeomorphism. But an homeomorphism preserves connectedness and \([0,1] \setminus \{1/2\}\) is not connected while \([0,1]^2 \setminus \{\varphi(1/2)\}\) is.

Nor can \(\varphi\) be piecewise continuously differentiable as the Lebesgue measure of \(\varphi([0,1])\) would be equal to \(0\).

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 \(I\).

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:

\(r \cdot (x+y)= r \cdot x + r \cdot y\) (\( \cdot\) is left-distributive over \(+\))

\((r +s) \cdot x= r \cdot x + s \cdot x\) (\( \cdot\) is right-distributive over \(+\))

\((rs) \cdot x= r \cdot (s \cdot x)\)

\(1_R \cdot x= x \)

\(+\) 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→

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→

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.

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→

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.

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 \(-P\).

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: