Norms, metrics, topology

Topological space
A topological space is the fundamental subject of the subdiscipline topology of mathematics. By introducing a topological structure on a set, it is possible to establish.


 * intuitive positional relations like "proximity" and
 * "convergence against" from the real numbers $${\mathbb R}$$ or from the $${\mathbb R}^{n}$$, respectively.

to many and very general structures (such as the topology of function spaces).

Definition: topology
A topology is a set system $$\cal T$$ consisting of subsets (open sets) of a basic set $$X$$ for which the following axioms are satisfied.

A set $$X$$ together with a topology $$\cal T$$ on $$X$$ is called topological space $$(X,\cal T)$$.
 * (T1) $$\emptyset, X \in \cal T$$
 * (T2) $$U \cap V \in \cal T$$ for all $$U,V \in \cal T$$.
 * (T3) For any index set $$I$$ and $$U_i \in \cal T$$ for all $$i \in I$$ holds: $$\bigcup_{i \in I} U_i \in \cal T$$.

Remark - closed sets
By defining all open sets in $$X$$ by the topology $$\mathcal{T}$$, all closed sets are also defined as complements of an open set $$U \in \mathcal{T}$$.
 * $$ A\subseteq X \mbox{ closed } :\Longleftrightarrow \exists_{U \in \mathcal{T}} : \, A = X\setminus U = U^c$$

Definition - open kernel of a set
Let $$M \subseteq X$$ be in the topological space $$(X,\cal T)$$, then the open core $$\stackrel{\circ}{M}$$ is defined as the "largest" open set contained in $$M$$:
 * $$\stackrel{\circ}{M} := \bigcup_{U\in \mathcal{T}, U\subseteq M} U $$

Remark - open core
Since in the definition $$\stackrel{\circ}{M} $$ is represented as the union of open sets $$U \in \mathcal{T}$$, $$\stackrel{\circ}{M} \in \mathcal{T}$$ open by axiom (T3).

Definition - connection of a set
Let $$M \subseteq X$$ be in the topological space $$(X,\cal T)$$, then the closure $$\overline{M}$$ is defined as the "smallest" closed set containing $$M$$:
 * $$\overline{M} := \bigcap_{U\in \mathcal{T}, U^c\supseteq M} U^c $$

Remark - open core
Since in the definition $$\stackrel{\circ}{M} $$ is represented as the intersection of closed sets $$U^c $$ with $$U\in \mathcal{T}$$, $$\overline{M}$$ is again open as the completion of any union of open sets, since it holds again according to (T3):
 * $$\overline{M} :=

\bigcap_{U\in \mathcal{T}, U^c\supseteq M} U^c =.

\bigg( \underbrace{\bigcup_{U\in \mathcal{T}, U^c\supseteq M} U }_{\in \mathcal{T}} \bigg)^c $$

Definition - Boundary of a set
Let $$M \subseteq X$$ be a set in the topological space $$(X,\cal T)$$, then the edge of a set from the closure of the set $$M$$ without the open core $$\stackrel{\circ}{M}$$ of $$M$$. The boundary is therefore defined as follows:
 * $$\partial{M} := \overline{M} \setminus \stackrel{\circ}{M} $$.

Remark
The sets $$X,\emptyset$$ are by definition open and mutually the two sets form the complements of each other. Thus these two sets are both open and closed at the same time. Therefore, the two sets have no boundary points.

Definition - Neighbourhood
Let $$x_o\in X$$ and $$U \subseteq X$$ be a set in a topological space $$(X,\cal T)$$, then the $$U$$ is called Neighbourhood of $$x_o$$ if there exists an open set $$\stackrel{\circ}{U}\in \cal T$$ with:
 * $$x_o \in \stackrel{\circ}{U} \subseteq U$$

The set of all Neighbourhood of $$x_o$$ with respect to the topology $$\mathcal{T}$$ is denoted by $$ \mathfrak{U}_{\mathcal{T}}(x_o)$$. $$\stackrel{\circ}{\mathfrak{U}}_{\mathcal{T}}(x_o)$$ denotes the set of all open neighbourhoods of $$x_o$$.

Definition - Neighbourhood basis
Let $$x_o\in X$$ and $$\cal{B} \subseteq \wp(X)$$ a set system in a topological space $$(X,\cal T)$$, then the \mathcal{B} is called the neighbourhood basis of $$x_o$$ if the following properties hold:
 * (B1) $$\mathcal{B} \subseteq {\mathfrak{U}}_{\mathcal{T}}(x_o)$$
 * (B2) for each environment $$U \in {\mathfrak{U}}_{\mathcal{T}}(x_o)$$ a neighborhood $$ B\in \mathcal{B}$$ with $$B \subseteq U $$.

Example
The set of open $$\varepsilon$$ neighbourhood $$(x_o-\varepsilon,x_o+\varepsilon) \subset \mathbb{R}$$ in $$(\mathbb{R}, |\cdot |)$$ with the Euclidean topology $$|\cdot |$$ generated by the amount $$\mathcal{T}_{\mathbb{R}}$$ is an neighbourhood basis of $$ {\mathfrak{U}}_{\mathcal{T}_{\mathbb{R}}}(x_o)$$.

Remark
The neighbourhood basis term helps to prove convergence statements for the neighbourhood basis only, and thus to obtain the statements for arbitrary neighbourhoods as well. In calculus one uses $$\varepsilon$$-neighbourhoods in definitions without explicitly addressing the topological aspect of the neighbourhood basis, that proofs in general arbitrary neighbourhoods and not only for the neighbourhood basis. Due to the fact that an arbitrary neighbourhood contains a set of the neighbourhood basis, the most of the convergence proofs can limit themselves to the neighbourhood basis.

Definition - base of topology
Let $$x_o\in X$$ and $$\cal{B} \subseteq \wp{(X)}$$ be a set system in a topological space $$(X,\cal T)$$, then the $$\mathcal{B}_{\mathcal{T}}$$ is called the basis of $$\mathcal{T}$$ if holds:
 * (BT1) $$\mathcal{B}_{\mathcal{T}} \subseteq {\mathcal{T}}$$
 * (BT2) for every open set $$U \in \mathcal{T}$$ there is an open set $$ B\in \mathcal{B}_{\mathcal{T}}$$ with $$B \subseteq U $$.

Example
The set of all open intervals $$(a,b) \subset \mathbb{R}$$ is a basis of the topology in the topological $$(\mathbb{R}, |\cdot |)$$ with the Euclidean topology $$|\cdot |$$ generated by the amount $$\mathcal{T}_{\mathbb{R}}$$.

Convergence in topological spaces
In calculus, the convergence of sequences is a central definition to define notions based on it, such as continuity, difference, and integrals. Sequences $$ (a_n)_{n\in \mathbb{N}} $$ with $$\mathbb{N}$$ as index set are unsuitable to define convergence in general topological spaces, because the index set $$\mathbb{N}$$ is not powerful enough concerning the neighbourhood basis. This is only possible if the topological space has a countable neighbourhood basis. Therefore one goes over either to Meshes or Filters

Example: topology on texts
Usually, one assumes that topologies are defined on mathematical spaces (e.g., number spaces, function spaces, (topological) groups, vector spaces, ...). However, the generality of the definition makes it also possible to define a topology on texts. This example was added because purely descriptively, e.g., texts in the German language This similarity of semantics, or syntax, is explored in more detail as an exercise in "Topology on Texts".
 * can have a similar statement and
 * use different words.

Describe similarity of words by metrics
From spoken words, represent the number of letters and the set of occurring letters as a table. How can you derive a distance of words from the tabulated list. make a suggestion for this. What are the properties of your proposed distance function. Is it a metric on the space of words?

Task - distance between words

 * Consider the words "bucket", "buket", "buckett". How can you express the differences of the words by a metric
 * Phonetic similarity words "bucket" and "pucket" have a phonetic similarity, but from the sequence of letters the spellings differ greatly. How can you notate similarity of spoken words (Speech Recognition) by a phonetic notation and in this notation of phonemes express a similarity of words as well.

Meaning of Properties topology

 * (T1) $$\emptyset, X \in \mathcal{T}$$ empty set and the basic set $$X$$ are open sets.
 * (T2) $$U \cap V \in \mathcal{T}$$ for all $$U,V \in \mathcal{T}$$: the average of finitely many open sets is an open set.
 * (T3) The union of any many open sets is again an open set.

Semantics: metric
A metric $$d$$ associates with $$d(x,y)$$ two elements $$x,y \in X$$ from a base space $$X$$ the distance $$d(x,y)$$ between $$x$$ and $$y$$.

Definition: Metric
Let $$X$$ be an arbitrary set. A mapping $$d\colon X\times X\to \mathbb{R}$$ is called a metric on $$X$$ if for any elements $$x$$, $$y$$ and $$z$$ of $$X$$ the following axioms are satisfied:


 * (M1) separation: $$d\left(x,y\right) = 0 \Leftrightarrow x = y$$,
 * (M2) symmetry: $$d\left(x,y\right) = d(y,x)$$,
 * (M3) triangle inequality: $$d\left(x,y\right) \leq d(x,z) + d(z,y)$$.

Illustration: metric triangle inequality


According to the triangle inequality, the distance between two points X,Y is at most as large as sum of the distances from X to Z and from Z to Y, that is, a detour via the point Z

Non-negativity
Non-negativity follows from the three properties of the metric, i.e. for all $$x,y \in X$$ holds. $$d(x,y)\geq0$$. The non-negativity follows from the other properties with:


 * $$0 = \frac{1}{2} d(x, x) \leq \frac{1}{2}(d(x, y) + d(y, x)) = $$.

$$= \frac{1}{2}(d(x, y) + d(y, x)) = \frac{1}{2}(d(x, y) + d(x, y)) = d(x, y).$$

Open sets in metric spaces

 * In a metric space $$(X,d)$$, one defines a set $$U \subset X$$ to be open (i.e. $$U \in \mathcal{T}_d$$) if for every $$u\in U$$ there is a $$\epsilon >0$$ that the $$\epsilon$$-sphere $$B_\epsilon^d(u): =\{\ x \in X | \ d(x,u)< \epsilon \}$$ lies entirely in $$U$$ (i.e. i.e. $$B_\epsilon^d(u) \subset U$$)
 * Show that with this defined $$\mathcal{T}_d$$, the pair $$(X,\mathcal{T}_d)$$ is a topological space (i.e., (T1), (T2), (T3) satisfied).

Norm on vector spaces
A norm is a mapping $$\|\cdot\|$$ from a vector space $$V$$ over the body $$\mathbb K$$ of the real or the complex numbers into the set of nonnegative real numbers $${\mathbb R}_0^{+}$$. Here the norm assigns to each vector $$x \in V$$ its length $$\| x \|$$.

Definition: Norm
Let $$V$$ be a $$\mathbb{K}$$ vector space and $$\mathbb{R}_0^{+}, \; x \mapsto \| x \|,$$ a mapping. If $$\|\cdot\|$$ satisfies the following axioms axioms N1,N2, N3, then $$\|\cdot\|$$ is called a norm on $$V$$.
 * (N1) Definiteness: $$\|x\| = 0 \;\Rightarrow\; x = \mathbb{0}_V$$ for all $$x \in V$$,
 * (N2) absolute homogeneity: $$\|\lambda\cdot x\| = |\lambda|\cdot\|x\|$$ for all $$x\in V$$ and $$\lambda\in \mathbb K$$.
 * (N3) Triangle inequality: $$\|x + y\| \leq \|x\| + \|y\|$$ for all $$x, y\in V$$.

Remark: N1
The property (N1) is actually an equivalence and it holds in any normed space. If $$\mathbf{0}_V \in V$$ is the zero vector in $$V$$ and $$0 \in \mathbb{K}$$ is the zero in the field $$\mathbb{K}$$, if $$V$$ is a $$\mathbb{K}$$ vector space).
 * (N1)' definiteness: $$\|x\| = 0 \;\Leftrightarrow\; x = \mathbf{0}_V$$ for all $$x \in V$$,
 * Since one uses a minimality principle for definitions for the defining property, one would not use a stronger formulation (N1)' in the definition for (N1), since the equivalence from the defining properties of the norm follow the properties of the vector space already for any normed space.
 * $$x\equiv \mathbf{0}_V \Rightarrow \|x\| = \|0\cdot \mathbf{0}_V\| = |0| \cdot \|mathbf{0}_V\| = 0 $$

Normed space / Metric space
A normed space $$(V,\|\cdot\| )$$ is also a metric space.
 * A norm $$\|\cdot\| $$ assigns to a vector $$v\in V $$ its vector length $$\|v\| $$.
 * The norm $$\|\cdot\| $$ can be used to define a metric via $$d(x,y):=\| x-y \| $$ that specifies the distance between $$x$$ and $$y$$.

Learning Task: generate metric from given norm
Let $$(V,\|\cdot\| )$$ be a normed space with norm $$\|\cdot\| : V \to \mathbb{R}_o^+$$. Show that the defined mapping $$d:V\times V \rightarrow \mathbb{R}$$ with $$d(x,y):=\| x-y \| $$ satisfies the properties of a metric.

Notation: norm

 * In the axiom (N2) $$ \| \lambda \cdot x \| = | \lambda | \cdot \|x\| $$, $$| \cdot |$$ denotes the amount of the scalar. "$$\cdot$$" sign: Outer linkage in vector space or multiplication $$(\mathbb{R},\cdot)$$.
 * $$\|x\|$$ indicates the length of the vector $$x\in V$$.
 * In (N3) $$\|x + y\| \leq \|x\| + \|y\|$$ for all $$x, y\in V$$. '"$$+$$"-sign denotes two distinct links (i.e., addition in $$(V,+)$$ and $$(\mathbb{R},+)$$, respectively.

Def: convergence in normalized space
Let $$(V,\|\cdot\|)$$ be a normalized space and $$(v_n)_{n\in\mathbb{N}} \in V^{\mathbb{N}}$$ a sequence in $$V$$ and $$v_o \in V$$:
 * $$ \lim_{n \to \infty}^{\|\cdot\|} v_n = v_o \ :\Longleftrightarrow \ \forall_{\epsilon > 0} \exists_{n_\epsilon \in \mathbb{N}} \forall_{n \geq n_\epsilon} \ : \ \|v_n - v_o\| < \epsilon $$

Def: convergence in metric space
Let $$(X,d)$$ be a metric space and $$(x_n)_{n\in\mathbb{N}} \in X^{\mathbb{N}}$$ a sequence in $$X$$ and $$x_o \in X$$:
 * $$ \lim_{n \to \infty}^d x_n = x_o \ :\Longleftrightarrow \ \forall_{\epsilon > 0} \exists_{n_\epsilon \in \mathbb{N}} \forall_{n \geq n_\epsilon} \ : \ d(x_n,x_o) < \epsilon $$

Def: Cauchy sequences in metric spaces
Let $$(X,d)$$ be a metric space and $$(x_n)_{n\in\mathbb{N}} \in X^{\mathbb{N}}$$ a sequence in $$X$$. $$(x_n)_{n\in\mathbb{N}}$$ is called a Cauchy sequence in $$X^{\mathbb{N}}$$:
 * $$ \forall_{\epsilon > 0} \exists_{n_\epsilon \in \mathbb{N}} \forall_{m,n \geq n_\epsilon} \ : \ d(x_n,x_m) < \epsilon $$

Equivalence: norms
Let two norms $$ \|\cdot\|_1 $$ and $$ \|\cdot \|_2 $$ be given on the $$\mathbb K$$ vector space $$V$$. The two norms are equivalent if holds:


 * $$\exists_{C_1,C_2 >0} \forall_{x \in V} \ : \ C_1 \|x\|_1 \leq \|x\|_2 \leq C_2 \|x\|_1$$.

Show that a sequence converges in $$\|\cdot\|_1 $$ exactly if it also converges with respect to $$\|\cdot\|_2$$.

Absolute value in complex numbers
Let $$z=z_1 + i\cdot z_2\in \mathbb{C}$$ be a complex number with $$z_1, z_2\in \mathbb{R}$$. Show that $$|z|:=\sqrt{z\cdot \overline{z}} $$ is a norm on the $$\mathbb{R}$$ vector space $$\mathbb{C}$$!

Historical Notes: Norm
This axiomatic definition of norm was established by Stefan Banach in his 1922 dissertation. The norm symbol in use today was first used by Erhard Schmidt in 1908 as the distance $$\|x-y\|$$ between vectors $$x$$ and $$y$$.

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Normen, Metriken, Topologie