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The notion of a [math]\sigma[/math]-Algebra is related to one of the most general constructions, that of a topology. To deal with Euclidean spaces and for the description of a natural notion of [math]\sigma[/math]-Algebra, we need to take a closer look at the topological point of view. Therefore we want to describe some point set aspects of topology, basically also introduced in analysis.

Definition (Topological space)

Let [math]X[/math] be a set. A topology on X is a family [math]\mathcal{O}[/math] of subsets of [math]X[/math] satisfying:

• The ground set is in [math]\mathcal{O}[/math], i.e. [math]X\in\mathcal{O}[/math],
• The empty set is in [math]\mathcal{O}[/math], i.e. [math]\varnothing\in\mathcal{O}[/math],
• (finite intersection) For [math]O_1,...,O_n\in\mathcal{O}[/math] with [math]n\geq1[/math] we get that [math]\bigcap_{i=1}^{n}O_i\in\mathcal{O}[/math],
• (arbitrary union) For [math]O_i\in\mathcal{O}[/math] with [math]i\in I[/math], where [math]I[/math] is any index set [math]I[/math], we get that [math]\bigcup_{i\in I}O_i\in\mathcal{O}[/math].

The elements of [math]\mathcal{O}[/math] are called open sets and the complements are called closed sets. Moreover, we call the tupel [math](X,\mathcal{O})[/math] a topological space.

[Hausdorff]

Let [math](X,\mathcal{O})[/math] be a topological space. The topology [math]\mathcal{O}[/math] on [math]X[/math] is said to be Hausdorff if and only if for all [math]x,y\in X[/math] with [math]x\not=y[/math] there is a [math]O_x\in\mathcal{O}[/math] with [math]x\in O_x[/math] and there exists a [math]O_y\in\mathcal{O}[/math] with [math]y\in O_y[/math], i.e. [math]O_x\cap O_y=\varnothing[/math].

Definition (Metric space)

Let [math]X[/math] be a set. A metric is a map [math]d:X\times X\longrightarrow \mathbb{R}_{+}[/math], which, for all [math]x,y,z\in X[/math], satisfies the following.

• (zero distance) [math]d(x,y)=0\Longleftrightarrow x=y,[/math]
• (symmetry) [math]d(x,y)=d(y,x),[/math]
• (triangle inequality) [math]d(x,y)\leq d(x,z)+d(z, y),[/math]

We call a set [math]X[/math] endowed with a metric, written [math](X,d)[/math], a metric space. If we have a norm [math]\|\cdot\|[/math] on [math]X[/math], and we then consider a normed space [math](X,\|\cdot\|)[/math], we get the relation [math]d(x,y):=\|x-y\|[/math], which defines a distance. If [math](X,d)[/math] is a metric space, the topology on [math]X[/math] is associated with [math]d[/math] and is, for some arbitrary index set [math]I[/math], given by

[[math]] \mathcal{O}_X^d:=\left\{\bigcup_{i\in I}B_{r_i}(x_i)\mid x_i\in X, r_i\in\mathbb{R}_{+}\right\},B_{r_i}(x_i):=\left\{y\in X\mid d(x_i,y) \lt r_i\right\} [[/math]]

Definition (Basis and separability)

A topological space [math](X,\mathcal{O})[/math] is said to have a countable basis of open sets [math]\{w_n\}_{n\in\N}[/math] if for every open set [math]O\in\mathcal{O}[/math], there exists a countable index set [math]I\subset\mathbb{N}[/math], such that

[[math]] O=\bigcup_{n\in I}w_n. [[/math]]

Moreover, a metric space [math](X,d)[/math] is said to be separable, if it contains a sequence [math](x_n)_{n\in\mathbb{N}}[/math] which is dense in [math]X[/math], that is, for all [math]x\in X[/math] there exists a subsequence [math](x_{n_k})_{k\in\N}[/math] of [math](x_n)_{n\in N}[/math], such that [math]d(x,x_{n_k})\xrightarrow{k\to\infty}0[/math].

Proposition

A metric space is separable if and only if it has a countable basis of open sets.

Show Proof

We first prove the direction [math]\Longrightarrow[/math]. Therefore we can observe that [math]\mathscr{B}:=\{B_r(x_n)\}_{r_\in\Q_+,n\in\N}[/math] is a basis of open sets, and thus we can write every open set as [math]O=\bigcup_{b\in\mathscr{B}\atop b\subset O}b[/math]. Now let [math]x\in O[/math]. Then there exists an [math]\varepsilon \gt 0[/math], such that [math]B_\varepsilon(x)\subset O[/math] and there is a [math]n_0[/math] with [math]d(x_{n_0},x) \lt \frac{\varepsilon}{4}[/math] and thus [math]x\in B_{\frac{\varepsilon}{2}}(x_{n_0})\subset O[/math]. Now we prove the direction [math]\Longleftarrow[/math]. Let therefore [math]\{w_n\}_{n\in\N}[/math] of open sets. Then we can choose a [math]x_n\in w_n[/math] and check that the sequence [math](x_n)_{n\in \N}[/math] is a dense subset, which gives the claim.

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Definition (Product topology)

Let [math](X,\mathcal{O}_X)[/math] and [math](Y,\mathcal{O}_Y)[/math] be two topological spaces. The product topology for the product space [math]X\times Y[/math] is defined, with an arbitrary index set [math]I[/math], by the family of open sets

[[math]] \mathcal{O}_{X\times Y}:=\left\{\bigcup_{i\in I}O^X_i\times \mathcal{O}^Y_i\mid O^X\in \mathcal{O}_X,O^Y\in\mathcal{O}_Y\right\}. [[/math]]

Definition (Continuity)

Let [math](X,\mathcal{O}_X)[/math] and [math](Y,\mathcal{O}_Y)[/math] be two topological spaces. A map [math]f:(X,\mathcal{O}_X)\longrightarrow(Y,\mathcal{O}_Y)[/math] is continuous if and only if for all [math]O^Y \in\mathcal{O}_Y[/math], the image of [math]O^Y[/math] under [math]f^{-1}[/math] is open, that is

[[math]] f^{-1}(O^Y)=\{x\in X\mid f(x)\in O^Y\}\in\mathcal{O}_X [[/math]]

Definition (Canonical projection)

Let [math]X[/math] and [math]Y[/math] be two sets. Then we can define the canonical projections to be the surjective maps

[[math]] \begin{align*} \pi_X: X\times Y &\longrightarrow X\\ (x,y)&\longmapsto x \end{align*} [[/math]]

[[math]] \begin{align*} \pi_Y:X\times Y&\longrightarrow Y\\ (x,y)&\longmapsto y \end{align*} [[/math]]

A useful observation is that the product topology is defined in such a way that the canonical projections are continuous, that is

[[math]] \pi_X^{-1}(O^X)=O^X\times Y\in \mathcal{O}_{X\times Y},\pi_Y^{-1}(O^Y)=X\times O^Y\in\mathcal{O}_{X\times Y} [[/math]]

We can also define a metric on two metric spaces [math](X,d)[/math] and [math](Y,\delta)[/math] given by

[[math]] D_p((x,y),(x',y'))=(d^p(x,x')+\delta^p(y,y'))^{\frac{1}{p}}, [[/math]]

for [math]p\geq 1[/math].

Proposition

Let [math](X,\mathcal{O}_X)[/math] and [math](\mathcal{O}_Y)[/math] be two topological spaces. If [math](X,\mathcal{O}_X)[/math] and [math](Y,\mathcal{O}_Y)[/math] have a countable basis of open sets, then [math](X\times Y,\mathcal{O}_{X\times Y})[/math] also has a countable basis of open sets. Moreover, Let [math](X,d)[/math] and [math](Y,\delta)[/math] be two metric spaces. If [math](X,d)[/math] and [math](Y,\delta)[/math] are separable, then [math](X\times Y, D_p)[/math] is also separable.

Show Proof

First, let [math]\mathcal{U}_X=\{U_n\}_{n\in\N}[/math] be a basis of open sets on [math]X[/math] and [math]\mathcal{V}_Y=\{V_n\}_{n\in\N}[/math] a basis of open sets on [math]Y[/math]. Then [math]\{U_n\times V_m\}_{(n,m)\in\N^2}[/math] is a basis of open sets for [math]X\times Y[/math], which proves the first claim. We leave the second claim as an exercise for the reader.

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General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].