Cardinality, topology and measurable spaces

The lecture recalls how the sizes of infinite sets are compared and the basic notions of topology, and then begins real analysis with measurable spaces.

Recap of set theory

The recap follows Hrbacek and Jech. Its aim is to distinguish between sets with infinitely many elements.

Notation. $A \subset B$ means that $A$ is contained in $B$ or equal to it, while $A \subsetneq B$ means that $A$ is strictly contained in $B.$

Cardinality

Definition 1 (Comparison of cardinalities). Let $X$ and $Y$ be sets. We write ${\#X \le \#Y},$ and say that the cardinality of $X$ is less than or equal to the cardinality of $Y,$ if there exists an injective map ${f \colon X \to Y.}$ We write ${\#X = \#Y}$ if there exists a bijective map ${f \colon X \to Y.}$ Finally, ${\#X < \#Y}$ means that ${\#X \le \#Y}$ and ${\#X \neq \#Y.}$

Remark 2. Let $X$ and $Y$ be sets.
(i) If $X$ is nonempty, then ${\#X \le \#Y}$ if and only if there exists a surjective map ${f \colon Y \to X.}$
(ii) (Cantor–Bernstein) If ${\#X \le \#Y}$ and ${\#Y \le \#X,}$ then ${\#X = \#Y.}$

Definition 3 (Countable and uncountable sets). Let $X$ be a set.
(i) $X$ is countable if $X$ is finite or ${\#X = \#\mathbb{N}.}$
(ii) $X$ is uncountable if $X$ is infinite but ${\#\mathbb{N} < \#X.}$

Example 4. The natural numbers, the integers and the rationals have the same cardinality, ${\#\mathbb{N} = \#\mathbb{Z} = \#\mathbb{Q},}$ so $\mathbb{Z}$ and $\mathbb{Q}$ are countable. The real line is uncountable: ${\#\mathbb{R} > \#\mathbb{N}.}$

Definition 5 (Extended real line). The extended real line is the set

$$\overline{\mathbb{R}} = [-\infty, +\infty] := \mathbb{R} \cup \{+\infty\} \cup \{-\infty\},$$

with the following operations in $\overline{\mathbb{R}}{:}$

$$\begin{aligned} &a \in \mathbb{R} & &\Longrightarrow \quad a + \infty = +\infty, \quad a - \infty = -\infty, \\ &a > 0 & &\Longrightarrow \quad a \cdot (+\infty) = +\infty, \quad a \cdot (-\infty) = -\infty, \\ &a < 0 & &\Longrightarrow \quad a \cdot (+\infty) = -\infty, \quad a \cdot (-\infty) = +\infty, \\ & & &\phantom{\Longrightarrow \quad} 0 \cdot (+\infty) = 0 \cdot (-\infty) = 0. \end{aligned}$$

The topology on $\overline{\mathbb{R}}$ is given in Example 16, and its Borel $\sigma$-algebra in Remark 28.

The power set

Definition 6 (Power set). Let $X$ be a set. The power set of $X$ is the family of all subsets of $X,$

$$\mathcal{P}(X) = \{ A : A \subset X \}.$$

Remark 7. If ${\#X = n \in \mathbb{N},}$ then ${\#\mathcal{P}(X) = 2^n.}$

Notation. For a set $X,$ $2^X$ is the set of all functions from $X$ to the set with 2 elements $\{0, 1\}{:}$

$$2^X := \{ f \colon X \to \{0, 1\} \}.$$

Proposition 8. Let $X$ be a set.
(i) ${\#\mathcal{P}(X) = \#2^X.}$
(ii) ${\#\mathcal{P}(X) > \#X.}$ In particular, there exists no biggest infinity.

Proof (complete). (i) Define

$$\phi \colon \mathcal{P}(X) \to 2^X, \qquad A \mapsto f_A \colon X \to \{0, 1\}, \qquad f_A(x) := \begin{cases} 1 & x \in A, \\ 0 & x \notin A. \end{cases}$$

The map $\phi$ is a bijection. Indeed, its inverse sends ${f \in 2^X}$ to the set ${\{ x \in X : f(x) = 1 \}.}$

(ii) Suppose, by contradiction, that ${\#\mathcal{P}(X) \le \#X.}$ By Remark 2 (i), applied to the nonempty set $\mathcal{P}(X),$ there exists a surjective map ${f \colon X \to \mathcal{P}(X).}$ Define

$$U := \{ x \in X : x \notin f(x) \} \subset X.$$

Since $f$ is surjective, there exists ${y \in X}$ with ${f(y) = U.}$ Is ${y \in U}$?
Case 1. If ${y \in U,}$ then ${y \notin f(y)}$ by the definition of $U,$ and ${f(y) = U}$ gives ${y \notin U,}$ which is impossible.
Case 2. If ${y \notin U = f(y),}$ then ${y \in U}$ by the definition of $U,$ which is impossible.

Hence ${\#\mathcal{P}(X) \le \#X}$ is false. To see that this gives ${\#\mathcal{P}(X) > \#X,}$ note that ${x \mapsto \{x\}}$ is injective, so ${\#X \le \#\mathcal{P}(X),}$ and ${\#X = \#\mathcal{P}(X)}$ would give ${\#\mathcal{P}(X) \le \#X.}$ $\blacksquare$

Figure 1 is the bijection $\phi$ of part (i) on a set of 3 elements, and Figure 2 runs the construction of $U$ of part (ii).

A	fA(1)	fA(2)	fA(3)
∅	0	0	0
{3}	0	0	1
{2}	0	1	0
{1}	1	0	0
{2, 3}	0	1	1
{1, 3}	1	0	1
{1, 2}	1	1	0
{1, 2, 3}	1	1	1

1 ∈ A2 ∉ A3 ∈ A

Figure 1. An instance with ${X = \{1, 2, 3\}{:}}$ the ${2^3 = 8}$ subsets $A$ of $X$ and the values of $f_A = \phi(A).$ The buttons choose the elements of $A$ and highlight its row.

Play New f

Figure 2. An instance with ${X = \{1, \dots, 6\}}$ and a map ${f \colon X \to \mathcal{P}(X).}$ Row $f(x)$ is the function $f_{f(x)}$ of part (i), and row $U$ is the opposite of the diagonal, so $U$ and $f(y)$ differ at $y$ for every $y.$ The button draws a new $f.$

Countable sets

Notation. A countable union is written $\bigcup_{n=0}^{\infty} X_n,$ also $\bigcup_n X_n.$ A sequence is written $(x_n),$ and a family of sets is written ${\mathcal{F} = \{ X_\alpha \}_{\alpha \in A}.}$

Proposition 9. (i) ${\#\mathbb{R} = \#\mathcal{P}(\mathbb{N}) > \#\mathbb{N}.}$
(ii) If $X$ and $Y$ are countable, then ${X \times Y = \{ (x, y) : x \in X, y \in Y \}}$ is countable.
(iii) If ${X_0, X_1, X_2, \dots}$ is a sequence of sets such that $X_n$ is countable for every $n,$ then $\bigcup_{n=0}^{\infty} X_n$ is countable: a countable union of countable sets is countable.

Recap of topology

The recap follows Manetti. Topology gives an abstract theory of the open, closed and related subsets of some ambient space.

Topological spaces

Definition 10 (Topology). Let $X$ be a set. A topology on $X$ is a family of subsets ${\tau \subset \mathcal{P}(X)}$ such that:
(i) ${\emptyset, X \in \tau;}$
(ii) for every ${\mathcal{F} = \{ U_\alpha \}_{\alpha \in A} \subset \tau,}$ the union $\bigcup_{\alpha \in A} U_\alpha$ belongs to $\tau;$
(iii) if ${U_1, U_2, \dots, U_k \in \tau,}$ then ${\bigcap_{n=1}^{k} U_n \in \tau.}$
The sets in $\tau$ are called open, and the couple $(X, \tau)$ is called a topological space.

Example 11. On $\mathbb{R}^n$ the Euclidean topology $\tau_E$ is defined as follows. We say that ${U \in \tau_E}$ if for every ${x \in U}$ there exists ${r > 0}$ with ${B_r(x) \subset U.}$ Here $B_r(x)$ is the open ball of centre $x$ and radius $r.$ Two other topologies on $\mathbb{R}^n$ are ${\tau = \{ \emptyset, \mathbb{R}^n \}}$ and ${\tau = \mathcal{P}(\mathbb{R}^n).}$

Definition 12 (Closed set, closure, interior, boundary). Let $(X, \tau)$ be a topological space. A set ${V \subset X}$ is closed if ${X \setminus V}$ is open. For ${V \subset X{:}}$
(i) the closure $\overline{V}$ of $V$ is the smallest closed set containing $V;$
(ii) the interior ${\operatorname{int}(V) = \mathring{V}}$ of $V$ is the biggest open set contained in $V;$
(iii) the boundary of $V$ is ${\partial V = \overline{V} \cap \overline{X \setminus V}.}$

The closure and the interior exist. Indeed, $\overline{V}$ is the intersection of the closed sets containing $V,$ which is closed, and $\mathring{V}$ is the union of the open sets contained in $V.$

V

int(V)

closure of V

∂V

Figure 3. An instance in the plane with the Euclidean topology. The set $V$ contains the solid part of its edge and not the dashed part; its interior, its closure and its boundary.

Definition 13 (Compact set). Let $(X, \tau)$ be a topological space and ${K \subset X.}$ The set $K$ is compact if for every ${\mathcal{F} = \{ U_\alpha \}_{\alpha \in A} \subset \tau}$ such that ${\bigcup_{\alpha \in A} U_\alpha \supset K}$ there exists a finite subfamily ${\tilde{\mathcal{F}} = \{ V_1, \dots, V_m \} \subset \mathcal{F}}$ such that ${\bigcup_{n=1}^{m} V_n \supset K.}$

Play New family

Figure 4. $K$ is a triangle, the disks are open sets whose union contains $K,$ and a finite subfamily is then highlighted whose union still contains $K.$ The button draws a new family.

Definition 14 (Precompact set). A set ${V \subset X}$ is precompact if $\overline{V}$ is compact.

Theorem 15 (Heine–Borel). A set ${K \subset \mathbb{R}^n}$ is compact if and only if $K$ is closed and bounded, that is, contained in a ball $B_r(0).$

Example 16 (Topology on the extended real line). The following subsets of $\overline{\mathbb{R}}$ are open in $\overline{\mathbb{R}}{:}$ the intervals $(a, b)$ with ${a, b \in \mathbb{R};}$ every set that is open in $\mathbb{R},$ which is also defined to be open in $\overline{\mathbb{R}};$ and the sets $[-\infty, a)$ and $(a, +\infty]$ for every ${a \in \mathbb{R},}$ which are also defined to be open.

Figure 5. The three kinds of open set of Example 16 on the extended real line. A filled end belongs to the set, an empty one does not.

Definition 17 (Continuous map). Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be two topological spaces and ${f \colon X \to Y.}$ The map $f$ is continuous if ${f^{-1}(A) \in \tau_X}$ for every ${A \in \tau_Y},$ that is, if the preimage of any open set is open.

Metric spaces

Definition 18 (Metric space). A metric space is a couple $(X, d)$ where $X$ is a set and $d$ is a distance on $X,$ that is, a function ${d \colon X \times X \to [0, +\infty)}$ such that, for all ${x, y, z \in X,}$

$$\begin{aligned} &d(x, y) = 0 \iff x = y, \\ &d(x, y) = d(y, x), \\ &d(x, y) \le d(x, z) + d(z, y). \end{aligned}$$

Definition 19 (Topology induced by a distance). Let $(X, d)$ be a metric space. For ${x \in X}$ and ${r > 0,}$ define

$$B_r(x) := \{ y \in X : d(x, y) < r \}.$$

A set $U$ in $X$ is open if for every ${x \in U}$ there exists ${r > 0}$ such that ${B_r(x) \subset U.}$ These open sets form the topology induced by the distance $d.$

Play

Figure 6. An open set $U$ of the plane with the Euclidean distance, its edge dashed because it does not belong to $U,$ and a ball $B_r(x)$ contained in $U$ around a moving point $x.$

A sequence ${(x_n) \subset X}$ converges to ${x_0 \in X},$ written ${x_n \to x_0,}$ if ${\lim_{n \to \infty} d(x_n, x_0) = 0.}$

Lemma 20. Let $(X, d)$ be a metric space.
(i) A set ${V \subset X}$ is closed if and only if $V$ is sequentially closed: if ${(x_n) \subset V}$ and ${x_n \to x_0,}$ then ${x_0 \in V.}$
(ii) A set ${K \subset X}$ is compact if and only if $K$ is sequentially compact: every ${(x_n) \subset K}$ has a subsequence $(x_{n_k})$ such that ${x_{n_k} \to x_0 \in K.}$
(iii) If $V$ is closed, $K$ is compact and ${V \subset K,}$ then $V$ is compact.

Play

Figure 7. Lemma 20 (i): a closed set $V,$ its edge solid because it belongs to $V,$ and a sequence of points of $V$ that converges to $x_0.$ The limit $x_0$ lies in $V.$

Lemma 21. Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces and ${f \colon X \to Y.}$ The following are equivalent.
(i) $f$ is continuous.
(ii) For every ${x_0 \in X}$ and every ${\varepsilon > 0}$ there exists ${\delta > 0}$ such that ${d_X(x, x_0) < \delta}$ implies ${d_Y(f(x), f(x_0)) < \varepsilon.}$
(iii) For every ${x_0 \in X}$ and every ${(x_n) \subset X}$ with ${x_n \to x_0,}$ that is, ${d_X(x_n, x_0) \to 0,}$ the sequence ${(f(x_n)) \subset Y}$ is such that ${f(x_n) \to f(x_0),}$ that is, ${d_Y(f(x_n), f(x_0)) \to 0.}$

Measurable spaces

Real analysis begins here, following Folland, with the families of sets on which it is built.

σ-algebras

Definition 22 ($\sigma$-algebra, measurable space). Let $X$ be a set. A measurable space is a couple $(X, \mathcal{M})$ where $\mathcal{M}$ is a $\sigma$-algebra, that is:
(i) ${\mathcal{M} \subset \mathcal{P}(X);}$
(ii) ${\emptyset \in \mathcal{M};}$
(iii) for every ${A \in \mathcal{M},}$ ${X \setminus A \in \mathcal{M};}$
(iv) for every ${(A_n) \subset \mathcal{M},}$ ${\bigcup_n A_n \in \mathcal{M}.}$

Example 23. For every set $X,$ $\mathcal{P}(X)$ and ${\mathcal{M} = \{ \emptyset, X \}}$ are $\sigma$-algebras. Most of the times a topology is not a $\sigma$-algebra, for example the Euclidean topology. Indeed, on $\mathbb{R}$ it contains $(0, 1)$ but not ${\mathbb{R} \setminus (0, 1).}$

Remark 24. Let $(X, \mathcal{M})$ be a measurable space.
(i) If ${E, F \in \mathcal{M},}$ then ${E \cap F \in \mathcal{M}}$ and ${E \setminus F \in \mathcal{M}.}$
(ii) If ${(E_n) \subset \mathcal{M},}$ then ${\bigcap_n E_n \in \mathcal{M}.}$

Proof (sketch). Intersections are written through complements and unions:

$$E \cap F = X \setminus \big( X \setminus (E \cap F) \big) = X \setminus \big( (X \setminus E) \cup (X \setminus F) \big).$$

Details. The sets ${X \setminus E}$ and ${X \setminus F}$ belong to $\mathcal{M}$ by (iii) of Definition 22, their union by (iv) applied to ${X \setminus E, X \setminus F, \emptyset, \emptyset, \dots,}$ and its complement by (iii) again. Then ${E \setminus F = E \cap (X \setminus F),}$ and in the same way ${\bigcap_n E_n = X \setminus \bigcup_n (X \setminus E_n).}$ $\blacksquare$

X \ E

Next step (1 of 4)

Figure 8. The identity of the proof, term by term. The button shades ${X \setminus E,}$ then ${X \setminus F,}$ then their union, then its complement ${E \cap F.}$

Definition 25 (Restricted $\sigma$-algebra). Let $(X, \mathcal{M})$ be a measurable space and ${\Omega \subset X.}$ The restricted $\sigma$-algebra is

$$\mathcal{M}_\Omega = \mathcal{M}|_\Omega := \{ E \cap \Omega : E \in \mathcal{M} \}.$$

One checks that $(\Omega, \mathcal{M}_\Omega)$ is a measurable space. Indeed, ${\emptyset = \emptyset \cap \Omega,}$ ${\Omega \setminus (E \cap \Omega) = (X \setminus E) \cap \Omega}$ and ${\bigcup_n (E_n \cap \Omega) = (\bigcup_n E_n) \cap \Omega.}$

Generated and Borel σ-algebras

Theorem 26 (Generation of $\sigma$-algebras). Let $X$ be a set and ${\mathcal{F} \subset \mathcal{P}(X).}$ There exists a $\sigma$-algebra $\mathcal{M}$ on $X$ such that:
(i) ${\mathcal{F} \subset \mathcal{M};}$
(ii) for every $\sigma$-algebra $\mathcal{Q}$ on $X$ with ${\mathcal{F} \subset \mathcal{Q},}$ ${\mathcal{M} \subset \mathcal{Q}.}$
In words, $\mathcal{M}$ is the smallest $\sigma$-algebra containing $\mathcal{F}.$ It is denoted $\sigma_0(\mathcal{F}).$

Proof (sketch). Consider

$$W := \{ \mathcal{Q} \subset \mathcal{P}(X) : \mathcal{Q} \text{ is a } \sigma\text{-algebra}, \ \mathcal{F} \subset \mathcal{Q} \}.$$

Then ${W \neq \emptyset,}$ as ${\mathcal{P}(X) \in W.}$ Define

$$\mathcal{M} := \bigcap \{ \mathcal{Q} : \mathcal{Q} \in W \}. \tag*{$\blacksquare$}$$

Details. A set belongs to $\mathcal{M}$ when it belongs to every ${\mathcal{Q} \in W.}$ Each property of Definition 22 holds in every ${\mathcal{Q} \in W,}$ hence in $\mathcal{M};$ every ${\mathcal{Q} \in W}$ contains $\mathcal{F},$ which gives (i), and a $\sigma$-algebra $\mathcal{Q}$ with ${\mathcal{F} \subset \mathcal{Q}}$ is in $W,$ which gives (ii).

Figure 9. The proof of Theorem 26: inside $\mathcal{P}(X),$ three of the $\sigma$-algebras of $W,$ each containing $\mathcal{F}.$ The intersection $\sigma_0(\mathcal{F})$ of all the members of $W$ contains $\mathcal{F}$ and lies inside each of them.

Definition 27 (Borel $\sigma$-algebra). Let $(X, \tau)$ be a topological space. The Borel $\sigma$-algebra is $\sigma_0(\tau),$ denoted $\mathcal{B}(X).$

Remark 28. For $\mathbb{R}$ with the Euclidean topology,

$$\begin{aligned} \mathcal{B}(\mathbb{R}) &= \sigma_0\big( \{ \text{open sets} \} \big) \\ &= \sigma_0\big( \{ (a, b) : a < b \} \big) \\ &= \sigma_0\big( \{ [a, b] : a < b \} \big) \\ &= \sigma_0\big( \{ (a, +\infty) : a \in \mathbb{R} \} \big). \end{aligned}$$

Similarly,

$$\begin{aligned} \mathcal{B}(\overline{\mathbb{R}}) &= \sigma_0\big( \{ (a, +\infty] : a \in \mathbb{R} \} \big), \\ \mathcal{B}(\mathbb{R}^N) &= \sigma_0\big( \{ \text{open rectangles} \} \big), \end{aligned}$$

where an open rectangle is a set ${(a_1, b_1) \times (a_2, b_2) \times \dots \times (a_N, b_N).}$

Figure 10. An open rectangle of the plane, the case ${N = 2.}$ Its edges are dashed because they do not belong to it.