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An absolutely normal number is a normal number in every integer base $K\ge 2$. This requires every finite word in every such positional notation to have its appropriate limiting digit block frequency. Knowing base-ten normality alone does not establish absolute normality; the base is part of the statement.

Borel's 1909 result says that absolute normality holds almost everywhere for Lebesgue measure: within $[0,1)$ the exceptions form a Lebesgue null set. In particular, the set of absolutely normal numbers has Lebesgue measure one there. This is a statement about the size of an exceptional set, not a test for a specified constant. ^{[ref] [ref]}

A modern route uses the Birkhoff ergodic theorem and ergodicity of integer multiplication on the circle. For each fixed base and digit interval, the time proportion equals its Lebesgue measure almost everywhere. The bases and finite words form countable sets, so a countable intersection of these measure-one sets still has Lebesgue measure one. For an absolutely normal number, the word $11$ therefore has frequency $1/4$ in base two, $1/9$ in base three, and $1/100$ in base ten. The same symbols represent different intervals in different bases.

If an integer-valued monotone sequence $(a_n)$ is nondecreasing and bounded above by $M$, it can increase strictly at most $\lfloor M\rfloor -a_1$ times. Thus its values eventually remain constant. Such sequences are determined by a finite initial list followed by a constant tail, so their collection is a countable set.

Cantor's diagonal argument shows that the set of all infinite binary sequences is an uncountable set. Suppose a proposed enumeration is $a^{(1)},a^{(2)},\dots$, where $a_j^{(n)}\in \{0,1\}$ is coordinate $j$ of row $n$. Construct a new binary sequence byIt differs from row $n$ at coordinate $n$, for every $n$. It is consequently absent from the entire proposed list, regardless of how that list was chosen. The failure of an enumeration of all binary sequences follows from this precise point of disagreement, rather than from a claim that the list looks too small.

For the displayed window, the first six diagonal bits are $1,0,0,0,1,1$, so the new sequence starts $0,1,1,1,0,0$. The drawing shows only finitely many rows, but the rule defines every coordinate and excludes every row of an infinite list. This distinguishes the argument from the Cantor pairing function: that construction lists all pairs of finite indices, whereas this one concerns an entire infinite sequence as each object to be listed. Finite binary strings form a countable set; all infinite binary sequences do not.

The connection with power sets is exact: a subset $A\subseteq \mathbb{N}$ corresponds to its indicator function, with a one in position $n$ exactly when $n\in A$. Changing the diagonal membership bit is the same mechanism as the Cantor's theorem construction $D=\{x:x\notin f(x)\}$. This reusable method turns a purported complete list into an explicit omitted object.

In his paper ^[ref], presented in 1891 and printed in the 1890–91 volume in 1892, Cantor uses infinite sequences of two symbols for the diagonal construction and then gives the broader comparison of cardinalities. The binary notation here is a modern presentation of that construction.

For fixed positive integer period $n$, integer-valued periodic functions on $\mathbb{Z}$ are in bijection with ${\mathbb{Z}}^n$ and hence form a countable set. Their union over periods is a countable union of countable sets. In contrast, period-one integer-valued functions on $\mathbb{Q}$ can prescribe arbitrary binary values on its infinitely many distinct classes modulo $\mathbb{Z}$. The power set of a countably infinite set injects into them, proving uncountability.

There are at most continuum many binary relations on a countable set. For the matching lower bound, order the blocks $\{2n,2n+1\}$ by their index and independently choose the internal comparison in each block. A binary sequence determines those choices and distinct sequences give distinct labelled total orders. The resulting order types can all coincide; the count concerns relations, not isomorphism classes.

If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.

A countably generated abelian group is a countable set. Given generators $g_1,g_2,\dots$, every element is ${\sum }_{j=1}^m{a}_j{g}_j$ for some $m$ and some $a_j\in \mathbb{Z}$. For each fixed $m$, this is the image of a finite cartesian product of copies of the countable set $\mathbb{Z}$; taking a countable union of countable sets proves the assertion. The same proof works for a module generated by a countable set over any countable ring.

A family of pairwise disjoint real intervals of positive length is countable. Each interval contains a rational number in its interior by the density of the rational numbers. Fix an enumeration of the rational numbers and select the first member lying inside each interval. Disjointness makes this selection an injection into a countable set. The argument also applies to disjoint nonempty open subsets of the real line.

A finite subset $A$ of the natural numbers, numbered from one, has the binary code ${\sum }_{j\in A}{2}^{j-1}$. Distinct finite subsets give distinct codes, because the largest differing exponent exceeds the sum of all smaller powers of two. Thus the finite subsets form a countable set, although the entire power set is an uncountable set by the Cantor diagonal argument. An enumeration transfers this coding to any countable set.

Every family of pairwise disjoint nonempty open disks in the plane is countable. The density of the rational numbers gives each disk a point of the countable set ${\mathbb{Q}}^2$. Fixing an enumeration and taking the first point in each disk yields an injection from disks to labels. This argument requires no countable choice of representatives and is useful when geometric separation would otherwise produce an uncountable family.

A submodule $N$ of a free module with basis of a module $E$, generated by a countable set, is contained in the submodule generated by some countable subset $E^{\prime }\subseteq E$. Take $E^{\prime }$ to be the union of the finite coordinate supports in a free module of a countable generating family for $N$. This union is countable, and every generator belongs to $⟨E^{\prime }⟩$, which proves the inclusion. Combining this with basis partition gives a module direct sum places $N$ inside a countably generated direct summand.

A forcing order has the countable chain condition when every antichain in a forcing order is countable. Such forcing preserves cardinals and cofinalities at least ${\aleph }_1$.

Every countable set $B$ of real numbers is contained in a countable set closed under taking all real roots of a polynomial with coefficients in that set. Let $\varphi (A)$ consist of these roots of a polynomial, excluding the zero polynomial. There are countably many finite coefficient lists over a countable set, and each nonzero polynomial has finitely many roots of a polynomial; hence $\varphi (A)$ is countable. Define $B_0=B$ and $B_{n+1}=B_n\cup \varphi (B_n)$. The countable union of countable sets theorem shows that $X={\bigcup }_n{B}_n$ is countable. Every finite coefficient list in $X$ belongs to one stage $B_N$, so its real roots of a polynomial belong to $B_{N+1}\subseteq X$. Retaining $B_n$ at the next stage matters: $\varphi (A)$ need not contain $A$.

If a normal function on an ordinal begins at a countable value and preserves countability from $\Phi (\xi )$ to $\Phi (\xi +1)$, transfinite induction shows it maps all countable ordinals to countable ordinals. At countable limits this uses countable unions of countable sets, hence axiom of countable choice in the usual proof. Iteration from zero either stabilizes at a countable fixed point or has a countable limit fixed by continuity. Applying this to ${\omega }^{\alpha }$ gives the countable epsilon number ${ϵ}_0$.

A countable ordinal has a countable set as its underlying set.

These are the ordinals strictly below the first uncountable ordinal.

If a random variable has a probability density function, each singleton has probability zero because its density has zero integral there. For a countable set $D=\{d_1,d_2,\dots \}$, remove duplicates to make the events $\{X=d_i\}$ disjoint. Countable additivity gives $\mathbb{P}(X\in D)={\sum }_i\mathbb{P}(X=d_i)=0$. The proof does not extend to uncountable unions merely because each constituent has zero probability.

In the ordinary topology on infinite subsets, the family $D$ of cofinite sets is a countable set, a dense and a meagre set. Countability follows by indexing its members by their finite complements. Each singleton is closed with empty interior, while every basic cylinder with finite initial segment $s$ contains $s\cup \{n:n>\max s\}$. Consequently $D$ is not a nowhere dense set. This example uses the ordinary cylinder topology, rather than the Ramsey cone topology.

A discrete probability distribution is a probability distribution concentrated on a finite or countable set of possible values $S$: a random variable $X$ with this law satisfies $\mathbb{P}(X\in S)=1$. Its probability mass function assigns a probability to each value:Thus the probability of an event is obtained by adding masses, rather than integrating a probability density function. The possible values need not be equally spaced, or even integers. When the relevant moments are finite, its expected value and variance are

For a constructed example, assign values $0,2,5$ masses $0.5,0.3,0.2$. They sum to one; $\mathbb{P}(X\ge 2)=0.3+0.2=0.5$, $\mathbb{E}X=0(0.5)+2(0.3)+5(0.2)=1.6$, and $\mathbb{E}[X^2]=4(0.3)+25(0.2)=6.2$. Hence the variance is $6.2-1.6^2=3.64$. A mean can lie between possible values: no outcome here equals $1.6$.

The useful organizing question is what an experiment records. One yes/no indicator random variable gives a Bernoulli distribution; a success count in a fixed number of independent attempts gives a binomial distribution; the attempt number of the first success gives a geometric distribution. A Poisson distribution describes an event count under different assumptions. Naming the quantity first helps prevent selecting a familiar formula for the wrong experiment.

A random variable is discrete when some finite or countable set $D\subseteq \mathbb{R}$ has $\mathbb{P}(X\in D)=1$. Its probability mass function is $p_X(x)=\mathbb{P}(X=x)$; these nonnegative masses sum to one. An event determined by the value has probability $\mathbb{P}(X\in A)={\sum }_{x\in A\cap D}{p}_X(x)$. Values outside $D$ may still occur on a zero-probability event without changing this definition.

Countable support need not consist of integers: a random variable constantly equal to $3/2$ is discrete. Integer-indexed tail-sum formulas therefore require a separate integer-valued hypothesis. For the dice sum in random variable, the expected value is ${\sum }_{s=2}^{12}s{p}_S(s)=7$ and the variance is ${\sum }_{s=2}^{12}(s-7{)}^2{p}_S(s)=35/6$; the same answer follows by adding the variances of its independent summands.

Enumerate all intervals with rational endpoints, using the countability of the rational numbers and listing pairs of enumeration indices by increasing sum. If every point $b$ of $B$ is an isolated point, some such interval contains $b$ and no other point of $B$. Assign to $b$ the least index of such an interval. Distinct points cannot have the same index, so this gives an injective function into $\mathbb{N}$ and proves that $B$ is a countable set. The argument uses isolation in the subspace topology, not closedness: the discrete subset $\{1/n:n\ge 1\}$ illustrates that an accumulation point outside $B$ is allowed.