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For a function $p:A\to B$, order each power set by inclusion. The inverse image map $p^{\ast }(V)=p^{-1}(V)$ has a left adjoint ${\exists }_p(U)=p(U)$ and a right adjoint ${\forall }_p(U)=\{b\in B:p^{-1}(\{b\})\subseteq U\}$. The two adjunctions follow from $p(U)\subseteq V⟺U\subseteq p^{-1}(V)$ and $p^{-1}(V)\subseteq U⟺V\subseteq {\forall }_p(U)$. Empty fibers satisfy the second condition vacuously, so points outside the image of $p$ belong to ${\forall }_p(U)$ for every $U$.

Start with a well-founded universe of sets over an infinite set $A$ of urelements. The full symmetric group of $A$ acts recursively on sets. Retain the objects all of whose descendants have a finite support in a permutation action. This class satisfies set-extensionality, foundation, infinity, pairing, union, separation, replacement and its internal power set axiom. A definable functional image has the finite union of the supports of its domain and parameters as a support; uniqueness makes the image invariant. The family of two-element subsets of $A$ has no choice function: a transposition of two atoms outside the function's finite support fixes their pair and moves either possible choice. For untyped membership with extensionality merely restricted to nonempty objects, atoms are also vacuous subsets; add all atoms to each internal power-set object.

The cardinal numbers defined by ${\beth }_0={\aleph }_0$, ${\beth }_{\alpha +1}=2^{{\beth }_{\alpha }}$, and ${\beth }_{\lambda }={\sup }_{\alpha <\lambda }{\beth }_{\alpha }$ at limit ordinals are the beth numbers. In particular ${\beth }_1=|\mathcal{P}(\omega )|$, the cardinality of the continuum. This notation distinguishes iterated power sets from the enumeration by Aleph numbers of well-orderable infinite cardinal numbers.

For a nonnegative integer $n$ and commuting quantities $x,y$,In expanding the $n$ factors, obtain $x^k{y}^{n-k}$ by selecting the $k$ positions that contribute $x$. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such choices. For example,The coefficients are row five of Pascal's triangle, in reverse order under the displayed $x^k{y}^{n-k}$ convention; the row is symmetric. Setting $x=y=1$ gives $1+5+10+10+5+1=32$, the number of subsets of a five-element set. Each member has two inclusion choices, agreeing with the rule of product and the cardinality of the power set.

The Boolean lattice is the power set of a finite set ordered by inclusion. Its rank-$h$ level consists of all $h$-element subsets.

Every subset of a discrete space is open, so its Borel sigma-algebra is the full power 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.

There is no surjection from a set $X$ onto its power set $\mathcal{P}(X)$. Consequently $\mathcal{P}(X)$ has strictly greater cardinality than $X$.

For any candidate function $f:X\to \mathcal{P}(X)$, form $D=\{x\in X:x\notin f(x)\}$. This is a subset of $X$ by set comprehension. If $f(d)=D$, then $d\in D$ holds exactly when $d\notin D$, a contradiction. Thus $D$ is missing from the image of $f$. On the other hand, $x\mapsto \{x\}$ is an injection from $X$ into its power set, establishing the strict comparison of cardinalities. For $X=\mathbb{N}$, the omitted selection is the subset version of Cantor's diagonal argument.

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.

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.

If $\alpha$ is a countable ordinal and $V_{\alpha }\models \mathsf{ZFC}$, then $\alpha$ must be a limit ordinal. Choose a countable cofinal sequence $({\alpha }_n)$ in $\alpha$. The internal axiom of choice gives, for every $n$, a bijection between $V_{{\alpha }_n}$ and some ordinal below $\alpha$; that ordinal is externally countable, so every $V_{{\alpha }_n}$ is countable. Hence $V_{\alpha }={\bigcup }_n{V}_{{\alpha }_n}$ is countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem, a contradiction.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

Define the head-count random variable $X:\Omega \to \{0,1,2\}$ byThe four elementary outcomes are inputs to this function; its three numerical values are outputs. Each fiber of a function collects all outcomes giving one value. Under the fair independent model, adding the masses in each fiber gives the probability mass functionConsequently the output values are not equally likely, although the inputs are. Using the reduced sample space $\{0,1,2\}$ with these masses gives the same head-count probability distribution; assigning $1/3$ to each value would describe a different model.

A question about a random variable's value becomes an event on the original sample space by taking an inverse image:For a real-valued random variable, this is an event whenever $D$ is a Borel set, by measurability. In this finite example every subset of $\{0,1,2\}$ is allowed. In particular,This is the same “at least one head” event counted above, expressed through an observable.^[ref]

Every event determined by $X$ is a union of complete fibers. There are $2^3=8$ such events, whereas the full power set of $\Omega$ has $2^4=16$. In particular, the first-head event $A=\{HH,HT\}$ is not determined by $X$: it contains $HT$ but excludes $TH$, although both give $X=1$. Choosing an observable makes calculations simpler precisely by forgetting distinctions; the useful choice is one that retains the distinctions the question needs.

Set $U(x)=U_0({\delta }_x)$. On a finite measurable space, points in the same atom of a sigma-algebra give the same Dirac measure, so $U$ is a measurable function and constant on atoms. Every probability measure is a finite mixture of representative Dirac measures with its atom masses as weights. Affineness then proves the expected-utility representation, even when the sigma-algebra is smaller than the power set.

A finitely supported permutation fixes every element outside some finite subset. The finitely supported permutations of $\mathbb{N}$ are the union of the finite symmetric groups acting on $\{1,\dots ,k\}$, and hence form a countable set. All permutations of $\mathbb{N}$ form an uncountable set: independently swapping or fixing each pair $(2j-1,2j)$ injects the power set of $\mathbb{N}$ into this larger set.

Finite subsets of a well-quasi-ordering, compared by Hoare domination preorder, form a well-quasi-ordering. Enumerate each finite subset as a word and apply Higman lemma: a subsequence embedding with increased letters supplies a domination witness for every source element. This does not assert the same result for the full power set.