CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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$\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert}$ for finite $A$

Statement

Let $A$ be a finite set and $n := \lvert A\rvert$ . Then the power set $\mathcal{P}(A)$ is finite and

$\lvert\mathcal{P}(A)\rvert = 2^{\,n},$

the power being the $\mathbb{N}$ -valued exponentiation of Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $\mathbb{R}$ . Moreover $n < 2^{\,n}$ .

The last inequality is the quantitative form, for finite $A$ , of Cantor's theorem $A \prec \mathcal{P}(A)$ (Cantor's theorem: $A \prec \mathcal{P}(A)$ ), which holds for every set whatsoever. The two statements are consistent and the proof below derives the inequality from Cantor's theorem rather than leaving them side by side.

Facts & Assumptions

Given: A finite set $A$ with $n := \lvert A\rvert$ , and $2 = \{0,1\}$ as a von Neumann natural (The natural numbers $\mathbb{N}$ (von Neumann)). Write $2^{A}$ for the set of functions $A \to 2$ .

[L1]

$\lvert X^{Y}\rvert = \lvert X\rvert^{\lvert Y\rvert}$ for finite $X$ , $Y$ , and $X^{Y}$ is finite (The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $\lvert A^{B}\rvert = \lvert A\rvert^{\lvert B\rvert}$ ).

[L2]

Cardinality (The cardinality $\lvert A\rvert$ of a finite set): $\lvert m\rvert = m$ for a natural $m$ ; a bijection transports finiteness and cardinality; and for finite $X$ , $Y$ one has $\lvert X\rvert = \lvert Y\rvert$ if and only if $X \approx Y$ .

[L3]

Cantor's theorem: $A \prec \mathcal{P}(A)$ , that is, there is an injection $A \to \mathcal{P}(A)$ and no bijection (Cantor's theorem: $A \prec \mathcal{P}(A)$ , Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L4]

Pigeonhole, claim 2: if $q < p$ then there is no injection $p \to q$ (The pigeonhole principle on $\mathbb{N}$ ).

[L5]

Trichotomy: exactly one of $p < q$ , $p = q$ , $q < p$ holds (Trichotomy of the order on $\mathbb{N}$ , Order on the natural numbers).

[L6]

Maps (Injection, surjection, bijection): a map with a two-sided inverse is a bijection; a composite of injections is an injection; and $f^{-1}[T] = \{x : f(x) \in T\}$ .

Proof

technique · direct

1.1

The characteristic function. For $S \subseteq A$ define $\chi_S : A \to 2$ by $\chi_S(x) = 1$ when $x \in S$ and $\chi_S(x) = 0$ otherwise, and let $X : \mathcal{P}(A) \to 2^{A}$ be $S \mapsto \chi_S$ . Let $Y : 2^{A} \to \mathcal{P}(A)$ be $f \mapsto f^{-1}[\{1\}]$ . Both composites are the identity: $\chi_S^{-1}[\{1\}] = \{x \in A : \chi_S(x) = 1\} = S$ ; and for $f : A \to 2$ and $x \in A$ the value $f(x)$ is $0$ or $1$ , so $\chi_{f^{-1}[\{1\}]}(x) = 1$ exactly when $f(x) = 1$ and $= 0$ otherwise, that is $\chi_{f^{-1}[\{1\}]} = f$ . Hence $X$ is a bijection and $\mathcal{P}(A) \approx 2^{A}$ .

L6construct

2.1

Therefore $\mathcal{P}(A)$ is finite and $\lvert\mathcal{P}(A)\rvert = \lvert 2^{A}\rvert = \lvert 2\rvert^{\lvert A\rvert} = 2^{\,n}$ , using [L1] and $\lvert 2\rvert = 2$ from [L2].

step 1.1L1L2

3.1

The inequality. By [L3] there is an injection $A \to \mathcal{P}(A)$ ; composing with bijections $n \to A$ and $\mathcal{P}(A) \to 2^{\,n}$ , which exist by [L2] and step 2.1, gives an injection $n \to 2^{\,n}$ . So $2^{\,n} < n$ is impossible by [L4], and $n \le 2^{\,n}$ by [L5]. Also $n \ne 2^{\,n}$ : otherwise $\lvert A\rvert = \lvert\mathcal{P}(A)\rvert$ , hence $A \approx \mathcal{P}(A)$ by [L2], contradicting [L3].

step 2.1L2L3L4L5L6

4.1

The two assertions are step 2.1 and step 3.1, so $\lvert\mathcal{P}(A)\rvert = 2^{n}$ and $n < 2^{n}$ .

step 2.1step 3.1∎

Remarks

The finiteness of $\mathcal{P}(A)$ is part of the statement, and it is what makes $[A]^{k}$ finite in the next definition: a set of $k$ -element subsets is a subset of $\mathcal{P}(A)$ .
Cantor's theorem is not weakened here. $A \prec \mathcal{P}(A)$ holds for every set, finite or infinite, and needs no counting; what the finite case adds is the value of the gap, $2^{n}$ against $n$ . The inequality above is deduced from Cantor's theorem, so no independent argument can disagree with it.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 75 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Power set (Wikipedia) (standard reference, not scraped)
Cantor's theorem (Wikipedia) (standard reference, not scraped)
P. Halmos, Naive Set Theory, §13 (standard reference, not scraped)
J. Sylvestre, Elementary Foundations 12.02, Properties of finite sets and their cardinality (LibreTexts) (standard reference, not scraped)

∣P(A)∣=2∣A∣\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert}∣P(A)∣=2∣A∣ for finite AAA