This power set calculator lists every subset of your set — the empty set, all singletons, all pairs, up through the set itself — and counts them: a set with n elements has exactly 2ⁿ subsets.
Every Subset, Including the Two Everyone Forgets
The power set P(S) is the set of ALL subsets of S — and its construction teaches set theory's favorite vigilance: the empty set ∅ is a subset of everything (vacuously: it has no element that could fail to be in S), and S is a subset of itself. For S = {a, b, c} the power set has eight members: ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}. The counting law is the cleanest in combinatorics: |P(S)| = 2ⁿ, because building a subset is n independent yes/no decisions — equivalently, subsets are n-bit strings, and the calculator literally generates them by counting in binary. This is also why the binomial coefficients appear: the subsets of size k number C(n,k), and summing Pascal's row gives ΣC(n,k) = 2ⁿ — the power set is the binomial theorem's combinatorial portrait. Cantor's theorem then uses exactly this object to prove the infinite gets bigger: for ANY set, even infinite, |P(S)| > |S| — the power set of the natural numbers is uncountable, which is the doorway to the hierarchy of infinities and to the continuum hypothesis.
Practically, power sets are the universe of several everyday computations: the possible feature-subsets in a machine-learning selection problem (2²⁰ ≈ a million for 20 features — the reason greedy selection exists), the subsets of a shopping basket in market-basket analysis, every possible committee from a roster, and the state space of anything with n binary switches. They are also a perfect intuition pump for exponential explosion: each added element DOUBLES the listing — 10 elements, 1,024 subsets (printable); 20, a million (a long scroll); 30, a billion (no one is listing those); 64 elements give 1.8×10¹⁹ subsets — on the order of all the grains of sand on Earth's beaches. The calculator caps listings at 10 elements because the wall of text IS the lesson, while the count 2ⁿ keeps working at any n you care to name.
Worked Examples (Verified)
- {a, b, c} → 2³ = 8 subsets: ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}.
- {1, 2, 3, 4} → 16 subsets, of which C(4,2) = 6 have exactly two elements.
- Empty set → P(∅) = {∅} — exactly 1 subset (2⁰), and it is not itself empty.
- 10 elements → 1,024 subsets; 20 elements → 1,048,576.
Power Set Discipline
- Duplicates in your input aren't elements twice — a set's roster dedupes (the calculator removes repeats), because {a, a} = {a}.
- ∅ ≠ {∅}: the first has 0 elements, the second has 1 (the empty set as an element). Power sets make this distinction unavoidable — lean into it.
- Order inside subsets doesn't matter: {a, b} and {b, a} are the same subset, listed once.
- The listing grows as 2ⁿ — if you need subsets of a 25-element set, you need an algorithm (or sampling), not a printout.
- Cardinality questions (how many subsets with property X) usually reduce to binomial counts — count without listing whenever n is large.
Related Tools
- Combination Calculator — counting the size-k subsets directly
- Math Calculators — the full collection
Frequently Asked Questions
How do you find the power set of a set?
List every subset: for each element, branch into "exclude it" and "include it," doubling the family at each step — equivalently, write all binary strings of length n and let bit i decide element i's membership. For {a, b, c}: ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}. This calculator performs the enumeration for up to 10 elements and always reports the true count, 2ⁿ, even when the listing is capped.
Why does a set with n elements have 2^n subsets?
Two proofs for the price of one: (1) each of the n elements is independently in or out — 2 choices, n times, multiplied: 2ⁿ. (2) Count subsets by size: C(n,0) + C(n,1) + … + C(n,n), which the binomial theorem evaluates at (1+1)ⁿ = 2ⁿ. The proofs illuminate different structure — the first gives the binary-listing algorithm, the second ties power sets to Pascal's triangle (each row sums to the next power of two, visible in any listing grouped by subset size).
Is the empty set in every power set?
Yes — ∅ is a subset of every set, so it belongs to every power set. So does the set itself. Forgetting either is the most common power-set error in coursework. The extreme case is instructive: P(∅) = {∅}, a set containing one element — the empty set itself, wrapped once. That {∅} ≠ ∅ (one has an element, the other doesn't) is the distinction on which ordinals are built in set-theoretic foundations: 0 = ∅, 1 = {∅}, 2 = {∅, {∅}}, and arithmetic climbs from nothing.
What is the difference between a subset and a proper subset?
A subset may equal the whole set; a PROPER subset must be strictly smaller. Every set has 2ⁿ subsets but 2ⁿ − 1 proper subsets (exclude only S itself) and 2ⁿ − 2 proper non-empty subsets (exclude ∅ too, when the question asks). The vocabulary matters in counting problems: "subsets" silently includes both bookends unless the problem says proper, non-empty, or non-trivial — three qualifiers, three different counts, all standing next to the 2ⁿ this calculator reports.
How big can a power set get?
Catastrophically, on schedule: 2ⁿ doubles per element. 2¹⁰ ≈ 10³ (1,024 — listed happily), 2²⁰ ≈ 10⁶, 2³⁰ ≈ 10⁹, 2⁶⁴ ≈ 1.8×10¹⁹, 2¹⁰⁰ ≈ 1.3×10³⁰. Any algorithm that enumerates power sets dies by the mid-20s of n; problems phrased as "check all subsets" (subset-sum, exact feature selection, brute-force set cover) are NP-hard territory precisely because the power set outgrows every computer. The count remains exact and instant — it's the LISTING that physics forbids, which is why this calculator caps one and not the other.
What is Cantor's theorem and how does it use power sets?
Cantor proved that for any set S, the power set is strictly larger: no listing of S's elements can be matched one-to-one against all its subsets — the diagonal argument constructs a subset differing from the k-th proposed partner in the membership of the k-th element, so it appears nowhere in the matching. For finite sets it's the obvious 2ⁿ > n; for infinite sets it manufactures bigger infinities endlessly — |ℕ| < |P(ℕ)| < |P(P(ℕ))| < … — the ladder this page's modest object climbs forever.
How is the power set related to binary numbers?
They're the same list. Number the elements 0…n−1; every subset is the set of positions where a binary number has 1s. Counting 0 to 2ⁿ−1 in binary enumerates the power set in a canonical order (this calculator's method), and the correspondence explains the doubling: adding an element appends one bit, doubling the numbers. It also makes power sets tangible in computing: a bitmask IS a subset, and iterating masks is how programmers walk all subsets of small collections.
Is this power set calculator free?
Yes — free, instant, no sign-up. Enter up to 10 comma-separated elements for the full power-set listing and the 2ⁿ count, grouped exactly as the binary construction builds them.