Power Set Calculator

Background.

A power set calculator takes a set and hands back every single subset of it — all of them, including the empty set at one end and the original set itself at the other. Type a, b, c and you get the eight members of P({a, b, c}): { }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, and {a, b, c}. The count is never a coincidence and never needs to be found by trial: it is always exactly 2 raised to the number of elements, because building a subset amounts to walking down the list of elements and making one independent yes-or-no decision about each one. Three elements, three binary decisions, 2 × 2 × 2 = 8 possible outcomes, and each distinct outcome is a distinct subset.

That doubling is the single most important thing to understand about power sets, and it is also why this calculator caps the input at 20 elements. A twenty-element set already has 1,048,576 subsets. Thirty would have over a billion. Fifty would have more subsets than there are grains of sand on Earth. Cantor's theorem generalises exactly this observation — the power set of any set, finite or infinite, is strictly larger than the set itself — and it is the reason there is no largest infinity. On this page the practical consequence is more modest: full listings are printed for sets of ten elements or fewer, truncated after the first forty subsets, while the counts stay exact all the way up to twenty.

Alongside the size, the calculator returns three counts that come up constantly in discrete mathematics homework and are easy to get wrong by one. The number of proper subsets is 2^n − 1, not 2^n − 2: a proper subset is any subset that is not equal to S, so the empty set is still included and only S itself is excluded. The number of subsets that contain one particular fixed element is 2^(n−1) — pick an element, insist it is in, and the other n − 1 elements are still free to be in or out, so exactly half the power set qualifies. And the sum of the sizes of all the subsets is n · 2^(n−1), which falls straight out of that last fact: each of the n elements shows up in 2^(n−1) different subsets, so the total number of element-slots across the whole power set is n times that.

A few input conventions matter and are applied silently, so they are reported back to you. Sets have no duplicates, so typing 2, 2, 3 gives the two-element set {2, 3} and the calculator tells you how many repeats it collapsed. Sets also have no intrinsic order, so {a, b} and {b, a} are the same subset and appear once; the listing is nevertheless printed in a fixed, readable order — by subset size first, then by the order you typed the elements — so that two people running the same input see the same output. Elements are compared as text exactly as typed, which means 1 and 1.0 count as two different elements. Commas, semicolons and line breaks all work as separators, and a pair of enclosing braces is stripped if you paste {a, b, c} straight out of a textbook.

What is power set calculator?

The power set of a set S, written P(S) or sometimes 2^S, is the set whose members are all of the subsets of S. It always contains at least two things — the empty set and S itself — and those two coincide only when S is empty, in which case P(∅) = {∅} has exactly one member. In axiomatic set theory the existence of P(S) is not derived from anything simpler; it is asserted outright by the Power Set axiom of Zermelo–Fraenkel set theory, one of the handful of axioms that say which sets exist at all. For a finite set with n elements, |P(S)| = 2^n. The notation 2^S is a deliberate pun on that fact: a subset can be encoded as a function from S into the two-element set {0, 1}, and the number of such functions is 2^n.

How to use this calculator.

1. 01Type the elements of your set into the box, separated by commas, semicolons or new lines. Enclosing braces are optional.
2. 02Read the headline number: that is |P(S)| = 2^n, the total number of subsets including the empty set and S itself.
3. 03Check the 'S as read' line. If the element count is not what you expected, you have either a duplicate (which the calculator collapses and reports) or a stray separator.
4. 04Read the full listing under 'P(S) — every subset'. Subsets are ordered by size, then by the order you typed the elements, so the empty set is always first and the whole set is always last.
5. 05Use 'Proper subsets' when a question asks for subsets other than S itself, and 'Subsets containing any one element' when a question fixes one member and asks how many subsets include it.
6. 06For sets larger than ten elements the listing stops but every count remains exact up to the twenty-element ceiling.

The formula.

Every subset of S is completely determined by, and completely determines, one yes-or-no decision per element: is this element in the subset or not? Those n decisions are independent of one another, so the multiplication principle gives 2 × 2 × … × 2 = 2^n distinct decision patterns, and distinct patterns give distinct subsets. This is why the count is exponential rather than polynomial, and why adding a single element to S doubles the number of subsets rather than adding a few. Encoding the decisions as an n-bit binary string makes the correspondence exact: 000 is the empty set, 111 is S, and each of the 2^n binary strings names exactly one subset.

The proper-subset count follows immediately. A proper subset is any subset that is not equal to S, so exactly one member of the power set — S itself — is excluded, leaving 2^n − 1. A frequent mistake is to subtract two, on the assumption that the empty set is also excluded; it is not. The empty set is a subset of every set, and it is a proper subset of every set except the empty set itself.

For the number of subsets containing one fixed element, hold that element's decision at 'in' and let the other n − 1 elements decide freely. That leaves 2^(n−1) patterns, which is exactly half of 2^n. The same argument with the decision held at 'out' gives the other half, which is a useful sanity check: the subsets containing a given element and the subsets missing it partition the power set into two equal halves.

The sum of all subset sizes uses that half-and-half result directly. Ask how many times each element appears anywhere in the power set: 2^(n−1) times, by the previous paragraph. There are n elements, so the total number of element-appearances across all subsets is n · 2^(n−1). Equivalently, this is the sum over k of k·C(n, k), and the two expressions agree for every n — a standard binomial identity.

No rounding occurs anywhere in this calculator. Every quantity it reports is an exact non-negative integer, computed in arbitrary-precision decimal arithmetic and converted to a plain number only when it is returned. The listing order is a display convention rather than a mathematical claim: sets are unordered, so any ordering of P(S) is equally correct, and this page fixes one so that results are reproducible.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Weisstein, Eric W. "Power Set." MathWorld — Wolfram Research. States directly: "The order of a power set of a set of order n is 2^n." This is the source for the headline count. Retrieved 2026-07-29.
2. [2]Levin, Oscar. Discrete Mathematics: An Open Introduction, 3rd edition, §0.3 "Sets" — defines pow(A), works the example |C| = 3 with |pow(C)| = 8, and states that the empty set is a subset of every set. It poses the general relationship between |A| and |pow(A)| as a question for the reader rather than asserting 2^n, so the closed form here is taken from MathWorld and from the binomial identity below. Open access; retrieved 2026-07-29.
3. [3]Guichard, David. Combinatorics and Graph Theory, Whitman College, §1.3 "Binomial Coefficients" — derives the identity from the binomial theorem: "If we then substitute x = 1 we get 2^n = Σ(i=0..n) C(n, i), that is, row n of Pascal's Triangle sums to 2^n." Open access; retrieved 2026-07-29.
4. [4]Halmos, Paul R. Naive Set Theory. Springer Undergraduate Texts in Mathematics, 1974 reprint of the 1960 Van Nostrand edition, §5 "Complements and Powers" — the Power Set axiom and the notation P(S). Print reference, cited bibliographically; not retrieved during authoring.
5. [5]Enderton, Herbert B. Elements of Set Theory. Academic Press, 1977, Chapter 2 — the Power Set axiom in Zermelo–Fraenkel set theory. Print reference, cited bibliographically; not retrieved during authoring.
6. [6]Cantor, Georg. "Über eine elementare Frage der Mannigfaltigkeitslehre." Jahresbericht der Deutschen Mathematiker-Vereinigung 1 (1891): 75–78 — Cantor's theorem, that the power set of any set is strictly larger than the set itself, which is why the element ceiling on this page exists. Bibliographic reference; not retrieved during authoring.