How this instrument works
The power set of a set S, written P(S), is not another version of S — it is the set of every subset S has, gathered into one collection. That collection always includes two subsets that are easy to overlook: the empty set ∅, which is vacuously a subset of anything, and S itself, since every set is trivially a subset of itself. For a set with n elements, P(S) has exactly 2ⁿ members — you are no longer counting the elements of S, you are counting the ways of picking a piece of it.
The exponent falls straight out of an inductive argument, not a lookup table. Start with the empty set: it has exactly one subset, itself, so |P(∅)| = 2⁰ = 1. Now add one element to a set that already has a known power set — every existing subset can either stay as it is or gain the new element, and nothing else can happen, so the collection of subsets exactly doubles with each element added. Four elements folded in one at a time from the empty set walk the count 1, 2, 4, 8, 16 — five numbers for four additions, because the first 1 is the starting point before any element joins.
A genuinely surprising consequence shows up when S is infinite: Cantor's theorem proves that P(S) is always strictly larger than S, even then, because no way of pairing elements of S with subsets of P(S) can ever reach every subset — a short diagonal argument builds a subset that any proposed pairing must miss. Applied to the natural numbers, this is the reasoning behind saying the real numbers form a bigger infinity than the counting numbers do, and it means there is no biggest set at all — take the power set of anything and something larger appears. Closer to hand, n itself must stay a whole number no smaller than zero here, since it counts elements in a set — there's no such thing as −2 elements or half an element to take subsets of.
- Enter the size of your set into the Number of elements, n field — any whole number from 0 up.
- Read the result in Number of subsets, 2ⁿ — the exact count of every subset that set contains.
- Start at n = 0 to see the baseline: the empty set's only subset is itself, so the count reads 1.
- Increase n one step at a time and watch Number of subsets, 2ⁿ double at every step, matching the doubling argument above.
- For larger sets, n accepts values up to 60 — past roughly n = 20 the count already runs past a million.
Worked example — a 4-element set's power set
Let S = {a, b, c, d}, a set of 4 elements. Enter n = 4 into Number of elements, n and Number of subsets, 2ⁿ reads 16 — the doubling walk runs 1 (just ∅), 2 (add a), 4 (add b), 8 (add c), 16 (add d), one doubling for each of the four elements folded in. Those 16 subsets break down by size as 1 + 4 + 6 + 4 + 1: the empty set, the four singletons {a}, {b}, {c}, {d}, the six pairs such as {a,b} and {b,d}, the four triples such as {a,b,c}, and S itself, {a,b,c,d}.
Each of those 16 subsets also matches a unique 4-bit binary code once the elements are ordered a, b, c, d and each bit marks included or not: {a,d} is 1001, the empty set is 0000, and the full set S is 1111. Counting in binary from 0000 to 1111 visits every one of the 16 subsets exactly once with none repeated and none left out, which is exactly why the total has to be a power of two — it is nothing more than every possible 4-bit string, and there are 2⁴ = 16 of those.
Questions
What exactly belongs to the power set of a set?
Every subset of the original set, with no exceptions — including the empty set ∅, which is a subset of everything, and the set itself, since any set is trivially a subset of itself. A set with n elements has exactly 2ⁿ such subsets; for n = 4 that's 16, running from ∅ up to the full 4-element set.
Why does the power set's size double every time one element is added?
Because each existing subset splits into exactly two once a new element joins the set: the same subset unchanged, and that subset with the new element folded in. Nothing else can happen to it, so the whole collection of subsets doubles — |P(S ∪ {x})| = 2 · |P(S)| — the same doubling, applied n times starting from the empty set's single subset, that produces 2ⁿ.
Is the power set of an infinite set still bigger than the set itself?
Yes, and this is Cantor's theorem: for any set, finite or infinite, its power set has strictly larger cardinality than the set does. A short diagonal argument shows why — any attempted pairing of elements to subsets always misses at least one subset — and the same reasoning is what proves the real numbers outnumber the natural numbers.
How is the power set different from the count of proper subsets?
A proper subset is any subset except the set itself, so there are 2ⁿ − 1 of those, not 2ⁿ — a common slip is forgetting to exclude S when a question specifically asks for proper subsets. Excluding the empty set too, for non-empty proper subsets, drops the count by one more, to 2ⁿ − 2.
Can I list all the subsets instead of just counting them?
Yes — order the n elements, then read off n-bit binary numbers from 0 up to 2ⁿ − 1; each bit position marks whether that element is in or out of the subset. For n = 4 that's 0000 through 1111, sixteen binary strings, one for each of the sixteen subsets, with no repeats.
Why can't n be negative or a fraction here?
n counts the elements of an actual set, and a set can't hold a negative or fractional number of things, so this field only accepts whole numbers from 0 up. That differs from raising 2 to a general real-number exponent, which is defined smoothly for negative and fractional powers — the restriction here comes from what n represents, not from the arithmetic of 2ⁿ itself.