What Is the Cardinality of a Power Set?
For a finite set $A$ with $n$ elements, the power set has exactly $2^n$ elements:
$$|P(A)| = 2^{n}, \quad \text{where } n = |A|.$$
The number of elements in a set is its cardinality, so this rule says the power set's cardinality is two raised to the original set's cardinality.
Why $2^n$? (the one-line proof)
To build any subset of $A$, you walk through the $n$ elements one at a time and make a single binary decision for each: in or out. Two choices, made independently, $n$ times over:
$$\underbrace{2 \times 2 \times \cdots \times 2}_{n \text{ times}} = 2^{n}.$$
Every distinct sequence of in/out choices produces a distinct subset, and every subset corresponds to exactly one such sequence, so the count of subsets is exactly $2^n$. That correspondence is the proof; the formula is just its arithmetic.
Properties of the Power Set
The empty set is always in $P(A)$. $\varnothing \subseteq A$ for every $A$, so $\varnothing$ is always a member of the power set.
$A$ itself is always in $P(A)$. A set is a subset of itself, so $A \in P(A)$.
$|P(A)| = 2^n$ is always even for $n \geq 1$, and never zero, because even an empty set has a power set.
Bigger sets have strictly bigger power sets. $|P(A)| = 2^n > n$ for every $n$, so a set can never be the same size as its power set. (This is the finite shadow of Cantor's theorem, which extends the result to infinite sets.)
Power sets exist for infinite sets too, the power set of $\mathbb{N}$ is infinite, and provably "larger" than $\mathbb{N}$ itself.
Examples of Power Set
Example 1
Find the power set of the empty set $\varnothing$.
The empty set has $n = 0$ elements.
By the rule, $|P(\varnothing)| = 2^0 = 1$.
Its only subset is $\varnothing$ itself.
$$P(\varnothing) = {\varnothing}.$$
Final answer: $P(\varnothing) = {\varnothing}$, a singleton whose one element is the empty set.
Example 2
A student writes the power set of $A = {a}$ as $P(A) = {a}$.
This is the most common first attempt, so it is worth seeing why it breaks. The instinct is to "wrap the elements in braces" and stop.
Check it against the formula. $A$ has $n = 1$ element, so $|P(A)|$ should be $2^1 = 2$, but ${a}$ has only one member, and that member is $a$, an object, not a set. The elements of a power set must be subsets.
The two subsets of ${a}$ are $\varnothing$ and ${a}$.
$$P(A) = {\varnothing,\ {a}}.$$
Final answer: $P({a}) = {\varnothing, {a}}$, with $2$ elements. The empty set is the subset students drop most often.
Example 3
Write the power set of $A = {1, 2, 3}$ and confirm its size.
$n = 3$, so expect $|P(A)| = 2^3 = 8$.
List subsets by size:
size 0: $\varnothing$
size 1: ${1}$, ${2}$, ${3}$
size 2: ${1, 2}$, ${1, 3}$, ${2, 3}$
size 3: ${1, 2, 3}$
$$P(A) = {\varnothing,\ {1},\ {2},\ {3},\ {1,2},\ {1,3},\ {2,3},\ {1,2,3}}.$$
Count: $1 + 3 + 3 + 1 = 8$. ✓
Final answer: $8$ subsets, matching $2^3$.
Example 4
A set has $64$ subsets. How many elements does it have?
Set $2^n = 64$.
Since $64 = 2^6$, we get $n = 6$.
Final answer: the set has $6$ elements. (Working backwards from the cardinality is a standard exam variant.)
Example 5
Find the power set of $B = {x, y}$ and identify which elements of $P(B)$ are themselves singletons.
$n = 2$, so $|P(B)| = 2^2 = 4$.
$$P(B) = {\varnothing,\ {x},\ {y},\ {x, y}}.$$
The singleton members are ${x}$ and ${y}$, each is a set with exactly one element.
Final answer: $P(B) = {\varnothing, {x}, {y}, {x,y}}$; the singletons inside it are ${x}$ and ${y}$.
Example 6
For $A = {1, 2, 3, 4}$, how many subsets contain the element $1$?
Total subsets: $2^4 = 16$.
Fix $1$ as "in." The remaining three elements ($2, 3, 4$) are each independently in or out: $2^3 = 8$ ways.
So $8$ of the $16$ subsets contain $1$.
Final answer: $8$ subsets contain the element $1$, exactly half, which makes sense, since every element is "in" for half of all subsets.
Why Counting Subsets Matters
"Why would anyone want every subset of a set?" Because a power set is the mathematical model of every possible combination of choices, and that shape appears everywhere once you look.
Every yes/no configuration is a subset. Toppings on a pizza, switches in a circuit, features turned on in a settings menu, each configuration is a subset of the full option set, so the number of configurations is $2^n$.
It is the engine behind truth tables. $n$ logical variables have $2^n$ rows in their truth table, one row per subset of "which variables are true."
It is how mathematics discovered different sizes of infinity. Cantor proved that $P(A)$ is always strictly larger than $A$, even when $A$ is infinite, which means there is no largest infinity. The historical account of Cantor's theorem is worth the detour once the finite case is solid.
The doubling is also a cautionary tale in computing: a brute-force search over "all subsets" of $n$ items is a $2^n$ algorithm, which is fine for $n = 10$ and hopeless for $n = 60$. The same formula that makes power sets elegant on paper makes naive subset-search intractable in practice.
Where the Power Set Trips Students Up
Mistake 1: Forgetting $\varnothing$ and $A$ itself
Where it slips in: any "list the power set" question.
Don't do this: list only the "real" subsets, the singletons and pairs, and leave out the empty set or the full set. The rusher writes $P({1,2}) = {{1}, {2}}$ and loses two of the four.
The correct way: the empty set and the original set are always subsets, so they are always in the power set. A quick check: if your list has fewer than $2^n$ entries, you dropped one of these two.
Mistake 2: Writing elements instead of subsets
Where it slips in: the first power set a student ever writes.
Don't do this: write $P({a, b}) = {a, b}$. The instinct is to copy the elements across.
The correct way: the elements of a power set are sets, so each must wear its own braces. ${a}$, not $a$, is a member of $P({a, b})$. The tell: a power set's elements should look like ${\dots}$, not bare objects.
Mistake 3: Using $n^2$ or $2n$ instead of $2^n$
Where it slips in: the cardinality calculation.
Don't do this: the memorizer recalls "something with 2 and $n$" and writes $2n$ or $n^2$. For $n = 3$ that gives $6$ or $9$, both wrong; the answer is $8$.
The correct way: it is $2^{n}$ (two to the power of $n$), because each of the $n$ elements is independently in or out. When unsure, fall back on the in/out argument rather than the memorised symbol.
Conclusion
The power set $P(A)$ is the set of all subsets of $A$, including $\varnothing$ and $A$ itself.
Its cardinality is $|P(A)| = 2^n$, where $n$ is the number of elements in $A$.
The $2^n$ rule comes from an independent in/out choice for each of the $n$ elements.
Elements of a power set are sets, so each is written in its own braces, ${a}$, not $a$.
Power sets grow explosively, which is why they model "all combinations" and why brute-force subset search is expensive.
A Practical Next Step
Work through these to lock it in: (1) write $P({p, q, r})$ and confirm it has $8$ members; (2) a set has $32$ subsets, how many elements does it have?; and (3) of all subsets of ${1, 2, 3, 4, 5}$, how many contain the element $5$? If the in/out counting still feels abstract, redraw the tree diagram from the top of this article for your own three-element set. To build set theory with a teacher, explore Bhanzu's algebra tutor, help with algebra, or math classes online. Want a live Bhanzu trainer to walk your child through subset counting? Book a free demo class.
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