Here is a question that trips up almost every student on first contact: the power set of a set A is the set of all subsets of A. The empty set \varnothing has no elements, so it shouldn't have any subsets either. So \mathcal{P}(\varnothing) should also be empty. Right?
Wrong — and the wrongness is instructive. \mathcal{P}(\varnothing) has exactly one element. The answer comes from applying the definition of subset very carefully, and once you see it, it is both surprising and completely forced.
The short answer
The power set of the empty set is the one-element set whose one element is the empty set. Cardinality 1, not 0.
This matches the general formula: if A has n elements, then \mathcal{P}(A) has 2^n elements. For A = \varnothing, we have n = 0, so \mathcal{P}(\varnothing) has 2^0 = 1 element. The formula is correct — it is the intuition "no elements means no subsets" that was wrong.
Why the empty set has a subset
Recall the definition: A \subseteq B means "every element of A is also an element of B." Now ask: is \varnothing \subseteq \varnothing?
The statement "every element of \varnothing is also an element of \varnothing" has nothing to check. There are no elements of \varnothing to begin with. In logic, a universal statement with no cases to verify is automatically true — it is called vacuously true. So yes, \varnothing \subseteq \varnothing.
Why vacuous truth is the correct convention: imagine a rule "every student in the empty classroom is wearing a red shirt." With no students, there is no student wearing a non-red shirt to contradict the rule. So the rule holds — vacuously. This is not a quirk; it is what consistency forces. If you declared the statement false instead, you would also be declaring false the mirror rule "every student in the empty classroom is not wearing a red shirt," which would make both an assertion and its opposite false at the same time.
So \varnothing is a subset of \varnothing. In fact, \varnothing is a subset of every set, by the exact same argument — the statement "every element of \varnothing is in B" has nothing to check and is vacuously true for any B. This is why when you list the power set of any set, the empty set is always the first subset in the list.
Listing the subsets of \varnothing
So what are the subsets of \varnothing? A subset of \varnothing is a set all of whose elements are in \varnothing — that is, a set that has no elements outside \varnothing. Since \varnothing itself has no elements, the only set that fits is the set with no elements at all: \varnothing itself.
There is exactly one such set. So:
- Subsets of \varnothing: just \varnothing.
- The power set of \varnothing is the set of these subsets: \{\varnothing\}.
- |\mathcal{P}(\varnothing)| = 1, matching the formula 2^0 = 1.
Why \{\varnothing\} is not the same as \varnothing
This is where the real confusion usually hides. Both \varnothing and \{\varnothing\} use the "empty set" symbol, so they look similar on the page. But they are genuinely different objects.
- \varnothing — the empty set. Zero elements. Cardinality 0.
- \{\varnothing\} — the set that contains the empty set as its single element. One element. Cardinality 1.
Picture it as boxes. \varnothing is an empty box — just four cardboard walls around nothing. \{\varnothing\} is a box that contains one smaller box, and that smaller box happens to be empty. The outer box has one item in it (the inner box). It is not empty. Having an empty thing inside a container is not the same as having nothing at all.
So \mathcal{P}(\varnothing) = \{\varnothing\} is a set with one element, and that one element is the empty set. Written formally:
Iterating the power set
For fun and for future reference, here is what happens if you apply the power-set operation repeatedly, starting from the empty set:
- \mathcal{P}(\varnothing) = \{\varnothing\} — 1 element
- \mathcal{P}(\mathcal{P}(\varnothing)) = \mathcal{P}(\{\varnothing\}) = \{\varnothing, \{\varnothing\}\} — 2 elements
- \mathcal{P}(\mathcal{P}(\mathcal{P}(\varnothing))) = \mathcal{P}(\{\varnothing, \{\varnothing\}\}) = \{\varnothing, \{\varnothing\}, \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\} — 4 elements
Each iteration doubles the count: 1, 2, 4, 8, 16, 32, \dots. Starting from nothing, you generate an infinite hierarchy of sets purely by repeatedly taking power sets. In the axiomatic set theory used to build the foundations of mathematics (Zermelo-Fraenkel), this is literally one of the starting moves — begin with \varnothing, take power sets, and you already have enough set-theoretic material to encode every natural number, integer, rational, and real.
Why the confusion happens
The confusion comes from a silent slip: reading "\mathcal{P}(\varnothing)" as "the power of nothing," and thinking that "nothing raised to anything is still nothing." But \mathcal{P}(A) is not an arithmetic operation on A — it is a set construction. It builds a new set by collecting all subsets of A. The new set is never empty, because the empty set is itself always a subset, and therefore always a member of the power set.
Or, using the formula: |\mathcal{P}(A)| = 2^{|A|}. For |A| = 0, this gives 2^0 = 1, not 0. The "doubling pattern" of subset-counting has a fixed overhead of one (the empty-set subset), and that overhead is never bypassed.
In practice
In a JEE problem, you will sometimes see statements like "How many subsets does a set with 0 elements have?" The correct answer is 1 — not 0 — and the unique subset is \varnothing itself. Writing 0 here costs a mark, and the reasoning is the one above: the empty set has exactly one subset, namely the empty set, which is why \mathcal{P}(\varnothing) = \{\varnothing\} and not \varnothing.
The deeper moral: empty ≠ nothing. The empty set is a genuine object in mathematics, and when you put it inside another set, that container has something in it — the empty set itself.
Related: Sets — Introduction · Does Every Set Have to Contain at Least One Element? · Power Set Explorer: All 8 Subsets of a 3-Element Set · How Many Subsets Does a Set with n Elements Have?