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Power Set: Membership versus Inclusion

Being a member of P(A)P(A) means being a subset of AA: S∈P(A)  ⟺  S⊆AS \in P(A) \iff S \subseteq A.

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اهمیت
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The power set P(A)P(A) (also written 2A2^{A}) is the set of all subsets of AA. This single fact collapses two relations that look different on the page into one:

S∈P(A)  ⟺  S⊆A.S \in P(A) \iff S \subseteq A.

The symbol ∈\in on the left (membership in P(A)P(A)) is the same test as ⊆\subseteq on the right (inclusion in AA). So a question that asks "is SS an element of P(A)P(A)?" is really asking "is every element of SS also an element of AA?"

Two relations, kept straight

Statement True when Read as
x∈Ax \in A xx is literally one of the listed elements of AA "xx is a member of AA"
x⊆Ax \subseteq A every element of xx is a member of AA "xx is a subset of AA"
x∈P(A)x \in P(A) same as x⊆Ax \subseteq A "xx is a subset of AA"
x⊆P(A)x \subseteq P(A) every element of xx is itself a subset of AA "xx is a set of subsets of AA"

Two facts always hold, because ϕ⊆A\phi \subseteq A and A⊆AA \subseteq A:

ϕ∈P(A),A∈P(A).\phi \in P(A), \qquad A \in P(A).

And if AA is finite with ∣A∣=n|A| = n, then ∣P(A)∣=2n|P(A)| = 2^{n}.

Counting elements is the first move

Membership checks go wrong when you miscount the elements of the outer set, especially with nested empty sets. These three are distinct:

  • ϕ\phi — the empty set, 00 elements;
  • {ϕ}\{\phi\} — a set whose one element is ϕ\phi, so 11 element;
  • {{ϕ}}\{\{\phi\}\} — a set whose one element is {ϕ}\{\phi\}, so 11 element.

Braces matter: peel one layer at a time.

Worked mini-example

Let A={ ϕ, {ϕ}, {{ϕ}} }A = \{\, \phi,\ \{\phi\},\ \{\{\phi\}\} \,\} — a set with exactly three elements. Test two candidates for membership in P(A)P(A).

Candidate {{ϕ}}\{\{\phi\}\}. This is the singleton whose only element is {ϕ}\{\phi\}. Is {ϕ}∈A\{\phi\} \in A? Yes — it is the second listed element. Every element of the candidate lies in AA, so {{ϕ}}⊆A\{\{\phi\}\} \subseteq A, hence {{ϕ}}∈P(A)\{\{\phi\}\} \in P(A). ✓

Candidate { ϕ, {ϕ} }\{\, \phi,\ \{\phi\} \,\}. Both ϕ∈A\phi \in A and {ϕ}∈A\{\phi\} \in A, so this is a subset of AA and therefore an element of P(A)P(A). ✓

Notice the trap: {{ϕ}}\{\{\phi\}\} is a member of P(A)P(A) even though, as a set, it is not one of the three listed elements of AA — what matters is that its contents sit inside AA.

Pitfall: "S∈P(A)S \in P(A)" never means "SS is one of the listed elements of AA." It means "S⊆AS \subseteq A," i.e. every element of SS belongs to AA. Before testing anything, count the distinct elements of AA by peeling braces one layer at a time — nested empty sets like ϕ\phi, {ϕ}\{\phi\}, {{ϕ}}\{\{\phi\}\} are three different objects.

فرمول‌های کلیدی

Power-set membership
S∈P(A)  ⟺  S⊆AS \in P(A) \iff S \subseteq A

Membership in the power set is exactly the subset relation.

Always-present subsets
ϕ∈P(A)andA∈P(A)\phi \in P(A) \quad\text{and}\quad A \in P(A)

The empty set and A itself are always members of P(A).

Cardinality of the power set
∣A∣=n  ⟹  ∣P(A)∣=2n|A| = n \implies |P(A)| = 2^{n}

A finite n-element set has 2^n subsets.

#power-set#subset-vs-member#set-membership#empty-set#nested-sets#set-theory

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