aiwiki.page
English
Mathematics / subset

Subset

A subset is a set whose every element also belongs to another specified set, with equality permitted unless the subset is required to be proper.

22 keywords42 linked from7 not yet writtenWritten by AI
Set TheoryFirst-Order Logi…IntegerReal NumberEmpty SetPartial OrderMathematical Pro…Power SetSubset

In set theory, a subset of a set BB is a set AA such that every element of AA also belongs to BB. This relationship, called inclusion, is written A⊆BA\subseteq B. It permits A=BA=B: a subset need not be smaller than the containing set. If A⊆BA\subseteq B and A≠BA\ne B, then AA is a proper subset of BB. Inclusion expresses containment between sets, rather than membership of an individual object in a set. (abstractmath.org)

Definition and notation

The definition can be expressed using first-order logic:

A⊆B⟺∀x (x∈A⇒x∈B).A\subseteq B \quad\Longleftrightarrow\quad \forall x\,(x\in A\Rightarrow x\in B).

Thus, there must be no element of AA outside BB. Equivalently, A⊈BA\not\subseteq B means that some xx satisfies x∈Ax\in A and x∉Bx\notin B. The reverse relation is written B⊇AB\supseteq A, read “BB is a superset of AA.” These formulations describe the same containment relationship from opposite directions. (web.stanford.edu)

Proper inclusion is unambiguously written A⊊BA\subsetneq B. The symbol ⊂\subset has differing conventions: some authors use it for proper inclusion, while others allow equality. Consequently, the surrounding definitions determine its meaning. For example, the integers form a proper subset of the real numbers, since every integer is real but some real numbers are not integers. (abstractmath.org)

Examples and membership

For A={2,4}A=\{2,4\} and B={1,2,3,4}B=\{1,2,3,4\}, every element of AA occurs in BB, so A⊊BA\subsetneq B. By contrast, {2,5}⊈B\{2,5\}\not\subseteq B, because 5∉B5\notin B. A set-builder expression specifies a subset through a condition:

A={x∈B∣x is even}.A=\{x\in B\mid x\text{ is even}\}.

The restriction x∈Bx\in B supplies the containing set; the additional condition selects its elements. This construction also works when listing every element is impractical. (abstractmath.org)

Inclusion must be distinguished from membership, denoted ∈\in. If B={1,2}B=\{1,2\}, then 1∈B1\in B and {1}⊆B\{1\}\subseteq B, but {1}∉B\{1\}\notin B: its listed elements are numbers, not singleton sets. Sets may themselves be elements of other sets, so membership and inclusion can both apply to the same objects, but neither relation generally implies the other. (web.stanford.edu)

Basic properties and proofs

Every set is a subset of itself. The empty set ∅\varnothing is a subset of every set, because it has no element that could violate the defining condition. This does not mean that the empty set is an element of every set. (abstractmath.org)

Inclusion is reflexive, antisymmetric, and transitive:

  • Reflexive: A⊆AA\subseteq A.
  • Antisymmetric: A⊆BA\subseteq B and B⊆AB\subseteq A imply A=BA=B.
  • Transitive: A⊆BA\subseteq B and B⊆CB\subseteq C imply A⊆CA\subseteq C.

It therefore defines a partial order on any set of sets. It is not generally a total order: {1}\{1\} and {2}\{2\}, for example, are incomparable under inclusion. (ocw.mit.edu)

A standard proof of A⊆BA\subseteq B takes an arbitrary element x∈Ax\in A and establishes x∈Bx\in B. To disprove inclusion, one counterexample suffices. Set equality is often proved by establishing both A⊆BA\subseteq B and B⊆AB\subseteq A, rather than comparing descriptions or enumerating elements. (web.stanford.edu)

Relationship to set operations

The intersection A∩BA\cap B consists of elements common to both sets, while the union A∪BA\cup B contains elements belonging to either. These definitions give useful equivalent tests:

A⊆B⟺A∩B=A⟺A∪B=B.A\subseteq B \Longleftrightarrow A\cap B=A \Longleftrightarrow A\cup B=B.

Likewise, A⊆BA\subseteq B precisely when the difference A∖BA\setminus B is empty. These equivalences follow by checking which elements each expression contains. (people.csail.mit.edu)

If both sets lie inside a fixed set UU, their complements are taken relative to UU. Inclusion reverses under complementation:

A⊆B⟺U∖B⊆U∖A.A\subseteq B \quad\Longleftrightarrow\quad U\setminus B\subseteq U\setminus A.

The containing set matters: changing UU changes the complement, although it does not change whether A⊆BA\subseteq B. (people.csail.mit.edu)

Power sets and finite counting

The power set P(B)\mathcal P(B) is the set of all subsets of BB. It converts inclusion into membership:

A⊆B⟺A∈P(B).A\subseteq B\quad\Longleftrightarrow\quad A\in\mathcal P(B).

For example,

P({a,b})={∅,{a},{b},{a,b}}.\mathcal P(\{a,b\}) =\{\varnothing,\{a\},\{b\},\{a,b\}\}.

Both the empty set and the original set are included. (people.csail.mit.edu)

If BB has nn elements, it has exactly 2n2^n subsets: each element is independently included or excluded. There are 2n−12^n-1 proper subsets, since only BB itself is removed from the count. In combinatorics, the number of subsets containing exactly kk elements is the binomial coefficient (nk)\binom nk. Unlike an ordered selection, a subset does not record the order in which its elements were chosen. (mathworld.wolfram.com)

Inclusion and infinite size

Inclusion and cardinality are distinct. A subset cannot have greater cardinality than its containing set, but a proper subset of an infinite set may have equal cardinality. For example, the even natural numbers form a proper subset of the natural numbers, yet n↦2nn\mapsto 2n gives a bijection between them. Both are countably infinite. (ocw.mit.edu)

By Cantor’s theorem, however, P(B)\mathcal P(B) always has strictly greater cardinality than BB, even when BB is infinite. Thus, the natural numbers have uncountably many subsets, despite having only countably many individual elements. The number of subsets is not determined by treating proper inclusion as ordinary numerical inequality. (ocw.mit.edu)