SUBSET

In mathematics, especially in set theory, a set ''A'' is a 'subset' of a set ''B'', if ''A'' is "contained" inside ''B''. The relationship of one set being a subset of another is called 'inclusion' or 'containment'.

Contents

Definitions

Symbols

Examples

Properties

Proposition 1

Proposition 2

Proposition 3

Proposition 4

Proposition 5

Other properties of inclusion

Definitions

If ''A'' and ''B'' are sets and every element of ''A'' is also an element of ''B'', then:
:
★ ''A'' is a subset of (or is included in) ''B'', denoted by ''A'' ⊆ ''B'',
:or equivalently
:
★ ''B'' is a 'superset' of (or includes) ''A'', denoted by ''B'' ⊇ ''A''.
If ''A'' is a subset of ''B'', but ''A'' is not equal to ''B'' (i.e. there exists at least one element of B not contained in ''A''), then
:
★ ''A'' is also a 'proper' (or 'strict') subset of ''B''; this is written as ''A'' ⊂ ''B''.
:or equivalently
:
★ ''B'' is a proper superset of ''A''; this is written as ''A'' ⊂ ''B''.
For any set ''S'', the inclusion relation ⊆ is a partial order on the set 2^''S'' of all subsets of ''S'' (the power set of ''S'').

Symbols

An easy way to remember the difference in symbols is to note that ⊆ and ⊂ are analogous to ≤ and <. For example, if ''A'' is a subset of ''B'' (written as ''A'' ⊆ ''B''), then the number of elements in ''A'' is less than or equal to the number of elements in ''B'' (written as |''A''| ≤ |''B''|). Likewise, for finite sets ''A'' and ''B'', if ''A'' ⊂ ''B'' then |''A''| < |''B''|.
Many authors do not follow the above conventions, but use ⊂ to mean simply subset (rather than proper subset). There is an unambiguous symbol, $subsetneq$ (or in Unicode), for proper subset. Some authors use both unambiguous symbols, ⊆ for subset and $subsetneq$ for proper subset, and dispense with ⊂ altogether. The corresponding remarks apply for supersets as well.

Examples

★ The set {1, 2} is a proper subset of {1, 2, 3}.

★ Any set is a subset of itself, but not a proper subset.

★ The empty set, written ∅, is also a subset of any given set ''X''. (This statement is vacuously true, see proof below) The empty set is always a proper subset, except of itself.

★ The set {''x'' : ''x'' is a prime number greater than 2000} is a proper subset of {''x'' : ''x'' is an odd number greater than 1000}

★ The set of natural numbers is a proper subset of the set of rational numbers and the set of points in a line segment is a proper subset of the set of points in a line. These are counter-intuitive examples in which both the part and the whole are infinite, and the part has the same number of elements as the whole (see Cardinality of infinite sets).

Properties

Proposition 1

The empty set is a subset of every set.
Proof: Given any set ''A'', we wish to prove that ø is a subset of ''A''. This involves showing that all elements of ø are elements of ''A''. But there are no elements of ø.
For the experienced mathematician, the inference " ø has no elements, so all elements of ø are elements of ''A''" is immediate, but it may be more troublesome for the beginner. Since ø has no members at all, how can "they" be members of anything else?
It may help to think of it the other way around. In order to prove that ø was ''not'' a subset of ''A'', we would have to find an element of ø which was not also an element of ''A''. Since there are no elements of ø, this is impossible and hence ø is indeed a subset of ''A''.

Proposition 2

The following proposition says that inclusion is a partial order.
If ''A'', ''B'' and ''C'' are sets then the following hold:
:reflexivity: ''A'' ⊆ ''A''
:antisymmetry: ''A'' ⊆ ''B'' and ''B'' ⊆ ''A'' if and only if ''A'' = ''B''
:transitivity: If ''A'' ⊆ ''B'' and ''B'' ⊆ ''C'' then ''A'' ⊆ ''C''

Proposition 3

The following proposition says that for any set ''S'' the power set of ''S'' ordered by inclusion is a bounded lattice, and hence together with the distributive and complement laws for unions and intersections (see The fundamental laws of set algebra), show that it is a Boolean algebra.
If ''A'', ''B'' and ''C'' are subsets of a set ''S'' then the following hold:
:existence of a least element and a greatest element:
::
★ ø ⊆ ''A'' ⊆ ''S'' (that ø ⊆ ''A'' is Proposition 1 above.)
:existence of joins:
::
★ ''A'' ⊆ ''A''∪''B''
::
★ If ''A'' ⊆ ''C'' and ''B'' ⊆ ''C'' then ''A''∪''B'' ⊆ ''C''
:existence of meets:
::
★ ''A''∩''B'' ⊆ ''A''
::
★ If ''C'' ⊆ ''A'' and ''C'' ⊆ ''B'' then ''C'' ⊆ ''A''∩''B''

Proposition 4

The following proposition says that, the statement "''A'' ⊆ ''B'' ", is equivalent to various other statements involving unions, intersections and complements.
For any two sets ''A'' and ''B'', the following are equivalent:
:
★ ''A'' ⊆ ''B''
:
★ ''A'' ∩ ''B''  =  ''A''
:
★ ''A'' ∪ ''B''  =  ''B''
:
★ ''A'' − ''B''  =   ø
:
★ ''B''′ ⊆ ''A''′
This shows that the relation of set inclusion can be characterized by either of the set operations of union or intersection, which means that the notion of set inclusion is axiomatically superfluous given either of those operations and equality.

Proposition 5

If the number of elements of the set A is ''n'', then the number of all subsets of A is equal to $2^n$.
The proof of this is an exercise in induction.

Other properties of inclusion

Inclusion is the canonical partial order in the sense that every partially ordered set (''X'', $preceq$) is isomorphic to some collection of sets ordered by inclusion. The ordinal numbers are a simple example—if each ordinal ''n'' is identified with the set [''n''] of all ordinals less than or equal to ''n'', then ''a'' ≤ ''b'' if and only if [''a''] ⊆ [''b''].
For the power set 2^''S'' of a set ''S'', the inclusion partial order is (up to an order isomorphism) the Cartesian product of ''k'' = |''S''| (the cardinality of ''S'') copies of the partial order on {0,1} for which 0 < 1. This can be illustrated by enumerating ''S'' = {''s''₁, ''s''₂, …, ''s''_''k''} and associating with each subset ''T'' ⊆ ''S'' (which is to say with each element of 2^''S'') the ''k''-tuple from {0,1}^''k'' of which the ''i''th coordinate is 1 if and only if ''s''_''i'' is a member of ''T''.