Subset
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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. Every set is a subset of itself.
More formally, 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,
- B is a superset of (or includes) A, denoted by B ⊇ A.
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|.
N.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.
For any set S, inclusion is a relation on the set of all subsets of S (the power set of S).
Examples
- The set is a proper subset of .
- The set of natural numbers is a proper subset of the set of rational numbers.
- The set is a proper subset of
- Any set is a subset of itself, but not a proper subset.
- The empty set, written ø, 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.
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.
The following proposition says that inclusion is a partial order.
PROPOSITION 2: 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: 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: For any two sets A and B, the following are equivalent:
- *A ⊆ B
- *A ∩ B = A
- *A ∪ B = B
- *A − B = ø
- *B′ ⊆ A′
Other properties of inclusion
The usual order on the ordinal numbers is given by inclusion.
For the power set of a set S, the inclusion partial order is (up to an order-isomorphism) the Cartesian product of |S| (the cardinality of S) copies of the partial order on , for which 0 < 1.
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