Subset

Set whose elements are all contained in another set

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Subset

Set whose elements are all contained in another set

In mathematics, a set A is a subset of a set B if and only if all elements of A are also elements of B; B is then a superset of A. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B. The relationship of one set being a subset of another is called inclusion (or sometimes containment). A is a subset of B may also be expressed as B includes (or contains) A or A is included (or contained) in B. A k-subset is a subset with k elements.

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From Wikipedia

In mathematics, a set A is a subset of a set B if and only if all elements of A are also elements of B; B is then a superset of A. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B. The relationship of one set being a subset of another is called inclusion (or sometimes containment). A is a subset of B may also be expressed as B includes (or contains) A or A is included (or contained) in B. A k-subset is a subset with k elements. When quantified, A ⊆ B {\displaystyle A\subseteq B} is represented as ∀ x ( x ∈ A ⇒ x ∈ B ) . {\displaystyle \forall x\left(x\in A\Rightarrow x\in B\right).} One can prove the statement A ⊆ B {\displaystyle A\subseteq B} by applying a proof technique known as the element argument:Let sets A and B be given. To prove that A ⊆ B , {\displaystyle A\subseteq B,} suppose that a is a particular but arbitrarily chosen element of A show that a is an element of B. The validity of this technique can be seen as a consequence of universal generalization: the technique shows ( c ∈ A ) ⇒ ( c ∈ B ) {\displaystyle (c\in A)\Rightarrow (c\in B)} for an arbitrarily chosen element c. Universal generalisation then implies ∀ x ( x ∈ A ⇒ x ∈ B ) , {\displaystyle \forall x\left(x\in A\Rightarrow x\in B\right),} which is equivalent to A ⊆ B , {\displaystyle A\subseteq B,} as stated above.

Text: Wikipédia, CC BY-SA 4.0. · Image: PolygonsSet.svg: kismalac derivative work: Stephan Kulla (St... (CC0) ·

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