For two _set_s A and B, if any of the collection A collection of elements are the elements of B, we say _set_ A contains the _set_ B, or B contains a collection of _set_s A, also said the collection is a _set_ A sub_set_ of B. If any of the collection A collection of elements are the elements of B, and _set_ at least one element of B does not belong to _set_ A, called the collection is a _set_ B, A sub_set_. Empty _set_ is a sub_set_ of any collection. Any one _set_ is a sub_set_ of itself. Empty _set_ is true of any non-empty sub_set_.
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Examples
We know that any natural number is a positive even number. That is, _set_ E is even an element of any _set_ of natural numbers N is an element. For two _set_s A and B, if any of the collection A collection of elements are the elements of B, then _set_ A sub_set_ of B called the collection. Denoted Read as "A with the B" (or B contains A). For example, the above If at least one element A does not belong to B, then A is not a sub_set_ of B can be written as Read as "A free in B" (or "B does not contain A").
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Nature
Proposition 1: The empty _set_ is a sub_set_ of any _set_. Proof: Given any _set_ A, to prove that Φ is a sub_set_ of A. This is required to give all the elements of Φ, the elements of A; However, Φ is no element. For experienced mathematicians, the inference "Φ has no elements, so all the elements of Φ are elements of A" is obvious; but for beginners, some trouble. Because there is no element of Φ, how to make "these elements" into other elements of the collection? Another way of thinking will help. In order to prove that Φ is not a sub_set_ of A, must find an element belonging to Φ, but it does not belong to A. Because there is no element of Φ, so it is not possible. Thus Φ must be a sub_set_ of A. This proposition Description: contains a partial order. Proposition 2: If A, B, C is _set_, then: Reflexivity: A ⊆ A Anti-symmetry: A ⊆ B and B ⊆ A if and only if A = B Transitive: If A ⊆ B and B ⊆ C then A ⊆ C The proposition that: for any _set_ S, S contains the sort by the power _set_ is a bounded lattice, and the combination of the above proposition, it is a Boolean algebra. Proposition 3: If A, B, C is a sub_set_ of _set_ S, then: There is a minimal element and a greatest element: Φ ⊆ A ⊆ S (that Φ ⊆ A is Proposition 1 above.) Existence and operation: A ⊆ A ∪ B If A ⊆ C and B ⊆ C then A ∪ B ⊆ C There is intersection: A ∩ B ⊆ A If C ⊆ A and C ⊆ B then C ⊆ A ∩ B The proposition that: expression "A ⊆ B" and the other using union, intersection and complement representation are equivalent, which includes the relations in the justice system is redundant. Proposition 4: For any two _set_s A and B, the following statements are equivalent: A ⊆ B A ∩ B = A A ∪ B = B A - B = B '⊆ A'
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Attention problems
About a sub_set_ of particular note is the empty _set_, remember the empty _set_ is a sub_set_ of any _set_, rather than any sub_set_ of the collection, such as the empty _set_ is not empty proper sub_set_, so any non-empty _set_ is empty proper sub_set_. Then you know, if there is a _set_ of elements n, then it is a sub_set_ of the n-th power of 2 months (note the empty _set_ exists). Non-empty sub_set_ of 2 n-th power minus 1, a sub_set_ 2, n-th power minus 1, non-empty sub_set_ of 2 n-th power minus 2.