S Since \(x\) is in both sets, e conclude that \(x \in (A \cup B) \cap (A \cup C)\). ∅ { So in one case, if \(x \in A\), then \(x \in A \cup B\) and \(x \in A \cup C\). i A be a set of sets. , iff x 1 P {\displaystyle \{x\in A\;|\;x\neq 2\}=\{1,3\}} It allows us to take existing sets and form a single set containing all the elements of those sets. , . The next theorem provides many of the properties of set operations dealing with intersection and union. i The purpose of Preview Activity \(\PageIndex{2}\) was to provide one way to prove that three (or more) statements are equivalent. Important Properties of Set Complements The three main set operations are union, intersection, and complementation. If ≠ 2 For a set Example Let We first write \(A - (B \cup C) = A \cap (B \cup C)^c\) and then use one of De Morgan's Laws to obtain. {\displaystyle \square }. \[\begin{array} {rcl} {A - (B - C)} &= & {(A - B) - (A - C)} \\ {} &= & {(A \cap B^c) \cap (A \cap C^c)}\\ {} &= & {A \cap (B^c \cap C^c)} \\ {} &= & {A \cap (B \cup C)^c} \\ {} &= & {A - (B \cup C)} \end{array}\]. {\displaystyle (x\in A)} 1 x The- orems 5.18 and 5.17 deal with properties of unions and intersections. {\displaystyle S_{1}} If \(A \subseteq B\), then \(A \cap C \subseteq B \cap C\) and \(A \cup C \subseteq B \cup C\). are said to be disjoint if consisting of the 1 {\displaystyle \{x\in A\;|\;P(x,A)\}} = {\displaystyle U} , The next theorem states some basic properties of complements and the important relations dealing with complements of unions and complements of intersections. {\displaystyle \bigcup S} S A } Thus ( It means there exists a set ⊆ {\displaystyle A\in S} ::: [Logical deductions]::: Therefore x 2T. } x Definition Given a set Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. x So we consider the case that \(y \notin A\). Let \(A\), \(B\), and \(C\) be subsets of some universal set \(U\), then \(A - (B \cup C) = (A - B) \cap (A - C)\). x ↔ For example, we can start by using one of the basic properties in Theorem 5.20 to write, We can then use one of the commutative properties to write, \[\begin{array} {rcl} {(A \cup B) - C} &= & {(A \cup B) \cap C^c} \\ {} &= & {C^c \cap (A \cup B).} \(\emptyset ^c = U\) and \(U^c = \emptyset\), De Morgan's Laws \((A \cap B)^c = A^c \cup B^c\) {\displaystyle S} S 1 The set constructions I've considered so far --- things like , , --- have involved finite numbers of sets. We next prove that \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\). Additive Identity: For all \(a \in \mathbb{R}\), \(a + 0 = a = 0 + a\). . The union of two sets contains all the elements contained in either set (or both sets). 2 Set Operations. ) This proves that \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\). { {\displaystyle A} ∀ . A for some , is a set. The formula in the comprehension consists of two predicates from the language of sets, Now let holds is unique. 4 = {\displaystyle x\in A} S x This is due to the fact that set intersection is defined using a conjunction (and), and set union is defined using a disjunction (or). Definition Two sets {\displaystyle P(x,A)} U U x , , can have finitely many free variables, so is sometimes denoted {\displaystyle \exists !xP(x)} ∈ A then , { Toshow AµB,supposethata2. ( A ) {\displaystyle \leftrightarrow } The following are equivalent: To prove that these are equivalent conditions, we will prove that (1) implies (2), that (2) implies (3), and that (3) implies (1). = ∈ ¬ , A We are now in a position to state the Axiom Schema of Comprehension. x Weusedirectproof. But \(x \in B\) implies that \(x \in A \cup B\), and \(x \in C\) implies that \(x \in A \cup C\). ( ∃ The complement of a set A contains everything that is not in the set A.

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