The Cartesian product of two sets is a set of ordered pairs. We also leave the term belonging undefined. If E denotes the set of all positive even numbers and O denotes the set of all positive odd numbers, then their union yields the entire set of positive integers, and their intersection is the empty set. Set theory - Set theory - Operations on sets: The symbol ∪ is employed to denote the union of two sets. The next question that comes to mind is: is the collection of subsets of a set, a set itself? First we show that \begin{equation} \forall C, (C\subseteq A\rightarrow C\subseteq B)\rightarrow A\subseteq B. The positive integers {1, 2, 3, …} are typically used for counting the elements in a finite set. For example, in the relation “is the same colour as,” each object bears the relation to itself as well as to some other objects. $$ By the principle of extension, this set is unique; and is called the Cartesian product of $A$ and $B.$ Thus, as desired, the Cartesian product of two sets is a set of ordered pairs. Let $x$ be an arbitrary element in $(A \cup B)’.$ Then \begin{alignat*}{2} & x\in (A \cup B)’ & \qquad & \\ & \quad \rightarrow [x\notin A \cup B] & & \text{by Definition of complement} \\ & \quad \rightarrow [\neg(x\in A \cup B)] & & \text{by Definition of $\notin$} \\ & \quad \rightarrow [\neg(x\in A \lor x\in B)] & & \text{by Definition of $\cup$} \\ & \quad \rightarrow [\neg(x\in A) \land \neg(x\in B)] & & \text{DeMorgan’s Law} \\ & \quad \rightarrow [x\notin A \land x\notin B] & & \text{by Definition of $\notin$} \\ & \quad \rightarrow [x\in A’ \land x\in B’] & & \text{by Definition of complement} \\ & \quad \rightarrow [x\in A’ \cap B’] & & \text{by Definition of $\cap$} \end{alignat*} Thus $x\in (A \cup B)’ \rightarrow x\in A’ \cap B’$ and consequently $(A \cup B)’ \subseteq x\in A’ \cap B’.$ In a similar fashion (simply reverse the implications) one may prove that $A’ \cap B’ \subseteq (A \cup B)’.$ Therefore, $(A \cup B)’=A’ \cap B’$. For example, suppose $a$ and $b$ are sets that are elements of the set $A,$ then by the principle of specification the set $\{a, b\}:=\{x\in A \mid x=a \text{ or } x=b \}$ exists, and by the principle of extension is unique. Sets are often also represented by letters, so this set might be E = {2, 4, 6, 8, 10, ...} . The Cartesian product of two sets A and B, denoted by A × B, is defined as the set consisting of all ordered pairs (a, b) for which a ∊ A and b ∊ B. If $A$ is a set, then $\emptyset \in \mathcal{P}(A)$ and $A\in \mathcal{P}(A).$, Proof. But in Calculus (also known as real analysis), the universal set is almost always the real numbers. 2. Thus all subsets of $A$ are proper subsets except the set $A$ itself, which is referred to as the improper subset of $A.$, Definition. There is a long interesting history of this problem and the reader is encourage to explore. Let $A$ be a subset of a universal subset $U.$ Then for all $x\in U,$ \begin{equation} \label{elesubst} x\in A \leftrightarrow \{x\}\subseteq A. Essential features of Cantorian set theory, Schemas for generating well-formed formulas, Schema for transfinite induction and ordinal arithmetic, Axiom for eliminating infinite descending species, Probability theory: The principle of additivity, Stanford Encyclopedia of Philosophy - Set Theory, MacTutor History of Mathematics Archive - A history of set theory. Thus we are lead to the definition of the intersection of a collection of sets $$ \bigcap_{X\in \mathcal{C}} X=\{x\mid x\in X \text{ for all $X$ in $\mathcal{C}$ }\} $$ where uniqueness is guaranteed by the principle of extension. Let $A$ and $B$ be sets. Given $n$ elements $a_1, a_2, \ldots, a_n,$ where $n\geq 3,$ we can define the ordered $n$-tuple $(a_1, a_2, \ldots a_n),$ in which $a_1$ i the first element, $a_2$ is the second element, and so on, and $a_n$ is the $n$-th element. Let $A$, $B$, and $C$ be sets. }$ Then by definition of $A\neq B,$ since $B$ contains an element not in $A,$ we have $A\neq B.$ Therefore we have $A\subseteq B$ and $A\neq B,$ and thus $A\subset B.$. Theorem. The intersection of a collection of sets is the set that contains all elements, each of which are in each of the sets in the collection, and no other elements. Then assume, independently, that $x\in B$ and show that $x\in A$ follows. A collection of sets is called disjoint if they have no elements in common. Theorem. $$ In a similar fashion (simply reverse the implications) one may prove that $(A\cap B)\cup (A\cap C) \subseteq A\cap (B \cup C).$ Therefore, $A\cap (B \cup C)=(A\cap B)\cup (A\cap C).$, The relative complement of $B$ with respect to a set $A$ is the set of elements in $A$ but not in $B.$, Definition. The symbol $\emptyset$ is the last letter in the Danish-Norwegian alphabet. \end{equation}, Proof. Symbolically: $$ \text{If $R = \{ x \mid x \notin x \},$ then $R \in R \leftrightarrow R \not \in R$. Let $A$ and $B$ be subsets of some universal set $U.$ $\qquad (1) $ $(A’)’=A$ $\qquad (2) $ $\emptyset ‘=U$ $\qquad (3) $ $U’=\emptyset $ $\qquad (4) $ $A \cap A’=\emptyset$ $\qquad (5) $ $A \cup A’ =U$ $\qquad (6) $ $A-B=A\cap B’$, Theorem. Let $S=\{a,b,c\},$ $T=\{1,a\},$ and $V=\{1,2,3,c\}.$ Find and compare the sets, $(S-T)-V$ and $S-(T-V).$, Solution. These example sentences are selected automatically from various online news sources to reflect current usage of the word 'set theory.' Let $A$ and $B$ be sets. A set is a group of objects. A set $A$ is called a subset of a set $B,$ denoted by $A\subseteq B,$ provided every element of $A$ is also an element of $B.$. \end{equation} Assume that for any set $C,$ if $C\subseteq A,$ then $C\subseteq B.$ Since $A\subseteq A,$ it follows that $A\subseteq B$ as needed. The intersection operation is denoted by the symbol ∩. Let $A$ and $B$ be subsets of some universal set $U.$ The relative complement of $B$ in $A$ is the set $A-B$ defined by $$A-B =\{x\mid x\in A \land x\notin B\}.$$ The relative complement of $B$ in $A$ is also denoted by $A\setminus B.$, When all sets under consideration are considered to be subsets of a given set $U,$ the absolute complement of $A$ is the set of all elements in $U$ but not in $A,$ and is denoted by $A’,$ that is, the complement of a set $A$ is defined by $$A’=\{x\in U \mid x\not\in A\}. If the occurrences of x and y are replaced by names of appropriate, specific objects, the result is a declarative sentence that is true or false. This pairing can be represented by the set {(x, 4), (z, 3), (w, 9)} of ordered pairs. (Note that the ordering relation is not symmetric.) We will say that a set is a collection of objects. Figure 1.16 pictorially verifies the given identities. (2): Let $x$ be an arbitrary element in $A\cap B.$ Then \begin{alignat*}{2} & x\in A \cap B & \qquad & \\ & \quad \rightarrow x\in A\land x\in B & & \text{by Definition of $\cap$} \\ & \quad \rightarrow x\in B\land x\in A & & \text{by commutativity of $\land$} \\ & \quad \rightarrow x\in B \cap A & & \text{by Definition of $\cap$} \end{alignat*} Thus $x\in A\cap B\rightarrow x\in B\cap A$ and consequently $$ A\cap B\subseteq B\cap A.$$ In a similar fashion (simply reverse the implications) one may prove that $B\cap A\subseteq A\cap B.$ Therefore, $A\cap B=B\cap A$. For a brief discussion of the reviews of (elementary) Halmos’ Naive Set Theory read this. Let $R$ be the set of all sets that are not members of themselves. (Specification) To every set $A$ and to every condition $P(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $P(x)$ holds. Prove or disprove that $$ \mathcal{P}(A\times B)=\mathcal{P}(A)\times \mathcal{P}(B), $$ for any sets $A$ and $B.$. For a brief discussion of … Be on the lookout for your Britannica newsletter to get trusted stories delivered right to your inbox. Thus, given a set $C$ we can form a collection of sets, denoted by $\mathcal{P},$ that contains the subsets of $C.$ We can use the principle of specification to ensure that this set will consist only of subsets of $C$ and the principle of extension to ensure that this set is unique. Let $A$ and $B$ be subsets of some universal set $U.$ $\qquad (1) $ $A\cap \emptyset =\emptyset $ $\qquad (2) $ (commutativity) $A\cap B=B\cap A$ $\qquad (3) $ (associativity) $A\cap (B \cap C) =(A\cap B)\cap C$ $\qquad (4) $ (idempotence) $A \cap A =A$ $\qquad (5) $ $A\subseteq B$ if and only if $A\cap B=A$.
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