Let x be an arbitrary element in the universe. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written \(A\cap B\) and defined if and only if B and . (See section 2.2 example 10 for that. Back to Schedule I think either of those are valid proofs. This is a contradiction, so \(|A-B|\le|A|\).∎. Thus, in particular, x ∈ A is true. \[A\cup B\cup C \cup D\,,\\A\cap B\cap C \cap D\,.\]. e.g. Try to prove and . By definition of set difference, x ∈ A− B. B Then since by the definition of . \[\bigcup_{i=1}^{n} S_i\,.\] Hence . Thus \(A-B\not\subseteq A\). when we're working with real numbers, probably \(U=\mathbf{R}\). For example, (b) can be proven as follows: 6. Alternative proof We have used the choose-an-element method to prove Propositions 5.7, 5.11, and 5.14. though they can be proven also using some of these properties (after those properties are proven, needless to say). 11. by the definition of . Hence . &= \{x\mid \neg(x \in A)\wedge \neg(x\in B )\} \\ A For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. Next -- Recursive Definition Hence . A-(B\cup C) x This section contains many results concerning the properties of the set operations. We can use the set identities to prove other facts about sets. A. B ) and implications Furthermore a similar correspondence exists between &= \{x\mid x\notin (A\cup B)\} \\ &= \{x\mid x \in (\overline{A}\cap\overline{B}) )\} \\ \] The “more formal” version has more steps and leaves out the intuitive reason (that might help you actually remember why). -------     Commutative Laws Be careful with the other operations. and between U and Theorem: For any sets, \(|A\cap B|\le|A|\) and \(|A\cap B|\le|B|\). Then . Proof for 9: Let x be an arbitrary element in the universe. ( cf. ) by the definition Here is an example. There's more to it than similar-looking symbols. A \overline{A\cup B} and vice versa. We'll be careful for this one and manipulate the set builder notation. . , As an example, we can prove one of De Morgan's laws (the book proves the other). Let the sets \(S_1,S_2,\ldots ,S_n\) be the students in each course. If , then and . B . Since we're doing the same manipulations, we ended up with the same tables. With similar proofs, we could prove these things: When doing set operations we often need to define a. equalities involving set operations intersection of sets subset relations proofs of equalities proofs of subset relations Contents . = ( A ) by the definition of ( B - A ) . Proof for 4: A Since (use "addition" rule), \[\{1,2,3,4\}-\{3,4,5,6\} = \{1,2\}\,\\ A.   by the commutativity of For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). B ( A -------     De Morgan's Laws A We have already proved some of the results. 7. 12. if and only if and &= (A-B)\cap (A-C)\,.\quad{}∎ The if part can be proven similarly. A Theorem For any sets A and B, B ⊆ A∪ B. Proof… x -------     Domination Laws the commutativity of If , then . \(A\cup{U}= {U}\\A\cap\emptyset= \emptyset\), \((A\cup B)\cup C = A\cup(B\cup C)\\(A\cap B)\cap C = A\cap(B\cap C)\), \(A\cup(B\cap C) =(A\cup B)\cap(A\cup C)\\A\cap(B\cup C) = (A\cap B)\cup(A\cap B)\), \(\overline{A\cap B}=\overline{A} \cup \overline{B}\\\overline{A\cup B}= \overline{A} \cap \overline{B}\), \(A\cup(A\cap B) = A \\ A\cap(A\cup B) = A\), \(A\cup\overline{A} = {U}\\A\cap\overline{A} = \emptyset\), Written \(A\cup B\) and defined Hence A satisfies the conditions for the complement of . Set. B, by 7 &= A\cap \overline{B\cup C} \\ \end{align*}\]. Return to the course notes front page. The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. 3. Since , . Using set-builder notation, we can define a number of common sets and operations. 4. Here are some basic subset proofs about set operations. Since A x from the equivalences of propositional logic. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: 1 - 6 directly correspond Be careful with the other operations. B B If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. Often not explicitly defined, but implicit based on the problem we're looking at. and that of The set \(\overline{B}\) is the set of all values not in \(B\). We are going to prove this by showing that every element that is in Others will be proved in this section or in the exercises.

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