prove that a intersection a is equal to a

One way to prove that two sets are equal is to use Theorem 5.2 and prove each of the two sets is a subset of the other set. You show that a is, in fact, divisible by b, b is divisible by a, and therefore a = b: 36 member and advisers, 36 dinners: 36 36. \end{aligned}\] We also find $\overline{A} = \{4,5\}$, and $\overline{B} = \{1,2,5\}$. That, is assume $\ldots$ is not empty. Toprove a set is empty, use a proof by contradiction with these steps: (1) Assume not. Generally speaking, if you need to think very hard to convince yourself that a step in your proof is correct, then your proof isn't complete. $ 1.Both pairs of opposite sides are parallel. Step by Step Explanation. Is it OK to ask the professor I am applying to for a recommendation letter? Hence (A-B) (B -A) = . What is the meaning of $A\subseteq B\cap C$? Then do the same for ##a \in B##. This position must live within the geography and for larger geographies must be near major metropolitan airport. Then a is clearly in C but since A \cap B=\emptyset, a is not in B. Proof. A\cup \varnothing & = \{x:x\in A \vee x\in\varnothing \} & \text{definition of union} Explain. (A B) is the set of all the elements that are common to both sets A and B. Here is a proofof the distributive law $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$. Thus, A B = B A. Thus $A \cup B$ is, as the name suggests, the set combining all the elements from $A$ and $B$. AC EC and ZA = ZE ZACBZECD AABC = AEDO AB ED Reason 1. The zero vector $\mathbf{0}$ of $\R^n$ is in $U \cap V$. Check out some interesting articles related to the intersection of sets. How to write intermediate proof statements inside Coq - similar to how in Isar one has `have Statement using Lemma1, Lemma2 by auto` but in Coq? It is clear that \[A\cap\emptyset = \emptyset, \qquad A\cup\emptyset = A, \qquad\mbox{and}\qquad A-\emptyset = A.\] From the definition of set difference, we find $\emptyset-A = \emptyset$. How would you prove an equality of sums of set cardinalities? Not sure if this set theory proof attempt involving contradiction is valid. As $A^\circ \cap B^\circ$ is open we then have $A^\circ \cap B^\circ \subseteq (A \cap B)^\circ$ because $A^\circ \cap B^\circ$ is open and $(A \cap B)^\circ$ is the largest open subset of $A \cap B$. These remarks also apply to (b) and (c). Hope this helps you. The students who like brownies for dessert are Ron, Sophie, Mia, and Luke. Exercise $\PageIndex{10}\label{ex:unionint-10}$, Exercise $\PageIndex{11}\label{ex:unionint-11}$, Exercise $\PageIndex{12}\label{ex:unionint-12}$, Let $A$, $B$, and $C$ be any three sets. The word "AND" is used to represent the intersection of the sets, it means that the elements in the intersection are present in both A and B. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. $ Job Description 2 Billion plus people are affected by diseases of the nervous system having a dramatic impact on patients and families around the world. (i) AB=AC need not imply B = C. (ii) A BCB CA. We have \[\begin{aligned} A\cap B &=& \{3\}, \\ A\cup B &=& \{1,2,3,4\}, \\ A - B &=& \{1,2\}, \\ B \bigtriangleup A &=& \{1,2,4\}. Consequently, saying $x\notin[5,7\,]$ is the same as saying $x\in(-\infty,5) \cup(7,\infty)$, or equivalently, $x\in \mathbb{R}-[5,7\,]$. Hence the intersection of any set and an empty set is an empty set. How many grandchildren does Joe Biden have? $x \in A \text{ or } x\in \varnothing The wire harness intersection preventing device according to claim 1, wherein: the equal fixedly connected with mounting panel (1) of the left and right sides face of framework (7), every mounting hole (8) have all been seted up to the upper surface of mounting panel (1). Show that A intersection B is equal to A intersection C need not imply B=C. rev2023.1.18.43170. Of the prove that a intersection a is equal to a of sets indexed by I everyone in the pictorial form by using these theorems, thus. Home Blog Prove union and intersection of a set with itself equals the set. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. For three sets A, B and C, show that. Their Chern classes are so important in geometrythat the Chern class of the tangent bundle is usually just called the Chern class of X .For example, if X is a smooth curve then its tangent bundle is a line bundle, so itsChern class has the form 1Cc1.TX/. must describe the same set. Thus, . Let ${\cal U} = \{\mbox{John}, \mbox{Mary}, \mbox{Dave}, \mbox{Lucy}, \mbox{Peter}, \mbox{Larry}\}$, \[A = \{\mbox{John}, \mbox{Mary}, \mbox{Dave}\}, \qquad\mbox{and}\qquad B = \{\mbox{John}, \mbox{Larry}, \mbox{Lucy}\}.\] Find $A\cap B$, $A\cup B$, $A-B$, $B-A$, $\overline{A}$, and $\overline{B}$. Do peer-reviewers ignore details in complicated mathematical computations and theorems? But then Y intersect Z does not contain y, whereas X union Y must. The statement should have been written as $x\in A \,\wedge\, x\in B \Leftrightarrow x\in A\cap B$., (b) If we read it aloud, it sounds perfect: \[\mbox{If $x$ belongs to $A$ and $B$, then $x$ belongs to $A\cap B$}.\] The trouble is, every notation has its own meaning and specific usage. $$ ST is the new administrator. And no, in three dimensional space the x-axis is perpendicular to the y-axis, but the orthogonal complement of the x-axis is the y-z plane. WHEN YOU WRITE THE UNION IT COMES OUT TO BE {1,2,3,4,5} Exercise $\PageIndex{8}\label{ex:unionint-08}$, Exercise $\PageIndex{9}\label{ex:unionint-09}$. No tracking or performance measurement cookies were served with this page. How do you do it? Thus, . we want to show that $x\in C$ as well. Solution For - )_{3}. Connect and share knowledge within a single location that is structured and easy to search. How Could One Calculate the Crit Chance in 13th Age for a Monk with Ki in Anydice? Forty Year Educator: Classroom, Summer School, Substitute, Tutor. If x (A B) (A C) then x is in (A or B) and x is in (A or C). \end{aligned}\] Express the following subsets of ${\cal U}$ in terms of $D$, $B$, and $W$. A union B is equal to a union if we are given that condition. Let us start with the first one. The result is demonstrated by Proof by Counterexample . or am I misunderstanding the question? It is called "Distributive Property" for sets.Here is the proof for that. Example: If A = { 2, 3, 5, 9} and B = {1, 4, 6,12}, A B = { 2, 3, 5, 9} {1, 4, 6,12} = . (b) Union members who voted for Barack Obama. Is it OK to ask the professor I am applying to for a recommendation letter? (Basically Dog-people). I've looked through the library of Ensembles, Powerset Facts, Constructive Sets and the like, but haven't been able to find anything that turns out to be useful. The properties of intersection of sets include the commutative law, associative law, law of null set and universal set, and the idempotent law. Prove that $A\cup \!\, \varnothing \!\,=A$ and $A\cap \!\, \varnothing \!\,=\varnothing \!\,$. Download the App! Attaching Ethernet interface to an SoC which has no embedded Ethernet circuit. The intersection of two sets is the set of elements that are common to both setA and set B. Lets provide a couple of counterexamples. For showing $A\cup \emptyset = A$ I like the double-containment argument. $$ Basis and Dimension of the Subspace of All Polynomials of Degree 4 or Less Satisfying Some Conditions. Could you observe air-drag on an ISS spacewalk? $$ . rev2023.1.18.43170. = {$x:x\in \!\, A$} = A, $A\cap \!\, \varnothing \!\,=$ {$x:x\in \!\, A \ \text{and} \ x\in \!\, \varnothing \!\,$} Conversely, $A \cap B \subseteq A$ implies $(A \cap B)^\circ \subseteq A^\circ$ and similarly $(A \cap B)^\circ \subseteq B^\circ$. How could magic slowly be destroying the world? If set A is the set of natural numbers from 1 to 10 and set B is the set of odd numbers from 1 to 10, then B is the subset of A. (a) People who did not vote for Barack Obama. Prove union and intersection of a set with itself equals the set. Find centralized, trusted content and collaborate around the technologies you use most. Intersection of a set is defined as the set containing all the elements present in set A and set B. And thecircles that do not overlap do not share any common elements. $S \cap T = \emptyset$ so $S$ and $T$ are disjoint. C is the point of intersection of the reected ray and the object. About; Products For Teams; Stack Overflow Public questions & answers; For any two sets A and B, the intersection, A B (read as A intersection B) lists all the elements that are present in both sets, and are the common elements of A and B. If so, we want to hear from you. If $A\subseteq B$, what would be $A-B$? Let A and B be two sets. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A\cap\varnothing & = \{x:x\in A \wedge x\in \varnothing \} & \text{definition of intersection} Theorem $\PageIndex{2}\label{thm:genDeMor}$, Exercise $\PageIndex{1}\label{ex:unionint-01}$. Write, in interval notation, $(0,3)\cup[-1,2)$ and $(0,3)\cap[-1,2)$. The Zestimate for this house is $330,900, which has increased by $7,777 in the last 30 days. Try a proof by contradiction for this step: assume ##b \in A##, see what that implies. Here c1.TX/ D c1. In particular, let A and B be subsets of some universal set. Operationally speaking, $A-B$ is the set obtained from $A$ by removing the elements that also belong to $B$. 100 - 4Q * = 20 => Q * = 20. Let us start with a draft. $ To prove that the intersection U V is a subspace of R n, we check the following subspace criteria: The zero vector 0 of R n is in U V. For all x, y U V, the sum x + y U V. For all x U V and r R, we have r x U V. As U and V are subspaces of R n, the zero vector 0 is in both U and V. Hence the . \\ & = A ki Orijinli Doru | Topolojik bir oluum. Next there is the problem of showing that the spans have only the zero vector as a common member. Let s \in C\smallsetminus B. Since $S_1$ does not intersect $S_2$, that means it is expressed as a linear combination of the members of $S_1 \cup S_2$ in two different ways. I get as far as S is independent and the union of S1 and S2 is equal to S. However, I get stuck on showing how exactly Span(s1) and Span(S2) have zero as part of their intersection. Circumcircle of DEF is the nine-point circle of ABC. The following table lists the properties of the intersection of sets. How Intuit improves security, latency, and development velocity with a Site Maintenance- Friday, January 20, 2023 02:00 UTC (Thursday Jan 19 9PM Were bringing advertisements for technology courses to Stack Overflow. The 3,804 sq. Proving two Spans of Vectors are Equal Linear Algebra Proof, Linear Algebra Theorems on Spans and How to Show Two Spans are Equal, How to Prove Two Spans of Vectors are Equal using Properties of Spans, Linear Algebra 2 - 1.5.5 - Basis for an Intersection or a Sum of two Subspaces (Video 1). The total number of elements in a set is called the cardinal number of the set. We rely on them to prove or derive new results. The symbol used to denote the Intersection of the set is "". For the subset relationship, we start with let $x\in U $. Find A B and (A B)'. Linear Discriminant Analysis (LDA) is a popular technique for supervised dimensionality reduction, and its performance is satisfying when dealing with Gaussian distributed data. (b) Policy holders who are either female or drive cars more than 5 years old. A^\circ \cap B^\circ = (A \cap B)^\circ\] and the inclusion \[ Prove that 5 IAU BU Cl = |AI+IBl + ICl - IAn Bl - IAncl - IBnCl+ IAnBncl 6. Let $A$ and $B$ be arbitrary sets. This websites goal is to encourage people to enjoy Mathematics! 1.3, B is the point at which the incident light ray hits the mirror. \end{aligned}\] Describe each of the following subsets of ${\cal U}$ in terms of $A$, $B$, $C$, $D$, and $E$. The chart below shows the demand at the market and firm levels under perfect competition. Then Y would contain some element y not in Z. Solution: Given: A = {1,3,5,7,9}, B = {0,5,10,15}, and U= {0,1,3,5,7,9,10,11,15,20}. Bringing life-changing medicines to millions of people, Novartis sits at the intersection of cutting-edge medical science and innovative digital technology. Location. 36 dinners, 36 members and advisers: 36 36. In symbols, $\forall x\in{\cal U}\,\big[x\in A\cup B \Leftrightarrow (x\in A\vee x\in B)\big]$. Looked around and cannot find anything similar. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Prove two inhabitants in Prop are not equal? A-B=AB c (A intersect B complement) pick an element x. let x (A-B) therefore xA but xB. This is a unique and exciting opportunity for technology professionals to be at the intersection of business strategy and big data technology, offering well-rounded experience and development in bringing business and technology together to drive immense business value. Post was not sent - check your email addresses! It only takes a minute to sign up. Thus, our assumption is false, and the original statement is true. Is the rarity of dental sounds explained by babies not immediately having teeth? For any two sets A and B, the union of sets, which is denoted by A U B, is the set of all the elements present in set A and the set of elements present in set B or both. You can specify conditions of storing and accessing cookies in your browser, Prove that A union (B intersection c)=(A unionB) intersection (A union c ), (a) (P^q) V (~^~q) prepare input output table for statement pattern, divide the place value of 8 by phase value of 5 in 865, the perimeter of a rectangular plot is 156 meter and its breadth is 34 Meter. $A\subseteq B$ means: For any $x\in{\cal U}$, if $x\in A$, then $x\in B$ as well. 'http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs'); A U PHI={X:X e A OR X e phi} 2.Both pairs of opposite sides are congruent. Then and ; hence, . The deadweight loss is simply the area between the demand curve and the marginal cost curve over the quantities 10 to 20. we need to proof that A U phi=A, A = {2, 4, 5, 6,10,11,14, 21}, B = {1, 2, 3, 5, 7, 8,11,12,13} and A B = {2, 5, 11}, and the cardinal number of A intersection B is represented byn(A B) = 3. $x \in A \wedge x\in \emptyset$ by definition of intersection. Let the universal set ${\cal U}$ be the set of people who voted in the 2012 U.S. presidential election. In this problem, the element $x$ is actually a set. (a) These properties should make sense to you and you should be able to prove them. \end{align}$. The deadweight loss is thus 200. Prove that if $A\subseteq C$ and $B\subseteq C$, then $A\cup B\subseteq C$. AC EC and ZA ZE Prove: ABED D Statement Cis the intersection point of AD and EB. Since a is in A and a is in B a must be perpendicular to a. Since we usually use uppercase letters to denote sets, for (a) we should start the proof of the subset relationship Let $S\in\mathscr{P}(A\cap B)$, using an uppercase letter to emphasize the elements of $\mathscr{P}(A\cap B)$ are sets. What is mean independence? Asking for help, clarification, or responding to other answers. If you just multiply one vector in the set by the scalar . In words, $A-B$ contains elements that can only be found in $A$ but not in $B$. Therefore, You listed Lara Alcocks book, but misspelled her name as Laura in the link. \\[2ex] So, if$x\in A\cup B$ then$x\in C$. Let us earn more about the properties of intersection of sets, complement of intersection of set, with the help of examples, FAQs. = {$x:x\in \!\, \varnothing \!\,$} = $\varnothing \!\,$. (adsbygoogle = window.adsbygoogle || []).push({}); If the Quotient by the Center is Cyclic, then the Group is Abelian, If a Group $G$ Satisfies $abc=cba$ then $G$ is an Abelian Group, Non-Example of a Subspace in 3-dimensional Vector Space $\R^3$. We should also use $\Leftrightarrow$ instead of $\equiv$. How would you fix the errors in these expressions? AB is the normal to the mirror surface. . find its area. Example 3: Given that A = {1,3,5,7,9}, B = {0,5,10,15}, and U = {0,1,3,5,7,9,10,11,15,20}. A great repository of rings, their properties, and more ring theory stuff. Indefinite article before noun starting with "the", Can someone help me identify this bicycle? $A\cap \varnothing = \varnothing$ because, as there are no elements in the empty set, none of the elements in $A$ are also in the empty set, so the intersection is empty. A B = { x : x A and x B } {\displaystyle A\cap B=\ {x:x\in A {\text { and }}x\in B\}} In set theory, the intersection of two sets and denoted by [1] is the set containing all elements of that also . The complement of A is the set of all elements in the universal set, or sample space S, that are not elements of the set A . The mathematical symbol that is used to represent the intersection of sets is ' '. the probability of happening two events at the . We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. For subsets $A, B \subseteq E$ we have the equality \[ This looks fine, but you could point out a few more details. If seeking an unpaid internship or academic credit please specify. The intersection of sets fortwo given sets is the set that contains all the elements that are common to both sets. Is this variant of Exact Path Length Problem easy or NP Complete, what's the difference between "the killing machine" and "the machine that's killing". Define the subsets $D$, $B$, and $W$ of ${\cal U}$ as follows: \[\begin{aligned} D &=& \{x\in{\cal U} \mid x \mbox{ registered as a Democrat}\}, \\ B &=& \{x\in{\cal U} \mid x \mbox{ voted for Barack Obama}\}, \\ W &=& \{x\in{\cal U} \mid x \mbox{ belonged to a union}\}. In simple words, we can say that A Intersection B Complement consists of elements of the universal set U which are not the elements of the set A B. Prove that $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$. Looked around and cannot find anything similar, Books in which disembodied brains in blue fluid try to enslave humanity. Follow on Twitter: The role of luck in success has a relatively minor, albeit consistent history in academic discourse, with a striking lack of literature engaging with notions of luck within occupational environments. Intersection of sets can be easily understood using venn diagrams. (a) What distance will it travel in 16 hr? How to determine direction of the current in the following circuit? For example, let us represent the students who like ice creams for dessert, Brandon, Sophie, Luke, and Jess. For example, take $A=\{x\}$, and $B=\{\{x\},x\}$. a linear combination of members of the span is also a member of the span. Comment on the following statements. Proof. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. It's my understanding that to prove equality, I must prove that both are subsets of each other. Theorem $\PageIndex{1}\label{thm:subsetsbar}$. Let $A$, $B$, and $C$ be any three sets. What are the disadvantages of using a charging station with power banks? However, you should know the meanings of: commutative, associative and distributive. It may not display this or other websites correctly. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange LWC Receives error [Cannot read properties of undefined (reading 'Name')]. Of course, for any set $B$ we have Therefore Math Advanced Math Provide a proof for the following situation. X/ is the anticanonical class,whose degree is 2 2g, where g is the genus . Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, I believe you meant intersection on the intersection line. The Centralizer of a Matrix is a Subspace, The Subspace of Linear Combinations whose Sums of Coefficients are zero, Determine Whether a Set of Functions $f(x)$ such that $f(x)=f(1-x)$ is a Subspace, The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero, The Subspace of Matrices that are Diagonalized by a Fixed Matrix, Sequences Satisfying Linear Recurrence Relation Form a Subspace, Quiz 8. Enter your email address to subscribe to this blog and receive notifications of new posts by email. When was the term directory replaced by folder? Why is sending so few tanks Ukraine considered significant? This means X is in a union. For example, consider $S=\{1,3,5\}$ and $T=\{2,8,10,14\}$. Elucidating why people attribute their own success to luck over ability has predominated in the literature, with interpersonal attributions receiving less attention. So, . Intersection and union of interiors. hands-on exercise $\PageIndex{4}\label{he:unionint-04}$. { "4.1:_An_Introduction_to_Sets" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.2:_Subsets_and_Power_Sets" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.3:_Unions_and_Intersections" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.4:_Cartesian_Products" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.5:_Index_Sets_and_Partitions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1:_Introduction_to_Discrete_Mathematics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "2:_Logic" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3:_Proof_Techniques" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4:_Sets" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "5:_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6:_Relations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7:_Combinatorics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8:_Big_O" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", Appendices : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "authorname:hkwong", "license:ccbyncsa", "showtoc:yes", "De Morgan\'s Laws", "Intersection", "Union", "Idempotent laws" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FCourses%2FMonroe_Community_College%2FMTH_220_Discrete_Math%2F4%253A_Sets%2F4.3%253A_Unions_and_Intersections, $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}$ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$, \[\begin{aligned} A\cap B &=& \{3\}, \\ A\cup B &=& \{1,2,3,4\}, \\ A - B &=& \{1,2\}, \\ B \bigtriangleup A &=& \{1,2,4\}.
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