Notes - Mathematics Olympiads - Set Theory

**Category : **11th Class

**Set Theory**

**A Set**

A set is the collection of things which is well-defined. Here well-defined means that group or collection of things which is defined distinguishable and distinct e.g. Let A is the collection of the group M = {cow, ox, book, pen, man}.

Is this group collection is a set or not? Actually this is not a set. Because it is collection of things but it cannot be defined in a single definition.

For example, A = {1, 2, 3, 4, 5,... n}

Here, collection A is a set because A is group or collection of natural numbers.

**e.g.** A = {a, e, i, o, u}= {x/x : vowel of English alphabet}

**Distinguish Between a Set and a Member**

**e.g.** A = {2, 5, 8, 7}

\[2\in A\]. It is read as 2 belongs to set A. Or 2 is the member/element of set A.

\[6\notin A\]- It means 6 doesn't belong to set A. A set is represented by capital letter of English/Greek Alphabet.

\[\alpha ,\beta ,\gamma ,\delta \]or A, B, C, U, S etc.

and its all elements is closed with { } bracket.

Hence what is the difference between 2 and {2}

\[\because \]2 can be element of a set whereas {2} represents a set whose one elements is 2.

**Note:** A set is the collection of **distinguish** and **distinct** things

A set can be written/represented in the two form

** **

**(a)** **Tabular form/Roaster form**

**(b) Set builder form**

**e.g.**

- \[A=\{a,e,i,o,u\}\to \]Tabular form/roaster form

= {\[x/x:\] a vowel of english letter} = It is said to be set-builder form

**Note:** \[x:x\]or \[x/x\]is read as x such that \[x\]

- A = {2, 4, 6, 8, 10} = {\[x/x\], is an even numbers \[\le \]10}

** **

**Type of Sets**

** **

**(i) Null Set:** A set which has no element, is said to be a null set and denoted by \[\phi \,or\,\{\}\]

**e.g.** **(a)** a set of three-eyed men

**(b)** Set of real solution of the equation \[{{x}^{2}}+1=0,\] \[{{x}^{2}}=-1\Rightarrow x=\sqrt{-1}=\]imaginary

** **

**(ii) Singleton Set:** A set having single element is said to be singleton set.

**e.g.** A = {2}

**(iii) Finite Set:** A set having definite and countable elements, is said to be finite set

**e.g.** \[A=\{a,e,i,o,u\}\]

**(iv) Infinite Set:** A set having uncountable and indefinite elements is said to be infinite set.

**e.g.** a set of stars.

**(v) Equal set:** Two or more than two sets are said to be equal sets if they have the same elements.

**e.g.** \[A=\{1,\,2,\,5,\,7,\,8\}\], \[B=\{2,\,5,8,1,7\}\]. So, \[A=B\]

**Operations on Sets**

- Union-operation
- Intersection operation
- Complement operation
- Difference operation

**e.g.** \[A=\{1,\,2,\,3,\,5\}\], \[B=\{2,\,5,7,\,8\}\]

\[A\,\bigcup B=\{1,\,2,3,5,7,8\}\], \[A\,\bigcap \,B=\{2,\,5\}\]

**Union Set**: Let A and B be two non-empty sets then A union B i.e. \[A\,\bigcup B\]is a set whose elements are the element of set A or B or both A and B.

**Intersection Set:**Let A and B be two non-empty sets. Then \[A\,\bigcap \,B\] is a set whose elements are the element of both A and B set.

**Universal Set:**A collection of given all set is said to be universal set generally universal set is denoted by\[\,\bigcup \].

**e.g.** \[\mathbf{U}=\{1,\,2,3,4,\,5,6,\,7,8,\,9,\,10\}\] Universal set, \[A=\{1,\,4,\,5,\,10\}\], \[B=\{2,\,3,6,\,7\}\]

\[\therefore \,\,A-B=\]difference between A and B\[=\{1,\,4,\,5,\,10\}\], \[\,B-A=\{2,\,3,6,7\}\]

A' = complement of \[A=\bigcup -A=\{2,\,3,\,6,\,7,8\}\], B' \[=\mathbf{U}-B\]

\[=\{1,\,2,3,4,\,5,6,\,7,8,\,9,\,10\}-\{2,\,3,\,6,\,7\}=\{1,\,4,\,5,\,8,\,10\}\]

\[A\,\Delta \,B=\]Symmetric difference between A & B \[=\mathbf{(A-B)}\,\,\mathbf{U}\,\,\mathbf{(B-A)}\]

\[=\{1,\,4,\,5,10\}\,\,\bigcup \,\,\{2,\,3,\,6,\,7\}=\{1,\,2,3,\,4,\,5,6,7,10\}\]

**Venn Diagram**

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**\[U=\{1,\,2,3,4,5,6,7,8,9,\,10\}\]**

\[A=\{2,\,3,\,4,\,6,7,8,\,10\}\]

\[B=\{1,3,\,5,7,\,\}\]

\[A\,\,\bigcup \,\,B=\{1,\,2,3,4,5,\,6,7,8,10\}\]

\[A\,\,\bigcap \,\,B=\{3,7\}\]

**Power Set:**Let A is a set. Total no. of all possible subsets of set A is denoted by\[{{2}^{n(A)}}\]. It is said to be power set. It is represented by**P(A)**

**e.g.** \[A=\mathbf{\{2,}\,\mathbf{3,}\,\mathbf{5\}}\], \[n(A)=3\]. Total number of subset \[={{2}^{n(A)}}={{2}^{3}}=8\]

**Possible Subset:**\[\{\phi ,\,\{2\},\,\{3\},\,\{5\},\,\{2,\,3\},\,\{3,\,5\},\,\{2,\,5\},\,\{2,3,\,5\}\}\]

**Definition (Subset):**Let A & B be two non-empty sets. We say that A is subset of B. It means that all elements of set A contains in set B and it is denoted by \[A\subset B\]

**e.g.** \[A=\{2,\,4,\,5,6\}\],\[B=\{1,\,2,4,\,5,\,6,7,\,8\}\], then \[A\subseteq B\] is true?, \[B\subseteq A\]

\[B\supseteq A\to B\]is a subset of A

**Power Set:**\[A=\{2,\,3,8\}\]

\[P(A)=\{\phi ,\,\{2\},\{3\},\{8\},\{2,3\},\{3,8\},\{2,8\},\{2,3,\,8\}\}\]

**Cartesian Product of Two sets:**Let A & B be two non-empty sets then cartesian product of A & B is written as \[A\times B\]& it is set of ordered pair of element of two sets.

Mathematically, it is written as

\[A\times B=\{(x,y);\,\,x\in A\,\And Y\in B\}\]

\[(x,y)\to \] Linear ordered pair

**e.g.** \[A=\text{ }\!\!\{\!\!\text{ 1,}\,\text{2,}\,\text{3 }\!\!\}\!\!\text{ }\], \[\text{B= }\!\!\{\!\!\text{ 2,}\,\text{5 }\!\!\}\!\!\text{ }\]

\[A\times B=\{1,\,2,\,3\}\times \{2,5\}=\{(1,\,2),\,(1,5),(2,\,2),(2,5),(3,\,2),\,(3,5)\}\]

\[B\times A=\{2,5\}\times \{1,\,2,3\}=\{(2,1),(2,2),(2,3),(5,1),(5,\,2),(5,\,3)\}\]worked on

** **

**Find x & y of**

\[(3x+y,x-1)=(x+3,2y-2x)\],

\[3x+y=x+3\Rightarrow 3x-x+y=3\]

\[\Rightarrow 2x+y=3\] ......... (1)

\[x-1=2y-2x\]

\[2y-2x-x=-1\Rightarrow 2y-3x=-1\] ....... (2)

Solving (1) & (2), we have

\[2x+y=3\],\[2y-3x=-1\]

subtracting (2) from (1) x 2, we have

\[4x+2y=6\], \[2y-3x=-1\]

\[7x=7\Rightarrow x=\frac{7}{7}=1\]

Putting the value of x in (1), we have

\[2\times 1+y=3\Rightarrow y=3-2=1\]

\[x=1\], \[y=1\]

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**Q**. If \[A=\{x:{{x}^{2}}-5x+6=0\}\]

\[B=\{2,\,4\}\And C=\{4,\,5\}\]

Find \[(A-B)\times (B-C)\]

**Sol:** \[{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5x+6=0}\]

\[\Rightarrow \,\,\,{{x}^{2}}-3x-2x+6=0\]

\[\Rightarrow \,\,\,x(x-3)-2(x-3)=0\]

\[\Rightarrow \,\,\,(x-3)(x-2)=0\]

\[\Rightarrow \,\,\,x-2,3\]

Now, \[A=\{x:{{x}^{2}}-5x+6=0\}=\{2,3\}\]

\[A-B=\{2,\,3\}-\{2,\,4\}=\{3\}\]

\[B-C=\{2,\,4\}-\{4,\,5\}=\{2\}\]

\[(A-B)\times (B-C)=\{3\}\times \{2\}=\{(3,\,2)\}\]

*play_arrow*Notes - Mathematics Olympiads - Set Theory

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