Sets, Relations and Functions (Part-2)


 SETS
Lesson 1

How are you, my dear friends? I hope all of you are doing well. With the class 12 pre-board exam coming up- and their joint Entrance examination preparation-most of my day is spent with my students. Preparing mock test papers, checking answer scripts, explaining their mistakes, and so on have kept me quite busy. As a result, my blog post got a little delayed. But as the saying goes, better late than never! The second part of set theory is Here.

In the last post we have discussed-what is set, different types of sets like finite set and infinite sets, null set, singleton set and equal sets.
Now we will discuss about Subsets, power set and Universal set. This is all about discussions in lesson 1. 

Subsets: If two sets `A` and `B` are such that every element of `A` is also an element of `B`, then `A` is called a subset of `B`. We write `A\subseteqB`.
If the set `A` is a subset of `B`, then `b` is called supper set of `A` and we write `A\supseteq B`.
For example, let `A=\{1,2,3\},B=\{1,2,3,4\}\ text{and}  C=\{1,2,5\}` .Clearly the each elements of `A` are also the element of `B` but not of `C` because `3` belongs to A but does not belong to `C`. Hence `A` is a subset of `B` but `A` is not a subset of `C` i.e., `A\subseteqB ` but `A` is not a subset of `C`.

Proper subsets: For the shake of convenience of discussion, we consider the null set `\phi` is a subset of every set and every set is a subset of itself. Thus if `A` is any set then `\phi\subseteqA` and `A\subseteqA` and `\phi` and `A` are called Improper subset of `A`. Except `\phi` and `A` if there are other subsets of set `A`, they are called Proper subsets of `A`. i.e, a set `A` is said to be proper subset of another set `B` if each element of `A` are also the element of `B` and in addition to this there should exists at least one element in set `B` which is not an element of `A`. In the above example `A` is a proper subset of `B` because each element of set `A` are also the elements of set `B` and we find that `4\inB` but `4\notinA`.  
The subsets of the set `A=\{a,b,c\}` are `\phi,\{a\},\{b\},\{c\},\{a,b\},\{b,c\},\{c,a\}\text {and} A` itself.
Remark: The total number of subsets (proper and improper) of a set having `n` number of elements is `2^n`.
Example of subsets: `Z,Z^+,Q,Q^+` are all proper subset of `R`; set of even numbers `\{2n:n\inZ\}`is a proper subset of `Z`; the set `\{x:x\inN  \text{and}  x>10\}` is a proper subset of `N`. 

Intervals as subsets of `R` The set `\{x:a<x<b\}` is called open interval, when `a,b\inR` and `a<b`. The open interval is written as `\left(a,b\right)` where `a,b` are not included in open interval. Incase `x` assumes all the real values in between `a` and `b` and both `a` and `b` are inclusive then the set  `\{x:a\leqx\leqb  \text{and}  x\inR\}` is called closed interval. The closed interval is written as `[a,b]`.
We can also have intervals closed at one end and open at the other, i.e., `\left[a,b\right)=\{x:a\leqx<b\}` is an open interval from `a \ text{to}  b` including `a` but excluding `b`. Again `\left(a,b\right]=\{x:a<x\leqb\}` is an open interval from `a` to `b` including `b` but excluding `a`.
The above notations provide an alternative way of designating the subsets of the set of real numbers. For example, `\left[0,\infty \right)` defines thee set of positive real numbers, while set `\left(0,\infty\right)` defines the set of negative real numbers.

Let us see, the intervals on number line.


Power Set: The power set of a given set `A` is the set of all subsets of it. suppose `A=\{a,b,c\}`. then its subsets are  `\phi,\{a\},\{b\},\{c\},\{a,b\},\{b,c\},\{c,a\}\text {and} A` itself. Now, the set consisting of the above subsets is called power set of set `A`. we write the power set of `A` as 
`P\left(A\right)=\{\phi,\{a\},\{b\},\{c\},\{a,b\},\{b,c\},\{c,a\},\{a,b,c\}\}`.

Remark: (a)The number of elements of a set is called cardinal number. If a set contains `x` number of elements, then we write `n\left(A\right)=x`. Thus for `n\left(A\right)=m`, `n\[P\left(A\right)\]=2^m`.
(b) A finite set having `n` elements has `2^n` subsets of which `2^n-1` are proper subsets.
(c) The elements of a power set are sets
(d) The singleton set `\{\phi\}` is the power set of`\phi` since `\phi` is a subset of any set.

Universal Set: In any discussion related to sets, when there exists a set that contains every set involved in that discussion as its subset, then that set is called a Universal Set.

For example, the set of natural numbers `N`, the set of whole numbers `Z`, the set of rational numbers `Q`, can all be considered as subsets of `R`. Hence, `R` can be considered the universal set in this case.
The universal set is generally denoted by `U` or `S`.

It should be noted that a universal set is not unique for all cases. For instance, the set of integers `Z` may serve as a universal set for the set of rational numbers `Q`, but the real number set `R` can also serve as a universal set. Let us look at a few more examples:

(i) The universal set of a set of right triangled triangle will be the set of all triangles.

(ii) The universal set of a set of isosceles triangles will also be the set of all triangles.

(iii) For a school, the set of all students at that school may be regarded as the universal set of the set of students of a particular class of that school.

Solved Example

1. Make correct statements by filling in the symbols `\subset` or ⊄ in the blank spaces:
(i)   `\{a,b,c\}........\{b,c,a,d,e\}`
(ii)  `\{\alpha,\beta,\gamma\}........\{\beta,\gamma,\delta\}`
(iii) `\{x:x  \text{is a student of class XI of your school}\}........\{x:x  \text{student of your school}\}`
(iv) `\{x;x  \text{is a circle in the plane}\}........\{x:x  \text{is a circle in the same plane with radius 1 unit}\}`

Solution (i): Let `A=\{a,b,c\}` and `B=\{b,c,a,d,e\}`. Here, `a,b,c\inA` also belongs to `B`. Hence  `\{a,b,c\}\subset\{b,c,a,d,e\}`.

(ii):  Let `A=\{\alpha,\beta,\gamma\}`  and  `B=\{\beta,\gamma,\delta\}`. Here, `\alpha\inA` but `\alpha\notinB`. Hence `A`⊄`B`

(iii): Let `A=\{x:x  \text{is a student of class XI of your school}\}` and  `B=\{x:x  \text{student of your school}\}`. Here, a student belongs to class XI is also belongs to the school. i.e., for each `x\in A\impliesx\inB`. Hence `A\subsetB`

(iv): Let `A=\{x:x  \text{is a circle in the plane}\}`  and  `B=\{x:x  \text{is a circle in the same plane with radius 1 unit}\}`. Here, `A` is a set of circles in a plane with different radius, but `B` is the set of circles in the same plane having radius 1 unit only. Thus `x\inA` does not imply `x\inB`. Hence `A`⊄`B`.

2. Examine whether the following statements are true or false.
(i)   `\{a,b\}` ⊄ `\{b,c,a\}`
(ii)  `\{a,e\}\subset\{x:x  \text{is a vowel in English alphabet}\}`
(iii)  `A=\{x:x  \text{is a isosceles triangle}\}`, `B=\{x:x  \text{is an equilateral triangles} \}`. Then `B\subsetA`
(iv)  `\{3\}\subset\{1,3,5\}`

Solution (i): All the elements of the first set are also the elements of second set, and the second set contains the element `c` which is not present in the first set. Hence the first set is a proper subset of the second set. So, the given statement is false.

(ii): Let `A=\{a,e\}` and  `B=\{x:x  \text{is a vowel in English alphabet}\}=\{a,e,i,o,u\}`. Clearly, `x\inA\impliesx\inB` and there is at least one element in set `B` which is not present in set `A`. Hence `A\subsetB`. So, the given statement is true.  

(iii): We, know that any two sides of any equilateral triangle must be equal and as a result of that an equilateral triangle is also an element of set of isosceles triangle. `\therefore B\subsetA`. Hence the statement is true.

(iv):  Clearly, `\{3\}\subset\{1,3,5\}` is true. [why?]

3. Let `A=\{1,2,\{3,4\},5\}`. Which of the following statements are incorrect and why?
(i)   `\{3,4\}\subsetA`   (ii)  `\{3,4\}\inA`   (iii)  `\{\{3,4\}\}\subsetA`   (iv)  `1\inA`   (v)  `1\subsetA`   (vi)   `\{1,2,5\}\subsetA`    (vii)  `\{1,2,5\}\inA`   (viii)  `\{1,2,3\}\subsetA`   (ix)  `\phi\inA`   (x)  `\phi\subsetA`   (xi)  `\{\phi\}\subsetA`.

Solution (i):  `\{3,4\}` is an element of `A`, so it cannot be a subset of `A`. Hence the statement is incorrect.

(ii)   As `\{3,4\}` is an element of set `A`, hence `\{3,4\}\inA` is a correct statement.

(iii):  Here, `\{\{3,4\}\}` is a set which contains a single element `\{3,4\}\inA`. So, `\{\{3,4\}\}\subsetA`.

(iv) & (v):  We see that 1 is an element of `A`. So, `1\inA` is correct statement and `1\subsetA` is incorrect statment.  

(vi) & (vii):  Here, `1,2,5` are elemnts of set A, so `\{1,2,5\}\subsetA`. Hence the statement is (vi) correct and (vii) is incorrect.

 (viii): Here `3\in \{1,2,3\}` but `3\notinA`. Hence `\{1,2,3\}\subsetA` is incorrect statement.

 (ix) & (x):  We know that null set is a subset every set. Hence `\phi\subsetA` and `\phi\notinA`. Hence statement (ix) is incorrect and (x) is correct. 

(xi): `\{\phi\}` is not a subset of A, `\phi\subsetA`. Hence, the statement (xi) is incorrect.

4.  Write down all the subsets of the following set.
(i)   `\{a\}`
(ii)  `\{1,2\}`
(iii) `{\alpha,\beta,\gamma\}`
(iv) `\phi` 

Solution (i): The subsets of `\{a\}` are `\phi` and `\{a\}`.

(ii): The subsets of `\{1,2\}`are `\phi`, `\{1,2\}`, `\{1\}`, `\{2\}`

(iii): The subsets of `{\alpha,\beta,\gamma\}` are `\phi`,`{\alpha,\beta,\gamma\}`, `\{\alpha\}`, `\{\beta\}`, `\{\gamma\}`, `\{\alpha,\beta\}`, `\{\beta,\gamma\}`, `\{\gamma,\alpha\}`

(iv): The null set has no element, i.e., `n\left(\phi\right)=0`, so number of subsets is `2^0=1` i.e., `\phi` itself. Thus, the subsets of `\phi` is `\phi` only.

5. Decide, among the following sets, which sets are subsets of one and another:
`A=\{x:x\inR  \text{and satisfy}  x^2-8x+12=0\}`, `B=\{2,4,6\}`, `C=\{2,4,6,8,...\}`, `D=\{6\}`  

Solution: Here,             `x^2-8x+12=0`
`\implies`                      `\left(x-2\right)\left(x-6\right)=0`
`\implies`                        `x=2,x=6` 
`\therefore`        `A=\{2,6}`. `B=\{2,4,6\}`, `C=\{2,4,6,8,...\}`, `D=\{6\}`  
`2,6\inA` also belongs to `B  \text{and}  C`, Thus, `A\subsetB` , ` A\subsetC`. 
As, `2,4,6\inB` also belongs to `C`. Thus, `B\subsetC` 
Again, `6\inD` also belongs to `A,B  \text{and}  C`. Thus `D\subsetA` `D\subsetB` and `D\subsetC`. 

6. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

(i)    If `x\inA` and `A\inB`, then `x\inB`
(ii)   If `A\subsetB` and `B\inC`, then `A\inC`
(iii)  If `A\subsetB` and `B\subsetC`, then `A\subsetC`
(iv)  If `A` ⊄ `B` and `B` ⊄ `C`, then `A` ⊄ `C`
(v)   If `x\inA` and `A`⊄`B`, then `x\inB`
(vi)  If `A\subsetB` and `x\notinB`, then `x\notinA`.    

Solution (i): Let `A=\{a,b\}` and `B=\{\{a,b\},c,d\}` . Here, `a,b\inA` and `A\inB` but `a,b\notinB`. Thus, `x\inA` and `A\inB` need not imply that `x\inB`. Hence the statement is false.  

(ii): Let  `A=\{1,2\}`,  `B=\{0,1,2,3\}` and `C=\{\{0,1,2,3\},4,5\}`. Here,  `1,2\inA` and also belongs to `B`. Thus `A\subsetB`. Again, we see that `B\inC`, but `{1,2\}notinC`,i.e., `A\notinC`. Thus, `A\subsetB` and `B\inC` need not imply that `A\inC`. Hence the statement is false.  

(iii): Let `x\inA`. `\becauseA\subsetB` , `\therefore` `x\inA\impliesx\inB`. Again `\because`  `B\subsetC`  `\therefore`  `x\inB\impliesx\inC`.
`\therefore`   from the above, we have, `x\inA\impliesx\inC`.   So, `A\subsetC`. Hence the statement is true. 

(iv): Let `A=\{0,1\}` and `B=\{1,2,3\}` and `C=\{0,1,2,5\}`. Here, `A` ⊄ `B` and `B` ⊄ `C` but `A\subsetC`. Thus, `A` ⊄ `B` and `B` ⊄ `C` need not imply that `A` ⊄ `C`. Hence the statement is False. 

(v): Let `A=\{0,1\}` and `B=\{1,2,3\}`. Here, `A` ⊄ `B` as `0\inA` but `0\notinB`. Thus, `x\inA` and `A` ⊄ `B` need not imply that `x\inB`. Hence the statement is False. 

(vi): `A\subsetB\impliesx\inA\impliesx\inB` and `x\notinB\impliesx\notinA` [Definition of subset]. Hence the statement is true. 

7. Write the following as intervals:
(i) `\{x:x\inR,-4<x\leq6\}`        (ii) `\{x:x\inR,-12<x<-10\}`  
(iii) `\{x:x\inR,0\leqx<7\}`        (iv) `\{x:x\inR,3\leqx\leq4}` 

Solution (i): `\{x:x\inR,-4<x\leq6\}=\(-4,6\}`    (ii) `\{x:x\inR,-12<x<-10\}=\(-12,-10\)`
(iii) `\{x:x\inR,0\leqx<7\}=\[0,7\)`                            (iv)  `\{x:x\inR,3\leqx\leq4}=\[3,4\]` 

8.  Write the following intervals in set-builder form:
(i) `\(-3,0\)`    (ii) `\[6,12\]`    (iii) `\(6,12\]`    (iv) `\[-23,5\)`

Solution (i): `\(-3,0\)=\{x:x\inR,-3<x<0\}`   (ii) `\[6,12\]=\{x:x\inR,6\leqx\leq12\}`
(iii) `\(6,12\]=\{x:x\inR,6<xleq12\}`                     (iv) `\[-23,5\)=\{x:x\inR,-23\leqx<5\}`

9. Prove that- 
(i)   "The null set is a subset of any set."
(ii)  " Every set is a subset of itself"
(iii) " If `A\subseteqB` and `B\subseteqC` then prove that `A\subseteqC`."
(iv) "If `A\subseteqB` and `B\subseteqA`, then prove that `A=B` and conversely."

Solution (i): Let `A` be an arbitrary set. If possible, suppose `\phi` ⊄`A`. Then, there is at least one element in`\phi` which is not in `A`. This fact contradicts the very definition of `\phi` that `\phi` has no element. Hence, our assumption is wrong. `\therefore\phi\subsetA` i.e., the null set is subset of any set.

(ii): Let `A` be an arbitrary set and `x\inA` be any element, `\becausex\inA\impliesx\inA\thereforeA\subseteqA`.
Note that: Every set is a subset of itself, but it is not a proper subset of itself.

(iii): Let, `x\inA` be any element.
`\becauseA\subseteqB,\thereforex\in\impliesx\inB`.....(i), and
`\becauseB\subseteqC,\thereforex\inB\impliesx\inC`....(ii).
`\therefore`    from (i) and (ii), we have, `x\inA\impliesx\inC,\thereforeA\subseteqC`. 

(iv): Let `x\inA` be any element and `y\inB` be any element.
`\becauseA\subseteqB`,    `\thereforex\inA\impliesx\inB`...............(i)
`\becauseB\subseteqA`,     `thereforey\inB\impliesy\inA`................(ii) 
`\therefore`    from (i) and (ii), we have `A=B`.
Conversely, `\becauseA=B\thereforex\inA\impliesx\inB` ........(a) and `y\inB\impliesy\inA`........(b)
From (a) it follows that `A\subseteqB` and from, (b) it follows that `B\subseteqA`.  

10. Prove that the null set is unique.
Solution: If possible, let `\phi` and `\psi` are two null sets. As we know that null set is a subset of any set, therefore `\phi\subseteqpsi` and `\psi\subseteqphi`.
`\therefore`    `\phi=\psi`.  Hence the null set is unique.

11.  Prove that if `A\subseteqB` and `B=C`, then `A\subseteqC`
Solution: Let `x\inA` be any element.
`\becauseA\subseteqB\thereforex\inA\impliesx\inB`.......(i)
`\becauseB=C`,    `\thereforex\in\impliesx\inC` and `x\in\impliesx\inB`.........(ii)
From (i) and (ii), we have `x\inA\impliesx\inC.\thereforeA\subseteqC`.

12. If `\{a,\{b\}\}`, find `P(A)`.
Solution: Let `x=\{b\}`. Then `A=\{a,x\}`
`\therefore`    `P(A)=\{\phi,\{a\},\{x\},\{a,x\}\}=\{\phi,\{a\},\{\{b\}\},\{a,\{b\}\}\}`.

Exercise-1(B) 

(MCQ)

A. Chose thee correct Answer 

1.  If `A=\{1,3,5,B\}` and `B=\{2,4\}`, then
(a) `4\inA`
(b) `\{4\}\subsetA`
(c) `B\subsetA`
(d) none of these.

2. If `A=\{1,2,3,4,5\}`, then the number of proper subsets of `A` is 
(a) 120
(b) 30
(c) 31
(d) 32 

3. Two finite sets have `m` and `n` elements. The number of subsets of the first set is 112 more than that of the second. The values of `m` and `n` are respectively 
(a) 4,7
(b) 7,4
(c) 4,4
(d) 7,7

4. Let `A` be the set of vowels of English alphabet. The total number of subsets of `A` is
(a) 10
(b) 30
(c) 32
(d) 25

5. Which one of the following is true?
(a) `\phi\in\{0\}`
(b) `0\in\phi`
(c) `\phi\subseteq\{0\}`
(d) `\{0\}\subseteq\phi`

B. Answer the following questions.

1. Let `A=\{a,e,i,o,u\}`. Then state with reason, which of the followings is true or false.
(i)   `\{a,e\}\inA`      (ii)   `\{a,e,i,o\}\subsetA`      (iii)   `a\inA`      (iv)   `\phi\subsetA`
(v)   `e\subsetA` 
[Ans: (i) False   (ii) True    (iii) True    (iv) True    (v) Flase.]

2. If `A\subseteqB,B\subseteqC \text{and} C\subseteqA`, then prove that, `A=B=C`.

3. Show that `n\(P\(P\(P\(\phi\)\)\)\)=4`

4. If `A` be a subset of `\phi`, prove that, `A=\phi`

5. If `A=B`, then show that, `A\subseteqB  \text{and}  B\subseteqA`. 

6. The sum of the elements of two finite sets is 9. If the total number of subsets of the first set is 56 ,more than the total number of subsets of the second set, then find the number of elements of the two sets. [Ans: 6 and 3] 

7. Given the sets `A=\{1,2,5\}`, `B=\{2,4,6\}` ans `C=\{0,2,4,6,8\}`, which of the following may be considered as universal set(s) for all the three sets `A,B` and `C`.
(i)   `\{0,1,2,3,4,5,6\}`    (ii)   `\phi`    (iii)   `\{0,1,2,3,4,5,6,7,8,9,10\}`    (iv)    `\{1,2,3,4,5,6,7,8\}`  
[Ans:  (iii) ]










 
 








 

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