Basic of Set theory Questions in English

(A) set is called finite if it contains a countable number of elements.
The set of months of a year consists of $12$ elements: $\{ \text{January, February, March, April, May, June, July, August, September, October, November, December} \}$.
Since the number of elements is fixed and countable $(12)$,the set is a finite set.

(B) The set $\{1, 2, 3, \ldots\}$ represents the set of all natural numbers.
Since the number of elements in this set is not countable and continues indefinitely,it is an infinite set.

(A) set is called finite if it has a countable number of elements.
The set $\{1, 2, 3, \ldots, 99, 100\}$ contains all natural numbers starting from $1$ up to $100$.
Since the number of elements in this set is exactly $100$,which is a definite and countable number,the set is a finite set.

(B) The set of positive integers greater than $100$ can be represented as $\{101, 102, 103, \dots \}$.
Since the number of elements in this set is not countable and continues indefinitely,it is an infinite set.

(A) set is called finite if it has a countable number of elements.
The set of prime numbers less than $99$ includes ${2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}$.
Since the number of elements in this set is fixed and countable,it is a finite set.

(B) The set of lines which are parallel to the $x$-axis is an infinite set.
This is because for every real number $c$,the line $y = c$ is parallel to the $x$-axis.
Since there are infinitely many real numbers $c$,there are infinitely many such lines.

(A) The set of letters in the English alphabet is a finite set because it contains a fixed number of elements,which is $26$.

(B) The set of numbers which are multiples of $5$ can be represented as $A = \{5, 10, 15, 20, 25, \dots \}$.
Since the number of elements in this set is unlimited,it is an infinite set.

(A) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is a very large number).

(B) circle is defined by its center $(h, k)$ and radius $r$. The equation of a circle passing through the origin $(0,0)$ is $x^2 + y^2 - 2hx - 2ky = 0$.
Since there are infinitely many choices for the center $(h, k)$ and the radius $r$,there are infinitely many circles that can pass through the origin $(0,0)$.
Therefore,the set of circles passing through the origin is an infinite set.

(A) $A=\{a, b, c, d\}$
$B=\{d, c, b, a\}$
The order in which the elements of a set are listed is not significant.
Since both sets contain the exact same elements,$\therefore A=B$.

(B) Given sets are $A=\{4, 8, 12, 16\}$ and $B=\{8, 4, 16, 18\}$.
Two sets are equal if they contain exactly the same elements.
It can be observed that $12 \in A$ but $12 \notin B$,and $18 \in B$ but $18 \notin A$.
Since the elements of $A$ and $B$ are not identical,$\therefore A \neq B$.

(A) $A = \{2, 4, 6, 8, 10\}$
$B = \{x : x \text{ is a positive even integer and } x \le 10\}$
Listing the elements of set $B$,we get:
$B = \{2, 4, 6, 8, 10\}$
Since all elements of set $A$ are present in set $B$ and vice versa,we conclude that $A = B$.

(B) Given:
$A = \{ x: x \text{ is a multiple of } 10 \} = \{ 10, 20, 30, 40, \ldots \}$
$B = \{ 10, 15, 20, 25, 30, \ldots \}$
Two sets are equal if they contain exactly the same elements.
It can be seen that $15 \in B$,but $15 \notin A$.
Since the elements of $A$ and $B$ are not identical,
$\therefore A \neq B$.

(B) Given sets are $A = \{ 2, 3 \}$ and $B = \{ x : x \text{ is a solution of } x^2 + 5x + 6 = 0 \}$.
To find the elements of set $B$,we solve the quadratic equation:
$x^2 + 5x + 6 = 0$
$x^2 + 3x + 2x + 6 = 0$
$x(x + 3) + 2(x + 3) = 0$
$(x + 2)(x + 3) = 0$
$x = -2$ or $x = -3$
Thus,$B = \{ -2, -3 \}$.
Comparing set $A$ and set $B$:
$A = \{ 2, 3 \}$
$B = \{ -2, -3 \}$
Since the elements of $A$ and $B$ are not the same,$A \neq B$.

(A) The set $A$ consists of the letters in the word $\text{FOLLOW}$.
Listing the distinct letters,we get $A = \{ F, O, L, W \}$.
The set $B$ consists of the letters in the word $\text{WOLF}$.
Listing the distinct letters,we get $B = \{ W, O, L, F \}$.
Since the order of elements in a set does not change the set,and both sets contain the exact same elements $\{ F, O, L, W \}$,the sets are equal.
Therefore,$A = B$.

(B=D, E=G) Two sets are equal if they contain exactly the same elements,regardless of the order in which they are listed.
Comparing the given sets:
$A = \{2, 4, 8, 12\}$
$B = \{1, 2, 3, 4\}$
$C = \{4, 8, 12, 14\}$
$D = \{3, 1, 4, 2\} = \{1, 2, 3, 4\}$
$E = \{-1, 1\}$
$F = \{0, a\}$
$G = \{1, -1\} = \{-1, 1\}$
$H = \{0, 1\}$
By comparing the elements,we observe that:
$B$ and $D$ contain the same elements $\{1, 2, 3, 4\}$,so $B = D$.
$E$ and $G$ contain the same elements $\{-1, 1\}$,so $E = G$.
Therefore,the equal sets are $B = D$ and $E = G$.

(A) Since the empty set $\phi$ is a subset of every set,we have $\phi \subset B$.

(B) To determine if $A \subset B$,we check if every element of $A$ is also an element of $B$.
Given $A = \{1, 3\}$ and $B = \{1, 5, 9\}$.
Since $3 \in A$ but $3 \notin B$,it follows that $A$ is not a subset of $B$.
Therefore,$A \not\subset B$.

(A) set $A$ is a subset of set $C$ (denoted as $A \subset C$) if every element of $A$ is also an element of $C$.
Since the elements of $A$ are ${1, 3}$ and these elements are present in $C = \{1, 3, 5, 7, 9\}$,it follows that $A \subset C$.

(A) Given sets are $B = \{1, 5, 9\}$ and $C = \{1, 3, 5, 7, 9\}$.
Since every element of set $B$ is present in set $C$,$B$ is a subset of $C$.
Therefore,$B \subset C$.

(N/A) Given sets are $A = \{a, e, i, o, u\}$ and $B = \{a, b, c, d\}$.
$(i)$ For a set $A$ to be a subset of $B$,every element of $A$ must be present in $B$.
In set $A$,the elements $\{e, i, o, u\}$ are present,but these elements are not present in set $B$.
Therefore,$A$ is not a subset of $B$.
(ii) For a set $B$ to be a subset of $A$,every element of $B$ must be present in $A$.
In set $B$,the elements $\{b, c, d\}$ are present,but these elements are not present in set $A$.
Therefore,$B$ is not a subset of $A$.

(N/A) No,it is not true.
Consider the sets $A = \{1\}$,$B = \{\{1\}, 2\}$,and $C = \{\{1\}, 2, 3\}$.
Here,$A \in B$ because the element $\{1\}$ is present in $B$.
Also,$B \subset C$ because every element of $B$ (which are $\{1\}$ and $2$) is present in $C$.
However,$A \not\subset C$ because the element $1$ is in $A$,but $1$ is not an element of $C$ (only the set $\{1\}$ is an element of $C$).

(A) Since every element of the set ${2,3,4}$ is also an element of the set ${1,2,3,4,5}$,we have:
${2,3,4} \subset {1,2,3,4,5}$

(B) set $A$ is a subset of set $B$ (denoted as $A \subset B$) if every element of $A$ is also an element of $B$.
In the given sets,the element $a$ is present in the first set ${a, b, c}$ but is not present in the second set ${b, c, d}$.
Therefore,$\{a, b, c\} \not\subset \{b, c, d\}$.

(A) Since every student of class $XI$ of your school is also a student of your school,the set of students of class $XI$ is a subset of the set of all students of your school.
Therefore,the correct symbol is $\subset$.

(B) The set on the left contains all circles in the plane,which includes circles of all possible radii (e.g.,$r=1, r=2, r=5$,etc.).
The set on the right contains only those circles in the same plane that have a radius of exactly $1$ unit.
Since there are many circles in the plane with radii other than $1$,the set of all circles is not a subset of the set of circles with radius $1$.
Therefore,the correct symbol is $\not\subset$.

(B) Since a triangle is a polygon with $3$ sides and a rectangle is a polygon with $4$ sides,no triangle can be a rectangle.
Therefore,${ x:x \text{ is a triangle in a plane} } \not\subset { x:x \text{ is a rectangle in the plane} }$.

(A) An equilateral triangle is a special type of triangle.
Since every equilateral triangle is a triangle,the set of all equilateral triangles in a plane is a subset of the set of all triangles in the same plane.
Therefore,the correct symbol is $\subset$.

(A) Since every even natural number (e.g.,$2, 4, 6, \dots$) is also an integer,the set of even natural numbers is a subset of the set of integers.
Therefore,${ x:x \text{ is an even natural number} } \subset { x:x \text{ is an integer} }$.

(B) The statement is False.
By definition,a set $A$ is a subset of set $B$ (denoted as $A \subset B$) if every element of $A$ is also an element of $B$.
Here,the set $\{a, b\}$ contains elements $a$ and $b$.
The set $\{b, c, a\}$ also contains elements $a$ and $b$.
Since every element of $\{a, b\}$ is present in $\{b, c, a\}$,it follows that $\{a, b\} \subset \{b, c, a\}$.
Therefore,the statement $\{a, b\} \not\subset \{b, c, a\}$ is False.

(A) The statement is True.
The set of vowels in the English alphabet is $\{a, e, i, o, u\}$.
Since both $a$ and $e$ are elements of the set of vowels,$\{a, e\}$ is a subset of the set of vowels.

(B) The statement is False.
For a set $A$ to be a subset of set $B$ (denoted as $A \subset B$),every element of $A$ must be an element of $B$.
Here,$2 \in \{1, 2, 3\}$,but $2 \notin \{1, 3, 5\}$.
Therefore,${1, 2, 3}$ is not a subset of ${1, 3, 5}$.

(A) True.
$A$ set $A$ is a subset of set $B$ if every element of $A$ is also an element of $B$.
Here,the set $\{a\}$ has only one element,$a$.
Since $a$ is also an element of the set $\{a, b, c\}$,the statement $\{a\} \subset \{a, b, c\}$ is true.

(B) False.
The elements of the set $\{a, b, c\}$ are $a, b,$ and $c$.
An element $x$ belongs to a set $A$ if $x \in A$.
Here,$\{a\}$ is a subset of $\{a, b, c\}$,not an element of it.
Therefore,the correct statement is $\{a\} \subset \{a, b, c\}$.

(A) The statement is True.
First,list the elements of the set $\{x : x \text{ is an even natural number less than } 6\} = \{2, 4\}$.
Next,list the elements of the set $\{x : x \text{ is a natural number which divides } 36\} = \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$.
Since every element of the first set (i.e.,$2$ and $4$) is present in the second set,the first set is a subset of the second set.

(A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The elements of set $A$ are $1$,$2$,$\{3, 4\}$,and $5$.
Since $\{3, 4\}$ is explicitly listed as an element within the set $A$,the statement $\{3, 4\} \in A$ is correct.

(N/A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
To check if $\{\{3, 4\}\} \subset A$,we must verify if every element of the set $\{\{3, 4\}\}$ is also an element of $A$.
The only element in the set $\{\{3, 4\}\}$ is $\{3, 4\}$.
Since $\{3, 4\} \in A$,it follows that $\{\{3, 4\}\} \subset A$.
Therefore,the statement $\{\{3, 4\}\} \subset A$ is correct.

(N/A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement $1 \in A$ is correct because $1$ is an element of the set $A$.

(A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement $1 \subset A$ is incorrect.
In set theory,an element of a set is denoted by the symbol $\in$. Since $1$ is an element of $A$,it should be written as $1 \in A$.
$A$ subset is formed by enclosing elements in braces. Therefore,the correct subset representation would be $\{1\} \subset A$.

(N/A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement ${1, 2, 5} \subset A$ is correct.
Reason: $A$ set $X$ is a subset of $A$ if every element of $X$ is also an element of $A$. Here,the elements of ${1, 2, 5}$ are $1, 2,$ and $5$. Since $1 \in A, 2 \in A,$ and $5 \in A$,the statement ${1, 2, 5} \subset A$ is true.

(A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The elements of set $A$ are $1$,$2$,${3, 4}$,and $5$.
The statement ${1, 2, 5} \in A$ is incorrect because ${1, 2, 5}$ is a subset of $A$ (i.e.,${1, 2, 5} \subset A$),but it is not an element of $A$.

(A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement $\{1, 2, 3\} \subset A$ is incorrect.
Reason: For a set to be a subset of $A$,every element of the set must be an element of $A$.
In the set $\{1, 2, 3\}$,the elements are $1, 2,$ and $3$.
While $1 \in A$ and $2 \in A$,the element $3$ is not an element of $A$ (only the set $\{3, 4\}$ is an element of $A$).
Therefore,$\{1, 2, 3\} \not\subset A$.

(N/A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement $\varnothing \subset A$ is correct.
Reason: The empty set $\varnothing$ is a subset of every set.

(A) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
The statement $\{\varnothing\} \subset A$ is incorrect.
For $\{\varnothing\}$ to be a subset of $A$,the element $\varnothing$ must be an element of $A$ (i.e.,$\varnothing \in A$).
However,the elements of $A$ are $1, 2, \{3, 4\},$ and $5$. None of these elements is the empty set $\varnothing$.
Therefore,$\varnothing \notin A$,which implies that $\{\varnothing\}$ is not a subset of $A$.

(C) The number of subsets of a set with $n$ elements is given by $2^n$.
Here,the set is $\{a\}$,which has $n = 1$ element.
Therefore,the number of subsets is $2^1 = 2$.
The subsets are the empty set $\varnothing$ and the set itself $\{a\}$.

(B) For a set with $n$ elements,the number of subsets is $2^n$.
Here,the set is $\{a, b\}$,which has $n = 2$ elements.
Therefore,the number of subsets is $2^2 = 4$.
The subsets are:
$1. \varnothing$ (the empty set)
$2. \{a\}$
$3. \{b\}$
$4. \{a, b\}$ (the set itself)

(N/A) The number of subsets of a set with $n$ elements is given by $2^n$. For the set $\{1, 2, 3\}$,$n = 3$,so there are $2^3 = 8$ subsets.
These subsets are: $\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\}, \{1, 2, 3\}$.

(A) The sample space of rolling a die is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ is 'getting a prime number'. The prime numbers in the set $S$ are $2, 3,$ and $5$. Thus,$A = \{2, 3, 5\}$.
Event $B$ is 'getting an odd number'. The odd numbers in the set $S$ are $1, 3,$ and $5$. Thus,$B = \{1, 3, 5\}$.

When two dice are thrown,the sample space $S$ is given by:
$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$
Event $A$ is getting an even number on the first die:
$A = \{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
The event $A$ or $B$ is the union of events $A$ and $B$,denoted as $A \cup B$.
Since every outcome in the sample space $S$ has either an even or an odd number on the first die,$A \cup B$ includes all possible outcomes.
Therefore,$A \cup B = S$.