1.5 Equal Sets

Equal sets

Two sets \(A\) and \(B\) are said to be equal if each element of \(A\) is an element of \(B\) and each element of is an element of \(A\). Thus, two sets \(A\) and \(B\) are equal if they have exactly the same elements but the order in which the elements in the two sets are written is immaterial.

If sets \(A\) and \(B\) are equal, then we write \(A=B\).
For example,
(i) \(A=\{4,5,6,7\}, \mathrm{B}=\{5,6,4,7\}\).
Here, \(A\) and \(B\) have exactly the same elements.
Therefore, \(A=B\).
(ii) \(A=\) Set of letters of the word LOYAL
\(B=\) Set of letters of the word ALLOY
Here, \(A=\{\mathrm{L}, \mathrm{O}, \mathrm{Y}, \mathrm{A}\}\) and \(B=\{\mathrm{A}, \mathrm{L}, \mathrm{O}, \mathrm{Y}\}\)
Therefore, \(A=B\).
Note that two sets \(A\) and \(B\) are equal if
\(x \in A \Rightarrow x \in B\) and \(x \in B \Rightarrow x \in A\).
(iii) Let \(\mathrm{A}=\{1,2,3,4\}\) and \(\quad \mathrm{B}=\{3,1,4,2\}\). Then \(\mathrm{A}=\mathrm{B}\).

Note: A set does not change if one or more elements of the set are repeated. For example, the sets \(\mathrm{A}=\{1,2,3\}\) and \(\mathrm{B}=\{2,2,1,3,3\}\) are equal, since each element of \(\mathrm{A}\) is in \(\mathrm{B}\) and vice-versa (the order in which elements in a set are represented does not matter). That is why we generally do not repeat any element in describing a set.

Example 8: Find the pairs of equal sets, if any, and give reasons:
Given sets:
\(A=\{0\}\)
\(B=\{x: x>15\) and \(x<5\}\)
\(C=\{x: x-5=0\}\)
\(D=\left\{x: x^2=25\right\}\)
\(E=\left\{x: x\right.\) is an integral positive root of the equation \(\left.x^2-2 x-15=0\right\}\)

Solution: Since \(0 \in A\) and 0 does not belong to any of the sets \(B, C, D\) and \(E\), it follows that, \(\mathrm{A} \neq \mathrm{B}, \mathrm{A} \neq \mathrm{C}, \mathrm{A} \neq \mathrm{D}, \mathrm{A} \neq \mathrm{E}\).

Since \(\mathrm{B}=\phi\) but none of the other sets are empty. Therefore \(\mathrm{B} \neq \mathrm{C}, \mathrm{B} \neq \mathrm{D}\) and \(\mathrm{B} \neq \mathrm{E}\). Also \(\mathrm{C}=\{5\}\) but \(-5 \in \mathrm{D}\), hence \(\mathrm{C} \neq \mathrm{D}\).

Since \(\mathrm{E}=\{5\}, \mathrm{C}=\mathrm{E}\). Further, \(\mathrm{D}=\{-5,5\}\) and \(\mathrm{E}=\{5\}\), we find that, \(\mathrm{D} \neq \mathrm{E}\). Thus, the only pair of equal sets is \(\mathrm{C}\) and \(\mathrm{E}\).

Singleton Set

A set having only one element is called a singleton set.

For example,
(i) \(A=\) The set of present President of India.
(ii) \(B=\{x: x\) is an even prime number \(\}\)

Set of Sets

A set \(S\) having all its elements as sets is called a set of sets, or a family of sets, or a class of sets.

For example.
(i) \(\{\{1,2\},\{2,5\},\{3,6,8\}\}\) is a set of sets having three elements \(\{1,2\},\{2,5\}\) and \(\{3,6,8\}\) which are themselves sets.
(ii) \(\{\phi\}\) is a singleton set of sets having null set \(\phi\) as its element.
(iii) \(S=\{\{1,2\}, 3,\{4\}:\}\) is not a set of sets as \(3 \in S\) is not a set.

Example 9: Are the following pairs of sets equal?
(i) \(A=\{x \mid x\) is prime factor of 6\(\}\);
\(B=\left\{x \mid x\right.\) is a solution of \(\left.x^2-5 x+6=0\right\}\)
(ii) \(A=\{x \mid x\) is a letter in the word REPLACED \(\}\);
\(B=\{y \mid y\) is a letter in the word PARCELED \(\}\)
(iii) \(A=\{x \mid x\) is a natural number, \(x>1\}\);
\(B=\{x \mid x\) is natural number, \(x \geq 1\}\)

Solution: (i)
\(
\begin{aligned}
A= & \{x \mid x \text { is prime factor of } 6\}=\{2,3\} \\
B= & \left\{x \mid x \text { is solution of } x^2-5 x+6=0\right\} \\
& x^2-5 x+6=0 \\
\text { or } & (x-2)(x-3)=0 \\
\text { or } & x=2,3 \\
\therefore & B=\{2,3\}
\end{aligned}
\)
Therefore, \(A=B\).
(ii)
\(
\begin{aligned}
A & =\{x \mid x \text { is a letter in the word REPLACED }\} \\
& =\{\mathrm{R}, \mathrm{E}, \mathrm{P}, \mathrm{~L}, \mathrm{~A}, \mathrm{C}, \mathrm{D}\} \\
B & =\{y \mid y \text { is a letter in the word PARCELED }\} \\
& =\{\mathrm{P}, \mathrm{~A}, \mathrm{R}, \mathrm{C}, \mathrm{E}, \mathrm{~L}, \mathrm{D}\}
\end{aligned}
\)
Therefore, \(A=B\).
(iii)
\(
\begin{aligned}
& A=\{x \mid x \text { is natural number, } x>1\}=\{2,3,4, \ldots\} \\
& B=\{x \mid x \text { is natural number, } x \geq 1\}=\{1,2,3, \ldots\}
\end{aligned}
\)
We observe that \(1 \in B\) but \(1 \notin A\). Therefore, \(A \neq B\).