
Which of the following is/are true?
I. If \[A\] is a subset of universal set $\bigcup $, then its complement \[{{A}^{c}}\] is also a subset of $\bigcup $.
II. If $\bigcup =\{1,2,3,...10\}$ and $A=\{1,3,5,7,9\}$ then ${{({{A}^{c}})}^{c}}=A$.
A. Only I is true
B. Only II is true
C. Both I and II are true
D. None of these
Answer
162.3k+ views
Hint: In this question, we are to find the true statement from the given statements. A fixed set is called a Universal set where all the sets under consideration are all subsets of a fixed set. By using this universal set, we can find the true statement.
Formula Used:Set: A collection of objects in which it is possible to decide whether a given object is belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
The universal set is denoted by $\bigcup $. So, a set \[A\] is said to be $A\subseteq \bigcup $ and its complement \[{{A}^{c}}\] is given by \[{{A}^{c}}=\bigcup -A\].
Complete step by step solution:Given that,
I. If \[A\] is a subset of universal set $\bigcup $, then its complement \[{{A}^{c}}\] is also a subset of $\bigcup $.
It is given that $A\subseteq \bigcup $ and we have \[{{A}^{c}}=\bigcup -A\]
From the definition of a universal set, we can write
$\begin{align}
& \bigcup -A\subseteq \bigcup \\
& \Rightarrow {{A}^{c}}\subseteq \bigcup \\
\end{align}$
Thus, the given statement is true.
II. If $\bigcup =\{1,2,3,...10\}$ and $A=\{1,3,5,7,9\}$ then ${{({{A}^{c}})}^{c}}=A$.
We have given that $A=\{1,3,5,7,9\}$
But we have
\[\begin{align}
& {{A}^{c}}=\bigcup -A \\
& \text{ }=\{1,2,3,4,5,6,7,8,9,10\}-\{1,3,5,7,9\} \\
& \text{ }=\{2,4,6,8,10\} \\
\end{align}\]
Then, we can write
$\begin{align}
& {{({{A}^{c}})}^{c}}=\bigcup -{{A}^{c}} \\
& \text{ }=\{1,2,3,4,5,6,7,8,9,10\}-\{2,4,6,8,10\} \\
& \text{ }=\{1,3,5,7,9\} \\
& \Rightarrow {{({{A}^{c}})}^{c}}=A \\
\end{align}$
Thus, the given statement is true.
Option ‘C’ is correct
Note: Here, since any set is the subset of the universal set, we get the complement of a complement set is always the set itself and the complemented set is also a subset of the universal set.
Formula Used:Set: A collection of objects in which it is possible to decide whether a given object is belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
The universal set is denoted by $\bigcup $. So, a set \[A\] is said to be $A\subseteq \bigcup $ and its complement \[{{A}^{c}}\] is given by \[{{A}^{c}}=\bigcup -A\].
Complete step by step solution:Given that,
I. If \[A\] is a subset of universal set $\bigcup $, then its complement \[{{A}^{c}}\] is also a subset of $\bigcup $.
It is given that $A\subseteq \bigcup $ and we have \[{{A}^{c}}=\bigcup -A\]
From the definition of a universal set, we can write
$\begin{align}
& \bigcup -A\subseteq \bigcup \\
& \Rightarrow {{A}^{c}}\subseteq \bigcup \\
\end{align}$
Thus, the given statement is true.
II. If $\bigcup =\{1,2,3,...10\}$ and $A=\{1,3,5,7,9\}$ then ${{({{A}^{c}})}^{c}}=A$.
We have given that $A=\{1,3,5,7,9\}$
But we have
\[\begin{align}
& {{A}^{c}}=\bigcup -A \\
& \text{ }=\{1,2,3,4,5,6,7,8,9,10\}-\{1,3,5,7,9\} \\
& \text{ }=\{2,4,6,8,10\} \\
\end{align}\]
Then, we can write
$\begin{align}
& {{({{A}^{c}})}^{c}}=\bigcup -{{A}^{c}} \\
& \text{ }=\{1,2,3,4,5,6,7,8,9,10\}-\{2,4,6,8,10\} \\
& \text{ }=\{1,3,5,7,9\} \\
& \Rightarrow {{({{A}^{c}})}^{c}}=A \\
\end{align}$
Thus, the given statement is true.
Option ‘C’ is correct
Note: Here, since any set is the subset of the universal set, we get the complement of a complement set is always the set itself and the complemented set is also a subset of the universal set.
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