What is the cardinality of the power set of the set {0, 1, 2}?
(a) 8
(b) 6
(c) 7
(d) 9
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Hint: We are given a set say S as {0, 1, 2}. We will first find the power set of S. The power set of any set contains all the subsets of the given set. So, we will find the subsets of S = {0, 1, 2}. Then we will make a set using a subset of S which will make our power set. At last, the cardinality of the power set is the total element of the power set, so that we count the elements and get our solution.
Complete step-by-step answer:
We are given a set S as {0, 1, 2}. We are asked to find the cardinality of the power set of S. We will first look for the power set of S and then find further things. Now, we know that the power set of any set is the collection of all possible subsets of the given set S.
Now, as we have S as {0, 1, 2}. So the possible subset of S are
\[\phi ,\left\{ 0 \right\},\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 0,1 \right\},\left\{ 1,2 \right\},\left\{ 0,2 \right\},\left\{ 0,1,2 \right\}\]
So, these are our subsets.
Now, the power set is a collection of all the subsets as elements. So, the power set will become
\[P\left( S \right)=\left\{ \left\{ 0 \right\},\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 0,1 \right\},\left\{ 1,2 \right\},\left\{ 0,2 \right\},\left\{ 0,1,2 \right\},\phi \right\}\]
Now, we look for the cardinality of the power set. The cardinality of the set is the total number of elements contained in that set. Our power set contains 8 elements, so we get that cardinality of the power set of S = {0, 1, 2} as 8.
So, the correct answer is “Option A”.
Note: This can be done in an alternate method. We have the set as {0, 1, 2}. We have to find the cardinality of the power set. We know that the power set is the collection of all the subsets of the given set and the total number of elements of the power set is given as
\[\text{Number of elements of power set}={{2}^{n}}\]
where n is the total elements in the given set.
Our set {0, 1, 2} has three elements. So, n = 3 implies that the number of elements in the power set is \[{{2}^{3}}\] that is 8.
The cardinality of the power set is the number of elements in the power set. From the above, we have the power set as 8 elements. Therefore, the cardinality of the power set of {1, 2, 0} is 8.
Complete step-by-step answer:
We are given a set S as {0, 1, 2}. We are asked to find the cardinality of the power set of S. We will first look for the power set of S and then find further things. Now, we know that the power set of any set is the collection of all possible subsets of the given set S.
Now, as we have S as {0, 1, 2}. So the possible subset of S are
\[\phi ,\left\{ 0 \right\},\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 0,1 \right\},\left\{ 1,2 \right\},\left\{ 0,2 \right\},\left\{ 0,1,2 \right\}\]
So, these are our subsets.
Now, the power set is a collection of all the subsets as elements. So, the power set will become
\[P\left( S \right)=\left\{ \left\{ 0 \right\},\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 0,1 \right\},\left\{ 1,2 \right\},\left\{ 0,2 \right\},\left\{ 0,1,2 \right\},\phi \right\}\]
Now, we look for the cardinality of the power set. The cardinality of the set is the total number of elements contained in that set. Our power set contains 8 elements, so we get that cardinality of the power set of S = {0, 1, 2} as 8.
So, the correct answer is “Option A”.
Note: This can be done in an alternate method. We have the set as {0, 1, 2}. We have to find the cardinality of the power set. We know that the power set is the collection of all the subsets of the given set and the total number of elements of the power set is given as
\[\text{Number of elements of power set}={{2}^{n}}\]
where n is the total elements in the given set.
Our set {0, 1, 2} has three elements. So, n = 3 implies that the number of elements in the power set is \[{{2}^{3}}\] that is 8.
The cardinality of the power set is the number of elements in the power set. From the above, we have the power set as 8 elements. Therefore, the cardinality of the power set of {1, 2, 0} is 8.