For the set X={m,n.p,q,r,s}, let R be the relation on P(X) (power set) given by A R B iff A and B have the same number of elements. List all the elements in {m}/R (equivalence class); in {m,n,p,q,r,s}/R. How many elements are in X/R? How many elements are in P(X)/R?
List the following elements in proper set notation. Place the elements in numerical order within the set.
* 0
* 1
* 123
* 4
* 34
To find the elements in the equivalence class of {m} under the relation R, we need to find all the subsets of X that have the same number of elements as {m}.
In this case, {m} is a singleton set, which means it contains only one element. So, we need to find all the subsets of X that also contain only one element. From the set X={m,n,p,q,r,s}, the subsets that satisfy this condition are:
{m}
So the elements in {m}/R (equivalence class) are {m}.
To find the elements in the equivalence class of {m,n,p,q,r,s} under the relation R, we need to find all the subsets of X that have the same number of elements as {m,n,p,q,r,s}.
In this case, {m,n,p,q,r,s} contains all the elements in X. So, we need to find all the subsets of X that also contain all the elements of X. From the set X={m,n,p,q,r,s}, the subset that satisfies this condition is:
{m,n,p,q,r,s}
So the elements in {m,n,p,q,r,s}/R (equivalence class) are {m,n,p,q,r,s}.
To find how many elements are in X/R, we need to count the number of distinct equivalence classes. Each equivalence class represents subsets of X that have the same number of elements. Since there are 6 elements in X, there can be equivalence classes with 0, 1, 2, 3, 4, 5, or 6 elements.
So, there are a total of 7 elements in X/R.
To find how many elements are in P(X)/R, we need to count the number of distinct equivalence classes, where P(X) represents the power set of X. The power set of X has 2^6 = 64 subsets.
Since each equivalence class represents subsets with the same number of elements, we can have equivalence classes with 0, 1, 2, 3, 4, 5, or 6 elements. So, there can be a maximum of 7 equivalence classes.
Therefore, there are a maximum of 7 elements in P(X)/R.
To find the elements in the equivalence class {m}/R, we need to find all the sets in P(X) that have the same number of elements as {m}.
In this case, {m} has one element. So, we need to find all the sets in P(X) that also have one element.
The elements in {m}/R are: {m}.
Moving on to {m,n,p,q,r,s}/R, we need to find all the sets in P(X) that have the same number of elements as {m,n,p,q,r,s}.
{m,n,p,q,r,s} has a total of six elements. So, we need to find all the sets in P(X) that also have six elements.
The elements in {m,n,p,q,r,s}/R are: {m,n,p,q,r,s}.
To find the number of elements in X/R, we need to find all the equivalence classes in X and count them.
In this case, an equivalence class is defined as a set in P(X) that contains the same number of elements.
To find the number of equivalence classes, we need to determine the number of possible sizes for sets in P(X). X has a total of 5 elements, so the possible sizes are 0, 1, 2, 3, 4, and 5.
In X, the empty set {} has 0 elements, and only {} is the subset of the other sets.
The sets with 1 element are {m}, {n}, {p}, {q}, {r}, and {s}.
The sets with 2 elements are {m,n}, {m,p}, {m,q}, {m,r}, {m,s}, {n,p}, {n,q}, {n,r}, {n,s}, {p,q}, {p,r}, {p,s}, {q,r}, {q,s}, and {r,s}.
The sets with 3 elements are {m,n,p}, {m,n,q}, {m,n,r}, {m,n,s}, {m,p,q}, {m,p,r}, {m,p,s}, {m,q,r}, {m,q,s}, {m,r,s}, {n,p,q}, {n,p,r}, {n,p,s}, {n,q,r}, {n,q,s}, {n,r,s}, {p,q,r}, {p,q,s}, {p,r,s}, and {q,r,s}.
The sets with 4 elements are {m,n,p,q}, {m,n,p,r}, {m,n,p,s}, {m,n,q,r}, {m,n,q,s}, {m,n,r,s}, {m,p,q,r}, {m,p,q,s}, {m,p,r,s}, {m,q,r,s}, {n,p,q,r}, {n,p,q,s}, {n,p,r,s}, {n,q,r,s}, {p,q,r,s}.
Finally, the set with 5 elements is {m,n,p,q,r}.
Counting all the sets, we have a total of 31 equivalence classes in X/R.
To find the number of elements in P(X)/R, we need to count all the equivalence classes in P(X).
The number of elements in P(X) is 2^5 = 32, as each of the 5 elements in X can either be present or absent in any set in P(X).
Since there are 31 equivalence classes in X/R, that means there is one equivalence class left in P(X)/R.
Therefore, there is 1 element in P(X)/R.