Given that the relation that aRb if and only if the smallest element of a is is equal to the smallest element in b?

Question

X is the set of all nonempty subsets of the set {1,2,3,4,5,6,7,8,9,10}. a,b are elements of X.

a) Find the number of elements in the equivalence class [{2,6,7}]?

My solution:

1) Choose 2 to be in the set, 1 way
2) 2^9, which includes any number to be in the set except 2

Ans: 1 * 2^9 = 512

b) Find the number of four-element sets, which are member of the equivalence class [{2,6,7}]?

My solution:

1) Choose 2 to be in the set, 1 way
2) Choose 2^3, which includes any 3 number to be in the set except 2

Ans: 1 * 2^3 = 8

Answer 1

It is clear that your relation is an equivalence relation.

The equivalence class of $[\{ 2,6,7\}]$ is the set of all subsets of $\{1,...,10 \}$ having least element $2$. Now, we say the following: Suppose $2$ is an element of our set, then any number from $3$ to $10$ can be in the equivalence class of this set. That is to say, there are $3^8=256$ elements.

As for $b$, we are now looking at $4$ element subsets. $2$ has to be one of the elements, that is fixed. The other $3$ elements can be chosen out of the elements $\{3,...,10 \}$ (we can't have $1$ as then that would be the least element). That is, we have to choose $3$ numbers out of $8$ without replacement, the ways of doing that is $\binom{8}{3} = 56$. Hence your answer will be $56$.

When we are choosing elements without replacement, we do not exponentiate. Hence answers like $2^3$ would not be likely in the case of $b$.

Answer 2

If I understand the question correctly, $a$ and $b$ are only considered to be in the same equivalence class if they have the same smallest element.

If this is the case, then for a) the answer would be $2^8=256$ since 1 cannot be in any of the equivalent elements.

For b) the answer would be ${8\choose 3}=56$ since you can have any 3 of the numbers from 3-10.