If $U=\{1,2,\ldots,1000\}$ and $A = \mathbb P(U) - \{ \emptyset \}$, the following relation $R$ is defined in $A$
$$XRY \Leftrightarrow (\min X = \min Y) \wedge (\max X = \max Y)$$
Calculate $\#[\{101,201\}]$
(The number of elements in the equivalence class of $\{101,201\}$)
What I did was simply use that:
$$\sum_{i=0}^{100} {100\choose i}=2^{100}$$
Is this right?
Every set in the equivalence class contains $101$ and $201$ as members, and may or may not contain others in the set $\{102,\ldots,200\}$. So you're just counting subsets of $\{102,\ldots,200\}$.
The number of members of the set $\{102,\ldots,200\}$ is $99$, so it would just be the number of subsets of a set of size $99$, i.e. it's $2^{99}$.