Distinct homeomorphism classes Let X be a three element set. For each of the following numbers n, determine the number of distinct
homeomorphism classes of topologies on X with exactly n open subsets (including the empty set and the
whole set). 
(a) 3
(b) 4
(c) 5
(d) 7
 A: (a) the non-empty proper open set can only have cardinality $1$ or $2$, and each possibility determines exactly one topology up to homeomorphism.
(b) there must be exactly two non-empty proper open sets, therefore at least one of them has cardinality $2$ (if there were two distinct open points, then their union would be a fifth open set). There cannot be a second open set of cardinality $2$, otherwise its intersection with the former would be an open point, thus a fifth open set. Therefore an open point and an open set of cardinality $2$. Whether or not the open point lies in the set of cardinality $2$ determines exactly one topology up to homeomorphism.
(c) there must be exactly three non-empty proper open sets. Therefore at least one has cardinality two. If there are two open points, then their union must be the aforementioned open set of cardinality $2$, and this is a topology. If there is a second open set of cardinality two, then its intersection with the former must be an open point and thus the fifth open set: this is another topology. It is clear that these pieces of information determine completely $2$ homeomorphism classes of topologies.
(d) $X$ has $8$ subsets, therefore let $U$ be the subset which is not in the topology. $U$ cannot have cardinality $2$, because otherwise its elements would be open points, against $U$ not being open. If $U$ is a point, then $U$ is the intersection of two subsets of cardinality $2$, which are open. Therefore $U$ is open. This means that there are $0$ topologies of cardinality $7$.
