# Number of topologies on a finite set with $k$ open sets upto homeomorphisms

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). $$1)3\\2)4\\3)5\\4)7$$

observation: Assume $$\tau_1$$ and $$\tau_2$$ are two homeomorphic topologies on $$X$$, then for every open set $$U$$ in $$\tau_1$$ there exists an open set $$U^{'}$$ such that $$|U| = |U^{'}|$$. Using this observation, I have solved the case when there are $$3$$ open sets. Such topology should be $$\{\phi,X,U\}$$ and $$|U| = 1$$ or $$2$$. Depends on the cardinality of $$U$$, the homomorphism class determined and so there are two homomorphism classes.

Next, the $$n=7$$ case is straight forward. Because there are no topologies with seven open sets. So here the answer is $$0$$.

I need some intuition to do the cases $$n=4$$ and $$5$$. Kindly share your thoughts. Thank you.

Take $$n=4$$. We automatically have $$\varnothing$$ and $$X$$. Let’s add an isolated point $$\{x\}$$. If we add a second isolated point $$\{y\}$$, $$\{x,y\}$$ will be open, and we’ll have too many open sets, so we must add a two-element set. There are just two possibilities: we add $$X\setminus\{x\}$$, or we add a set $$\{x,y\}$$ for some $$y\in X\setminus\{x\}$$. I’ll leave it to you to check whether those work.

Alternatively, we could start by adding a two-element set $$\{x,y\}$$. Then we can’t add a second two-element set (why not?), so we must add an isolated point, and we’re back in the first case.

Can you apply the same sort of reasoning to the case $$n=5$$?

In this answer I list all of the topologies on $$3$$ points.

• We see that $$n=3$$ gives two homeomorphism types, $$\{x\}$$ or $$\{x,y\}$$. ($$x,y,z$$ will be distinct points here and hereafter).

• $$n=4$$ also has two types, non-trivial sets $$\{x\}, \{x,y\}$$ or $$\{x\}, \{y,z\}$$.

• $$n=5$$ also two types, non trivial sets $$\{x,y\}, \{x,z\}, \{x\}$$ and $$\{x\},\{y\}, \{x,y\}$$.

• We have one for $$n=6$$ (see there), none for $$n=7$$, one for $$n=8$$ (discrete).

So in all: $$2,2,2,0$$ types for a to d, in order.