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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.

Want to learn
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2 Answers2

5

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 $n=5$?

Brian M. Scott
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4

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.

Henno Brandsma
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