How many of these are topologies?

Question

Let $X$ be a set with $3$ elements. The set of subsets of the power set of $X$ is $2^{2^3}$ elements. How many of these are topologies?

Is there a trick to this problem, or is it just a "plug and chug" casework thing?

Maybe it might help to look at finite topologies as preorders. — Martin Sleziak, Jan 17 '15 at 10:28
Another fact which might come handy is that every point has the smallest neighborhood. (Spaces with this property are called finitely generated spaces or Alexandrov spaces. — Martin Sleziak, Jan 17 '15 at 10:30
In http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2000&task=show_msg&msg=0041.0001 I give them all. — Henno Brandsma, Jan 17 '15 at 12:04

tomasz · Answer 1 · 2015-01-17T11:05:02.863

4

Hint: a topology on a finite set is determined by its minimal basis (which amounts to choosing a minimal open set containing each point).

edited Jan 17 '15 at 11:05

answered Jan 17 '15 at 08:57

tomasz

score 2 · Answer 2 · answered Jan 17 '15 at 08:48

2

I think it's really just systematic listing, and trial and error.

answered Jan 17 '15 at 08:48

user67953840

score 2 · Answer 3 · answered Jan 17 '15 at 22:54

Let the elements be $a$, $b$, and $c$. We find the number of topologies.

Case 1.

$\mathcal{T}$ has no singleton sets.

It could have no doubleton sets. Thus it is the trivial topology, so there is $1$ way to pick this.
It could have $1$ doubleton. There are $3$ ways to pick this.
It could have more than $1$. But then their intersection would be a singleton, which is impossible.

Case 2.

$\mathcal{T}$ has $1$ singleton set.

It could have no doubleton. There are $3$ ways to pick the singleton.
It could have $1$ doubleton. All of these work, so there are $3 \times 3 = 9$ ways $(3$ ways to pick single, $3$ double$)$.
It could have $2$ doubletons. Then their intersection must be the singleton $($since $X$ has $3$ elements, the intersection of any $2$ element set is a $1$-element one$)$. Thus once we fix the singleton, we have to select the doubletons containing it. So there are $3$ ways to select.
It could have all $3$ doubletons. But then it would have more than $1$ singleton, impossible.

Case 3.

$\mathcal{T}$ has $2$ singleton sets. Thus it must contain their union, a doubleton.

It could have $1$ doubleton. This works, so there are $3$ ways.
It could have $2$ doubletons. This also works. There are $3$ ways to pick singles, $2$ to pick extra double. So there are $3 \times 2 = 6$ ways.
It could have $3$ doubletons. But this is impossible, since it would have all $3$ singletons.

Case 4.

$\mathcal{T}$ has $3$ singletons. This is the discrete topology, so there is $1$ way.

So the total is $$1 + 3 + 3 + 9 + 3 + 3 + 6 + 1 = 29 \text{ topologies}.$$

3 Answers3