Topology questions

1. Mar 1, 2008

jacobrhcp

a) Is there an algorithm for the number of topologies on finite sets?

b) If two spaces are homeomorphic, are intersections of opens sent to intersections of opens? Are unions of opens sent to unions of opens?

I tried to find an algorithm in the first part, and tried to prove the second part, both cases I failed. Anyone knows if there is a way to do this? The lecture notes I use does not seem to speak of it.

For the first one, I convinced myself that if it exists, it was very hard to do... so it probably doesn't exist or isn't found yet, otherwise I'd expect my notes to mention it.

For the second one, I suspect it's very easy (I'm just reading the chapter on homeomorphisms) because homeomorphisms are called 'the isomorphisms of topology'

Last edited: Mar 1, 2008
2. Mar 1, 2008

Hurkyl

Staff Emeritus
Brute force doesn't work? (Or did you forget to consider it?)

I assume you mean finite intersections?

How did you try to prove it?

Last edited: Mar 1, 2008
3. Mar 1, 2008

jacobrhcp

by brute force you mean writing out all topologies for sets with 1,2,3 and 4 elements? because I did, numer of labeled topologies on a set with:

1 element: 1
2 elements: 4
3 elements: 29
4: elements: complicated

the point is I don't really see any pattern here just yet. I must say I did not count how many nonhomeomorphic ones there were. *starts counting* for 1 it's 1, for 2 it's 3, for 3 I'd have to count a bit longer.... but since it's bigger than 5 at least, it's not an easy way to do get an algorithm (EDIT: considering symmetry of labeled topologies I suspect it's 8).... and then still, how to prove it?

and no I did not mean finite, but I think you mean that it's only true for finite intersections =). Although I'd expect infinite intersections of opens to be sent to infinite intersections of opens, only you wouldn't be sure they were open after taking the infinite intersection.

I tried writing out the definitions and then starting like this:

'suppose f:X->Y is a homeomorphism. then if $$U_1 -> U_1'$$, $$U_2 -> U_2'$$ this means $$U_1 \cap U_2$$is sent to an open subset of Y, because f and $$f^{-1}$$ are continuous.

Now I wanted to show this is $$U_1' \cap U_2'$$. I imagine this has to do with continuity as well... because the 'bijective' part of homeomorphisms is certainly not enough for this to work.

So admittedly I did not get very far, but I don't know how to continue.

I feel you suspect this is a homework question, but I assure you it's not: I'm trying to learn myself topology from lecture notes and internet, since I am not able to attend the exercise classes I don't even know what the homework is, nor is there any exercise that looks remotely like these.

Last edited: Mar 1, 2008
4. Mar 1, 2008

Hurkyl

Staff Emeritus
Yes -- but without stopping at 4. If you want to know how many topologies there are on a particular finite set, you could simply write out all topologies for it. That qualifies as an algorithm! Asking for an 'efficient' algorithm, or a 'simple' arithmetic expression is another question entirely.

Well, to state the algorithm more precisely, you would:
1. Write down every subset of P(X).
2. For each subset of P(X), test if it's a topology.
3. Count how many subsets of P(X) were, in fact, topologies.

If you so desired, you could include additional steps to sort the topologies into homeomorphism classes.

It may seem like a dumb algorithm, but it is an algorithm -- and dumb algorithms like these often offer easy answers to interesting theoretical questions.

Actually, I meant the latter. I read something into your original post that wasn't there.

The notions of intersection and union are set-theoretic -- you already know that the homeomorphism maps open to open (and back), so are you sure 'bijective' isn't enough to finish the question?

Have you thought about how your conjecture specializes to the case where each space has the discrete topology?

Anyways, the main thing you seem not to have done is invoke the definition of "intersection".

I hadn't really thought about it; I'm very much in the habit of trying to point people in the right direction -- they learn more when they solve the problem themselves! Ideally I won't even do that much; it's better to teach a person how to find paths and evaluates them than it is to show them the right path!

5. Mar 1, 2008

jacobrhcp

haha, thanks a lot =)

so there is no simple algorithm found (which was what I actually was looking for)? =(... that's a shame.

anyways, for the proof of 'my' conjecture ;)

the definition of intersection is: $$U_1' \cap U_2':=$$x in X| x in $$U_1'$$and x in $$U_2'$$

and in the case of the discrete topology, the 'continuous' part is a bit of an overkill in a homeorphism, because bijections between discrete topologies are automatically continuous, right? So in that case, 'bijection' should be enough.

In general,

$$f(U_1 \cap U_2$$) = f({x in X| x in $$U_1'$$ and x in $$U_2'$$})={y in Y| y=f(x) in $$U_1'$$ and y=f(x) in $$U_2'$$}=$$U_1' \cap U_2'$$, right?... but even now I'm a bit lost as to why the second equality holds, and why we needed openness again for this to work (or did we?)?

Last edited: Mar 1, 2008
6. Mar 1, 2008

jacobrhcp

btw, I like the xkcd comic in your signature =)... what does xkcd stand for anyway?