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# Topology on Z watch

1. I'm doing a past exam question (Q12 here) and it's mostly fine, but I'm not too happy with how I answered part (c).

Essentially I need to determine whether defines a topology on . At first I thought it did, but after a bit of working I'm now pretty sure it isn't.

EDIT: See post #4 for updated working

To simplify things a bit, I've said that if is such that then there exists some collection such that for some , and hence we can write where . Therefore, sets of the form for and are a basis for (if it's a topology).

What I've done is, for let and ; then the are clearly basis sets. If then , since because are coprime, their lowest common multiple is ; and there does not exist a number smaller than . [I haven't really justified this; my number theory's not really up to scratch... any ideas?]

Now let where is the sth prime. Then is coprime to , and so . Letting , we can't choose such that , so . As the arbitrary union of basis sets doesn't lie in the collection, it can't be a topology.

I'm pretty sure this is right, but I feel that my reasoning is a bit clumsy, or at the very least that I've needed to define far too many things than I should have to come up with a badly behaved set. Any ideas about how I can clean this up would be greatly appreciated! [One of my tasks for this holiday is improving my rigour and reasoning.]

EDIT: Thinking about it, I didn't even need to justify that form a basis did I? They're open sets and that's enough Damn. But any further advice would still be appreciated!
2. I'm not clear on your definition of U_r; are you saying the a_r are fixed, or can vary?
3. (Original post by DFranklin)
I'm not clear on your definition of U_r; are you saying the a_r are fixed, or can vary?
I'm saying they're fixed but arbitrary.
4. I've refined it a bit. Now my working goes more like this:

Suppose is such that (r minimal) and suppose . Let ; and then , so , and s is also minimal.

Then , since and so , and is minimal (*) (i.e. if there exists s.t. then ).

So we can proceed inductively, and deduce that if denotes the ith prime and and , then , and because , we must have that is minimal here too. Let and note that as . Hence does not satisfy the condition, so , but since for all , is not closed under arbitrary unions (so does not define a topology).

I'm happier with this, but I still don't like not having justified that is minimal (*), so help there would be appreciated.
5. (Original post by nuodai)
..
The thing that's bothering me here is the passage from the finite to the infinite case. It's not obvious to me that just because , it's not possible that there's some that works for the union U. Consider the case where P_n is the set of all multiples of the nth prime. The set is a lot more complicated than the infiinte union (which is just Z).
6. (Original post by DFranklin)
The thing that's bothering me here is the passage from the finite to the infinite case. It's not obvious to me that just because , it's not possible that there's some that works for the union U. Consider the case where P_n is the set of all multiples of the nth prime. The set is a lot more complicated than the infiinte union (which is just Z).
I think by my (updated) definition, the in my last post are the same as your , so this poses a problem. But wouldn't the infinite union miss 1 and -1? Then if there were to exist such that , then since we'd have (contradiction).
7. (Original post by nuodai)
..
Good point. So I think this is actually the simplest approach:

Take U_p = pZ; then each U_p is in the topology. But the union U of all U_p = Z \ {-1, 1}. Then U isn't a member of T, since if it were, we could find k > 0 s.t. ... (as in your post above).
8. (Original post by DFranklin)
Good point. So I think this is actually the simplest approach:

Take U_p = pZ; then each U_p is in the topology. But the union U of all U_p = Z \ {-1, 1}. Then U isn't a member of T, since if it were, we could find k > 0 s.t. ... (as in your post above).
Describing it this way definitely seems like a better idea than letting and so on. Thanks for your help!

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