# Arithmetic progression topology, Z not compact

1. Apr 4, 2013

### ArcanaNoir

1. The problem statement, all variables and given/known data

The Dirichlet Prime Number Theorem indicates that if a and b are relatively prime, then the arithmetic progression $A_{a,b} = \{ ...,a−2b,a−b,a,a+b,a+2b,...\}$ contains infinitely many prime numbers. Use this result to prove that Z in the arithmetic progression topology is not compact

2. Relevant equations

A basis for The arithmetic progression topology is given by $B=\{A_{a,b}\mid a,b\in \mathbb{Z} \textrm{ and } b\ne 0 \}$.

3. The attempt at a solution
I don't know how to answer the question using the given result. My thoughts were to let the open cover be $\cup A_{0,p}$ where p is prime but that leaves out -1, 0, and 1.

2. Apr 4, 2013

### clamtrox

Actually 0 is also included in that cover, right?

When you write the cover like that, it's not obvious at all which arithmetic progression you should use to "pluck the hole". But maybe if you move it one step to left or right...

3. Apr 4, 2013

### ArcanaNoir

Does $$\cup A_{0,p} \cup A_{1,25}\cup A_{-1,35}$$ work? On the surface I think it does...

4. Apr 4, 2013

### I like Serena

There you go.
You found a branch of math where I can't help you anymore.
Perhaps you can explain it to me?

5. Apr 4, 2013

### ArcanaNoir

Oh crap!
You know what a topology is, right?

6. Apr 4, 2013

### I like Serena

Yes.
But I do not know yet what an arithmetic progression topology is.

7. Apr 4, 2013

### ArcanaNoir

Do you know what a basis for a topology is?

8. Apr 4, 2013

### I like Serena

I just looked it up.
When I tried to apply it to $\mathbb R$, I guess a basis for it is the the set of all open intervals.

9. Apr 4, 2013

### ArcanaNoir

Right. :) I don't much like the non-standard topologies. Finding the open sets can be non-intuitive.

10. Apr 4, 2013

### I like Serena

Ah well, much of mathematics is realizing how things work in obvious cases, and then try to generalize it to less obvious cases. ;)

11. Apr 4, 2013

### Dick

Take say $A_{1,5}$. That contains an infinite number of primes. Now take $A_{3,5}$. That also contains an infinite number of primes. All of the primes in the first sequence are 1 mod 5. All of the primes in the second sequence are 3 mod 5. Now here's the hint. None of the primes in the second sequence are contained in the first. Let the first set in the cover be $A_{1,5}$ and add another one, then use things like $A_{0,p}$ to "plug the holes" which you know exist. Hint concludes.

Last edited: Apr 4, 2013
12. Apr 4, 2013

### Dick

The basis for the topology over Z is all arithmetic sequences. This has nothing to do with open intervals. The sets in the topology just subsets of integers. Not reals.

Last edited: Apr 4, 2013
13. Apr 5, 2013

### ArcanaNoir

Thanks! I think I have it from here now :)

14. Apr 5, 2013

### ArcanaNoir

We were talking about standard topology on R, it was a side comment, not related to the problem. :)

15. Apr 5, 2013

### I like Serena

Oh crap!
My reputation on PF is dwindling already.

And I still don't know what an arithmetic progression topology is.

16. Apr 5, 2013

### Bacle2

Maybe use unions of finite intersection of the basic numbers to generate a few open sets.

17. Apr 5, 2013

### Dick

Oh, you can figure this out ILS. Look up defining a topology by a basis. The basis sets are just arithmetic sequences. If you are working over R, then the arithmetic progression topology is just the discrete topology. Try and prove that. Now over the integers Z, it's different. Just work through it. Understanding what it is is not as hard as proving it's not compact.

Is there some kind of reputation meter you are watching? Because your reputation is fine with me.

And anyway, the arithmetic sequence topology is not really something everyone knows. It's an nonstandard exotic topology used for making challenging topology problems.

Last edited: Apr 5, 2013