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    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 \mathcal{T} = \{ U \subseteq \mathbb{Z}\, :\, \exists k > 0\ \text{s.t. } \forall n,\ n \in U \Leftrightarrow n+k \in U \} defines a topology on \mathbb{Z}. 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 U \in \mathcal{T} is such that n \in U \Leftrightarrow n + k \in U then there exists some collection 0 \le a_1 < a_2 < \cdots < a_r < k such that n \in U \Leftrightarrow n \equiv a_i \pmod k for some a_i, and hence we can write U = U^{(1)} \cup \cdots \cup U^{(r)} where U^{(i)} = \{ n\, :\, n \equiv a_i \pmod k \}. Therefore, sets of the form \{ n\, :\, n \equiv a \pmod k \} for k \in \mathbb{N} and 0 \le a < k are a basis for \mathcal{T} (if it's a topology).

    What I've done is, for r \in \mathbb{N} let 0 \le a_r < r and U_r = \{ a_r + mr\, :\, m \in \mathbb{Z} \}; then the U_r are clearly basis sets. If (r,s)=1 then x \in U_r \cup U_s \Leftrightarrow x + rs \in U_r \cup U_s, since because r,s are coprime, their lowest common multiple is rs; and there does not exist a number smaller than rs. [I haven't really justified this; my number theory's not really up to scratch... any ideas?]

    Now let V_s = U_2 \cup U_3 \cup U_5 \cup \cdots \cup U_{p_s} where p_s is the sth prime. Then p_s is coprime to 2 \times 3 \times \cdots \times p_{s-1}, and so x \in V_s \iff x + 2 \times 3 \times \cdots \times p_s \in V_s. Letting V_{\infty} = \displaystyle \bigcup_{i=1}^{\infty} U_{p_i}, we can't choose k such that x \in V_{\infty} \iff x+k \in V_{\infty}, so V_{\infty} \not \in \mathcal{T}. 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 U^{(i)} form a basis did I? They're open sets and that's enough Damn. But any further advice would still be appreciated!
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    I'm not clear on your definition of U_r; are you saying the a_r are fixed, or can vary?
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    (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.
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    I've refined it a bit. Now my working goes more like this:

    Suppose U \in \mathcal{T} is such that n \in U \iff n + r \in U (r minimal) and suppose (s,r)=1. Let U_s = \{ ms\, :\, m \in \mathbb{Z} \}; and then n \in U_s \iff n + s \in U_s, so U_s \in \mathcal{T}, and s is also minimal.

    Then n \in U \cup U_s \iff n + rs \in U \cup U_s, since (r,s)=1 and so \text{lcm}(r,s) = rs, and rs is minimal (*) (i.e. if there exists k>0 s.t. n \in U \cup U_s \iff n + k \in U \cup U_s then rs \le k).

    So we can proceed inductively, and deduce that if p_i denotes the ith prime and U_{p_i} = \{ mp_i\, :\, m \in \mathbb{Z} \} and U^{(r)} = \displaystyle \bigcup_{i=1}^r U_{p_i} , then x \in U^{(r)} \iff x + p_1p_2\dots p_r \in U^{(r)}, and because (p_1 \dots p_{r-1}, p_r) = 1, we must have that p_1 \dots p_r is minimal here too. Let k_r = p_1 \dots p_r and note that k_r \to \infty as r \to \infty. Hence U= \displaystyle \bigcup_{i=1}^{\infty} U_{p_i} does not satisfy the condition, so U \not \in \mathcal{T}, but since U_{p_i} \in \mathcal{T} for all i \in \mathbb{N}, \mathcal{T} 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 rs is minimal (*), so help there would be appreciated.
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    (Original post by nuodai)
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    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 k_r \to \infty, it's not possible that there's some k_\infty that works for the union U. Consider the case where P_n is the set of all multiples of the nth prime. The set P_1 \cup P_2 ... \cup P_n is a lot more complicated than the infiinte union (which is just Z).
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    (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 k_r \to \infty, it's not possible that there's some k_\infty that works for the union U. Consider the case where P_n is the set of all multiples of the nth prime. The set P_1 \cup P_2 ... \cup P_n is a lot more complicated than the infiinte union (which is just Z).
    I think by my (updated) definition, the U_{p_i} in my last post are the same as your P_i, so this poses a problem. But wouldn't the infinite union miss 1 and -1? Then if there were to exist k>0 such that x \in \mathbb{Z} \backslash \{\pm 1 \} \iff x+k \in \mathbb{Z} \backslash \{\pm 1 \}, then since -1-k \in \mathbb{Z} \backslash \{\pm 1 \} we'd have -1-k+k=-1 \in \mathbb{Z} \backslash \{\pm 1 \} (contradiction).
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    (Original post by nuodai)
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    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).
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    (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 r \to \infty and so on. Thanks for your help!
 
 
 
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