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Subrings of Real numbers which are discrete

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Find all subrings of [itex]\mathbb{R}[/itex] which are discrete subsets


    2. Relevant equations
    For the purpose of our class, a ring is a ring with identity, not necessarily commutative.


    3. The attempt at a solution
    First suppose that [itex]S\subset \mathbb{R}[/itex] is a subring of [itex]\mathbb{R}[/itex]. Then, by definition we must have [itex]0\in S[/itex] and
    [itex]1\in S[/itex]. Since [itex]S[/itex] is closed under addition, [itex]\underbrace{1+1+...+1}_\text{n times}=n\in S[/itex]. That is, for all
    [itex]n\in \mathbb{N}[/itex], [itex]n\in S[/itex]. Likewise, since the additive inverse of every element of [itex]S[/itex] must also be in [itex]S[/itex], we
    have[itex]-n\in S[/itex] for all [itex]n\in \mathbb{N}[/itex]. We thus conclude that [itex]\mathbb{Z}\subset S[/itex].

    Now suppose that [itex]S[/itex] is also a discrete subset of [itex]\mathbb{R}[/itex]. That is, for every element [itex]s\in S[/itex] there exists
    [itex]r>0[/itex] such that for each [itex]x\in S\setminus{\{s\}}[/itex] we have [itex]|s-x|>r[/itex]. Note that since
    [itex]\mathbb{Z}\subset S[/itex],[itex]r<1[/itex].

    okay this is where I am right now and am somewhat (for the moment) stuck. Although I haven't clearly stated it yet, I am claiming that the only such subring is the integers. Using what I have already proven (the integers are contained in any subring of [itex]\mathbb{R}[/itex] and the assumption that S is discrete
    I wish to show that [itex]\min_{x\in S\setminus{\{s\}}}|s-x|=d=1[/itex], and I already know [itex]d\leq 1[/itex]. Perhaps if I assume [itex]d<1[/itex], I will arrive at a contradiction? Any thoughts?

    I apologize that the proof so far is a little rough, it is more a less just a sketch. Thanks everyone
     
  2. jcsd
  3. Jan 17, 2012 #2

    Dick

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    I think you are almost there. 0 is in S. If S is discrete then there is a s>0 in S such that s=min(|s| for s in S), right? There is a little bit of work to show that, but once you have then where would you go from there?
     
  4. Jan 17, 2012 #3
    Thanks Dick. I understand where the element s you are describing comes from and why it is in S i just am having a difficult time figuring out what I should do with it. well i think i might be getting an idea as I type this,
    if s<1, then s^2 for instance is also in S (S is a ring) and yet s^2<s<1, contradicting the definition of s?
     
  5. Jan 17, 2012 #4

    Dick

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    You know s<=1 because 1 is in S as well. Yes, that's the final contradiction. Now back up and tell me why min(|s| for s in S) must be in S. Think about the properties of greatest lower bounds and R and the discreteness of S.
     
  6. Jan 17, 2012 #5
    the set [itex]\{|s|:s\in S\}[/itex] is bounded from below since S is by assumption discrete (there exists some r>0 such that [itex]|0-s|>r[/itex] for all [itex]s\in S[/itex]) and thus has a greatest lower bound, although-that doesn't necessarily mean the set contains its infimum im missing some details here
     
  7. Jan 17, 2012 #6

    Dick

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    Suppose the set doesn't contain its infimum. That mean there must be elements of S REALLY close to the infimum. Lots of them. That's a hint.
     
  8. Jan 17, 2012 #7
    Oh i see, S would contain a cluster point which would contradict the fact that S is discrete
     
  9. Jan 17, 2012 #8

    Dick

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    Bingo!
     
    Last edited: Jan 17, 2012
  10. Jan 17, 2012 #9
    Okay thanks a lot : ).
     
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