Subrings of Real numbers which are discrete

[Theorem.]

Homework Statement

Find all subrings of \mathbb{R} which are discrete subsets

Homework Equations

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

The Attempt at a Solution

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

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

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 \mathbb{R} and the assumption that S is discrete
I wish to show that \min_{x\in S\setminus{\{s\}}}|s-x|=d=1, and I already know d\leq 1. Perhaps if I assume d<1, 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

 

[Dick]

Homework Statement

Find all subrings of \mathbb{R} which are discrete subsets

Homework Equations

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

The Attempt at a Solution

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

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

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 \mathbb{R} and the assumption that S is discrete
I wish to show that \min_{x\in S\setminus{\{s\}}}|s-x|=d=1, and I already know d\leq 1. Perhaps if I assume d<1, 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

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?

 

[Theorem.]

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?

 

[Dick]

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?

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.

 

[Theorem.]

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.

the set \{|s|:s\in S\} is bounded from below since S is by assumption discrete (there exists some r>0 such that |0-s|&gt;r for all s\in S) and thus has a greatest lower bound, although-that doesn't necessarily mean the set contains its infimum I am missing some details here

 

[Dick]

the set \{|s|:s\in S\} is bounded from below since S is by assumption discrete (there exists some r>0 such that |0-s|&gt;r for all s\in S) and thus has a greatest lower bound, although-that doesn't necessarily mean the set contains its infimum I am missing some details here

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.

 

[Theorem.]

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.

Oh i see, S would contain a cluster point which would contradict the fact that S is discrete

 

[Dick]

Oh i see, S would contain a cluster point which would contradict the fact that S is discrete

Bingo!

 

[Theorem.]

Bingo!

Okay thanks a lot : ).