# Homework Help: Subrings of Real numbers which are discrete

1. Jan 17, 2012

### Theorem.

1. The problem statement, all variables and given/known data
Find all subrings of $\mathbb{R}$ 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 $S\subset \mathbb{R}$ 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

2. Jan 17, 2012

### Dick

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?

3. Jan 17, 2012

### 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?

4. Jan 17, 2012

### Dick

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.

5. Jan 17, 2012

### Theorem.

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

6. Jan 17, 2012

### Dick

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.

7. Jan 17, 2012

### Theorem.

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

8. Jan 17, 2012

### Dick

Bingo!

Last edited: Jan 17, 2012
9. Jan 17, 2012

### Theorem.

Okay thanks a lot : ).