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Theorem.
- 237
- 5
Homework Statement
Find all subrings of [itex]\mathbb{R}[/itex] 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 [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