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Show that there are only two types of subgroups in $\mathbb{R}$ , either Discrete or Dense?

SSK
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    Separate nontrivial subgroups of $\mathbb R$ into two cases based on whether the subgroup has a minimal positive element or not. – Nishant Mar 29 '15 at 17:52
  • @Nishant ok. Thanks but I would be happy if you could assist me in helping me realize that how the subgroup which has minimal positive element are the discrete ones exactly and the ones which dont have minimal positive element are dense ones exactly. – SSK Mar 29 '15 at 17:56

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Let $G$ be a subgroup of $(\mathbb R, +)$. If $G=\{0\}$, then clearly $G$ is discrete. If $G$ is nontrivial, then it contains a positive element. Either it has a minimal positive element or it doesn't.

Assume $G$ contains a minimal positive element $d$, and let $x\in G$. We can write $x=nd+r$, where $n$ is an integer and $r\in [0, d)$ (this is just division with remainder). Since $G$ is a subgroup, and $x, d\in G$, we get that $r=x-nd\in G$. Since $r<d$, which is supposed to be the minimal positive element, $r=0$, and so $x=nd$. Therefore, $G$ just consists of all integer multiples of $d$, and thus $G$ is discrete (just surround each point in a subset of $G$ with an open $d/2$ ball to get that every subset of $G$ is open).

Now assume $G$ contains no minimal positive element. First, let $y$ be the infimum of the set of positive elements of $G$. If $y>0$, then $y\notin G$, but we can find $y_1\in (y, 2y), y_2\in (y, y_1)$ such that $y_1, y_2\in G$. Then $y_1-y_2$ is positive, less than $y$, and still in $G$, contradicting $y$ being the infimum of the set of positive elements. Thus, $y=0$.

Now, let $x\in\mathbb R, \varepsilon>0$. By what we proved above, there exists an element $g\in G$ such that $g\in (0, \varepsilon)$. Then, we can again write $x=ng+\delta$, where $\delta\in [0, \varepsilon)$. Thus, $x$ is within $\varepsilon$ distance of some element of $G$. Since this is true for all $\varepsilon>0$, $x\in\overline{G}$, and since this is true for all $x$, $G$ is dense in $\mathbb R$.

Nishant
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  • How will you show that a set of the form {$\mathbb{Z}+\mathbb{Z}\alpha$} where $\alpha$ is irrational number a dense set. In other words , it does not have a minimum positive element. – SSK Mar 29 '15 at 19:28
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    If that subgroup had a minimal positive element $x$, it would be an integer divisor of both $1$ and $\alpha$ (i.e., $nx=1, mx=\alpha$, where $n, m\in\mathbb Z$). Then, $1/n=\alpha/m\implies \alpha=m/n$, so $\alpha$ is rational. – Nishant Mar 29 '15 at 19:38
  • @Nishant can you please tell how can you guarantee the existence of $y_{1}$ belonging to ( y, 2y) and similarly $y_{2} $ belonging to ( y, $y_{1}$) ? I had same question and I was directed to your answer. –  Jun 04 '20 at 16:53
  • Since $y$ is the infimum of the set of positive elements of $G$, and $y\notin G$, this means that for every $z>y$, $(y, z)$ contains a positive element of $G$: if this weren't true, then every positive element of $G$ would be greater than or equal to $z$, and so the infimum would be at least $z$, rather than $y$. – Nishant Jun 06 '20 at 00:14
  • @Nishant can you please tell if y>0 , then why y doesn't belongs to G. I don't think infimum implies that as by infimum it is only clear that the element is greatest among its lower bounds? –  Jun 06 '20 at 12:00
  • Since we assume that there is no mimimal positive element of $G$, and since $y$ is the infinum of the positive elements of $G$, we know that $y\notin G$, since if it were, then it would be the minimal positive element of $G$. – Nishant Jun 15 '20 at 02:02