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Let's say $G$ is some additive subgroup of $\mathbb{R}$ that has at least two elements.

From what I understand, $G$ is then either dense in $\mathbb{R}$, or has some least positive element. What is the reason for this?

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Let $a=\inf\{g\in G; g>0\}$. (Note that this set is non-empty and thus $a\ge 0$, $a<\infty$.)

Suppose that $a\notin G$.

This implies that for each $\varepsilon>0$ there is $g\in G$ such that $a<g<a+\varepsilon$. The same argument gives the existence of $g'$ such that $a<g'<g<a+\varepsilon$. Thus we found $h_\varepsilon=g-g'$ which belongs to $G$ and fulfills $0<h_\varepsilon<\varepsilon$.

Now it is relatively easy to see that $\bigcup\limits_{\varepsilon>0}\{z\cdot h_\varepsilon; z\in\mathbb Z\}$ is dense in $\mathbb R$ and it is subset of $G$.

(If you choose some interval $(a-\varepsilon,a+\varepsilon)$, which has length $2\varepsilon$, then it must contain some element of the form $z\cdot h_\varepsilon$, since the distance between two neighboring elements of this form is less than $\varepsilon$.)


An analogous claim (or at least one of them - there are other possible generalizations) in higher dimensions would be:

An additive subgroup of $\mathbb R^n$ is discrete if and only if it is generated by linearly independent vectors $a_1,\dots,a_m\in\mathbb R^n$, $m\le n$.

In the case $n=1$ we get only one generator.

Proof of this claim can be found e.g. in Stewart, Tall: Algebraic Number Theory and Fermat's Last Theorem 6.1 (in Section 6.1 Lattices) and probably in many other texts dealing with lattices in $\mathbb R^n$.

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  • $\begingroup$ I might be naive in pointing this out, but isn't it $h_{\epsilon}=g-g^{'}$ rather than $h_{\epsilon}=g^{'}-g$? $\endgroup$ – Kam May 27 '18 at 12:50
  • $\begingroup$ Thanks, corrected. $\endgroup$ – Martin Sleziak May 27 '18 at 13:02
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    $\begingroup$ @gen As I do not want to digress too much from the original question here, I have left a few comments in the group theory chatroom. $\endgroup$ – Martin Sleziak Mar 2 '19 at 5:54
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Let $G$ be an additive subgroup of $\mathbb{R}$. Suppose that there does not exists the least positive element, i.e. $\inf\{|x|:x\in G-\{0\}\}=0$. Then we can prove that $G$ is dense in $\mathbb{R}$ as follows: suppose that $y\in\mathbb{R}$, for any $\epsilon>0$, there exist $x\in G$ such that $|x|<\epsilon$. We can assume that $x>0$, otherwise, we can take $-x$ which belongs to $G$ since $G$ is an additive group. Then there exists an integer $n$ such that $$nx\leq y<(n+1)x,$$ which implies that $$|y-nx|<x<\epsilon,$$ where $nx\in G$ since $G$ is an additive group. This shows that $G$ is dense in $\mathbb{R}$.

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I once used this fact to solve a problem in the American Mathematical Monthly. Here's the proof I gave, which is rather similar to Paul's.

We first show that if $G$ has a limit point, then it is dense. Fix $\epsilon > 0$ and let $y$ be a limit point of $G$. Choose $x_1, x_2 \in G$ with $0 < |{x_1-y}| < \epsilon/2$, and $|{x_2 - y}| < |{x_1 - y}|$. Then $x := x_1 - x_2 \in G$ with $0 < |{x}| < \epsilon$. $G$ now contains all integer multiples of $x$, so every interval of length at least $\epsilon$ contains an element of $G$. $\epsilon$ was arbitrary so $G$ is dense.

As the contrapositive of this, if $G$ is not dense then it has no limit points. In this case, let $a := \inf \{ x \in G, x > 0\}$. (This set is nonempty since $G$ is nontrivial and hence contains at least one positive number.) Since neither $0$ nor $a$ are limit points of $G$, $a > 0$ and $a \in G$. By construction, $a$ is the least positive element of $G$. (Moreover, $G$ is generated by $a$.)

For those curious about the Monthly problem, it's #11345, volume 115 number 2, page 166, February 2008:

11345. Proposed by Roger Cuculière, France. Find all nondecreasing functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x+f(y)) = f(f(x)) + f(y)$ for all real $x$ and $y$.

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    $\begingroup$ Someone mention in the other question first chapter of Aliprantis, Burkinshaw: Principles of Real Analysis as a possible reference. Did you find it there? (I did not.) $\endgroup$ – Martin Sleziak Dec 11 '11 at 6:31
  • $\begingroup$ @MartinSleziak: I didn't look. $\endgroup$ – Nate Eldredge Dec 13 '11 at 19:48
  • $\begingroup$ I'm curious which Monthly problem can be solved by using this fact. Can you give a reference? Thank you very much. $\endgroup$ – Eric Sun Oct 19 '18 at 4:44
  • $\begingroup$ @EricYewenSun: I added it to the answer. $\endgroup$ – Nate Eldredge Oct 19 '18 at 14:16
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We assume $G$ has no least positive element. Take $x_n \rightarrow 0^+$. Take a decreasing subsequence $(y_n)$ of $(x_n)$.

After each $y_i$, we insert additional terms equal to $y_i - x_j < y_{i+1}$, preserving monotonicity [possible, since we can make $x_j$ as small as we want], until the sum of all these terms exceeds $\frac{1}{i}$. We obtain $(z_i)$.

$S=\sum_{i=1}^{\infty} (-1)^i z_i$ is a convergent alternating series that is absolutely divergent by comparison with the harmonic series, hence by the Riemann series theorem we can rearrange terms to make it sum to any real number. Now, taking the partial sums of such a rearrangement, we obtain a sequence tending to any real number. Hence $G$ is dence in $\mathbb{R}$.

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    $\begingroup$ You have probably a typo $(-)^i$ instead of $(-1)^i$ and $S$ instead of $G$ in the first sentence. Maybe it would be good to mention that $G$ is closed under finite sums, therefore the partial sums $s_n$ are elements of $G$ that converge to $S$. (Someone who is inexperienced might not see right away how your argument implies the density of $G$.) $\endgroup$ – Martin Sleziak Dec 10 '11 at 14:40
  • $\begingroup$ Thanks, fixed. I tend to write $(-)$ in my own work, but here it is indeed probably better to stick to $(-1)$. $\endgroup$ – Peteris Dec 10 '11 at 14:44

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