Let $(A,+)$ be a subgroup of $(\mathbb{R},+)$ and $A \neq \left\{0\right\}$. Consider the set $A_+ = \left\{a \in A: a>0\right\}$.

  1. Prove that $\alpha = \inf A_+$ is in $\mathbb{R_+}$ (non-negative real numbers). (have proved)
  2. Show that whenever $\alpha > 0$ then it is in $A_+$.
  3. Deduce $A = \alpha \mathbb{Z}$.
  4. Show that whenever $\alpha =0$, $A$ is dense subset of $\mathbb{R}$.

This is an interesting problem which is an intersection between Analysis and Algebra. I hope everyone can sit here and discuss it with me. I will also try to put the solution below whenever I have an idea.

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    $\begingroup$ @user10354138 Sorry, it is my typo. $H$ should be $A$ and I have already edited it. $\endgroup$ Commented Sep 23, 2023 at 13:04
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    $\begingroup$ I still see an $H$ there. $\endgroup$
    – jjagmath
    Commented Sep 23, 2023 at 13:08
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    $\begingroup$ Is $\Bbb R_+$ the set of positive reals or the set of non-negative reals? $\endgroup$
    – jjagmath
    Commented Sep 23, 2023 at 13:10
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    $\begingroup$ @jjagmath I have edited the typo. And $\mathbb{R}_+$ is the set of positive reals numbers. $\endgroup$ Commented Sep 23, 2023 at 13:18
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    $\begingroup$ Then, if you have proved $\alpha>0$ in part 1, why part 2 says "whenever $\alpha>0$"? and then part 4 makes no sense since $\alpha$ is never $0$. $\endgroup$
    – jjagmath
    Commented Sep 23, 2023 at 13:24

3 Answers 3


For the question 1:

Firstly, I show that $A_+$ is non-empty using the assumption that $(A,+)$ is a subgroup of $(\mathbb{R},+)$. Now, $A_+$ is non-empty and bounded from below by 0. Then there exists $\alpha \ge 0$ s.t $\alpha = \inf A_+$.

Finally, we prove that $\alpha >0$ by the definition of infimum by choosing $\varepsilon = \alpha/2$.

Edited: We cannot choose that $\varepsilon = \alpha/2$ since we are not sure that $\alpha>0$ and $\mathbb{R}_+$ in this problem means non-negative. We just finally prove that $\alpha \ge 0$.


For $2)$, suppose $\alpha > 0$ and that $\alpha\notin A$. Then there must be some $\beta\in A_+$ such that

$$\alpha < \beta < \alpha + \frac{\alpha}{2}$$

And some $\beta' \in A_+$ such that

$$ \alpha < \beta' < \beta $$

But since $A$ is a subgroup this means that

$$\beta - \beta' \in A$$

.And since $\beta' < \beta$, we also have that

$$ \beta - \beta' > 0$$

So $\beta - \beta' \in A_+ $. But since $ \beta' > \alpha $, we have that

$$ 0< \beta - \beta' < \frac{\alpha}{2} < \alpha$$

Contradicting our assumption that $\alpha$ is a lower bound of $A_+$.

For $3)$ we can do a similar thing, if $A\neq \alpha \mathbb{Z}$ then there there must be some element of $A$ between $\alpha$ and $2\alpha$ (why?) from which $\alpha$ can be subtracted to obtain the same contradiction as before.

$4)$ follows if you observe that for any $\varepsilon>0$ there is some $x\in A$ such that $0<x<\varepsilon$, then supposing $A$ is not dense there must be two positive real numbers $a,b$ such that

$$ (a,b)\cap A = \emptyset $$

If we then let $u$ be some element in $A_+$ such that $u<a$ then we can find some

$$ 0< x < \frac{b-a}{2}$$

And define

$$ n= \left\lfloor \frac{a-u}{x} \right\rfloor +1 $$

Such that $u+ nx \in (a,b)\cap A$, which is a contradiction

  • $\begingroup$ In your solution to question 2, why can you conclude that $\beta - \alpha \in A$? At the beginning, you have just supposed that $\alpha \notin A$. In addition, the "infimum inequality" does not need strictly less $\alpha < \beta$ (the equality still be accepted). $\endgroup$ Commented Sep 23, 2023 at 20:46
  • $\begingroup$ @TungNguyen sorry that was a mistake... I have corrected it now. As for the infimum inequality, you can prove that it the infimum of a set is not included in it, then there must be a member of the set that is arbitrarily close and above the infimum. Does this answer your question? $\endgroup$
    – Carlyle
    Commented Sep 23, 2023 at 21:04

For the question 2:

Since $\alpha=inf A_+$, there exists a sequence $\left(a_n\right)$ in $A_+$ converges to $\alpha$.

A is a group then $-a_n\in A$. We have $\left|a_m-a_n\right|\in A_+$. Note that $\left(a_n\right)$ is also a Cauchy sequence therefore $\left|a_m-a_n\right|<\alpha,\forall m,n\ge N$. So $\left|a_m-a_n\right|$ must be $0$ for all $m,n\ge N$ which means $a_n=a_N,\forall n\ge N$. Finally $\alpha=\lim a_n = a_N \in A_+$


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