# About infimum of a subgroup of $(\mathbb{R},+)$

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.

• @user10354138 Sorry, it is my typo. $H$ should be $A$ and I have already edited it. Commented Sep 23, 2023 at 13:04
• I still see an $H$ there. Commented Sep 23, 2023 at 13:08
• Is $\Bbb R_+$ the set of positive reals or the set of non-negative reals? Commented Sep 23, 2023 at 13:10
• @jjagmath I have edited the typo. And $\mathbb{R}_+$ is the set of positive reals numbers. Commented Sep 23, 2023 at 13:18
• 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 $0$. Commented Sep 23, 2023 at 13:24

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 $$0, then supposing $$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 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

• 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). Commented Sep 23, 2023 at 20:46
• @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? 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 $$\alpha=\lim a_n = a_N \in A_+$$