Infimum of a subgroup $= 0$ implies...? Let $H$ be a subgroup of $\Bbb{R}$ (under addition).
Assume that for all $b \in \Bbb{R}_{>0}$, there exists $h \in H$ such that $0 < h < b$. Show that for all $x$, $y \in \Bbb{R}$ such that $x < y$, there is $h \in H$ such that $x < h < y$.

This problem even stumped my professor. I know that the problem is basically implying that $\inf\{h \in H\}=0$ and using it to prove the following inequality, and I understand it conceptually, but I'm not sure how to prove it using just arbitrary $x$, $y$, and $h$. Any help would be appreciated.
 A: HINT: For $h\in H$, let $Mult(h)=\{z\cdot h: z\in\mathbb{Z}\}$ be the set of multiples of $h$. For $x\in\mathbb{R}$ and $h\in H$, let


*

*$Small_h(x)$ be the greatest $k\in Mult(h)$ such that $k\le x$; and

*$Big_h(x)$ be the least $k\in Mult(h)$ such that $k>x$.
Note that $Big_h(x)=Small_h(x)+h$.
Suppose $Big_h(x)\not\in (x, y)$. What can you say about $h$ in terms of $y-x$?
A: Under multiplication:
Not true.  Let H = {$2^n| n \in \mathbb Z$}.  For any $b > 0$ there is a $1/2^n$ such that $0 < 1/2^n < b$, but there is no $2^n in [5, 7]$.
Under addition: 
Let $0 < h < y-x$.  By Archimedian principle find $n \in \mathbb Z$ such that $nh \le x$ but $(n + 1)h > x$.  Then $(n + 1)h = nh + h < x + (y-x) = y$ so $x < (n+1)h < y$.  $(n+1)h \in H$.
A: The question has already been answered by fleablood and Noah Schweber, so I want to remark on why their solution looks the way it does.
Maybe you think the solution looks inelegant. You wonder, why can't we solve the problem using just a few algebraic tricks, like we did so often in high school? If we're really clever, maybe we can add one inequality to another inequality in such a way that the answer pops right out? Something like:
$$\begin{eqnarray}
0&<&h&<&y-x \\
0&<&\frac{h}{2}&<&\frac{y-x}{2} \\
&&\vdots&& \\
&&&&\mathrm{(insert\; miracle)} \\
&&\vdots&& \\ x&<&3h&<&y
\end{eqnarray}$$
Nope, and here's why.
Consider the following generalization of your problem. We will replace the real numbers with a generic group $R$. For the question to mean roughly the same thing as before, $R$ should be a linearly ordered Abelian group. Then the straightforward generalization reads like this:


*

*Let $(R,+,0,<)$ be a linearly ordered Abelian group. Let $H$ be a subgroup of $R$ (under $+$). Assume that for all $b \in R_{>0}$, there exists $h \in H$ such that $0 < h < b$. Is it true that for all $x$, $y \in R$ such that $x < y$, there is $h \in H$ such that $x < h < y$?


This time the answer is no! Here's a simple counterexample: Let $R=\mathbb Z\times\mathbb Q$, the lexicographic product of the integers with the rational numbers. (This just means that elements of $R$ are compared in the dictionary order.) Let $H=\{0\}\times\mathbb Q$, the set of pairs whose first coordinate is zero. Then $H$ satisfies the hypotheses of the question: For $b=(b_1,b_2)\in R_{>0}$, if $b_1=0$ then take $h=(0,b_2/2)$ and if $b_1>0$ then take $h=(0,1)$. Now let $x=(1,0)$ and $y=(1,1)$. Then $H$ does not have any element between $x$ and $y$; indeed $H<x$ altogether!
The existence of this counterexample means that we cannot answer the original question using just the laws of ordered groups! We need to bring in some additional information about the real numbers $\mathbb R$.
Now, look at the answers again. They use the fact that $\mathbb R$ is not just any old ordered group; it is an Archimedean ordered group. The proof goes through just the same for any Archimedean group, which is strong evidence that the Archimedean property is the right property to call upon.
