Let $S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$ Prove that $S$ is dense in $\Bbb R.$

Question

Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that $S$ is dense in $\Bbb R.$

My definition/knowledge of density is limited to only these two statements:

• A set $G$ is dense in $\Bbb R,$ iff for every $x,y\in \Bbb R$ , there exists a $g$ in $G,$ such that $x<g<y.$

• A set $G$ is dense in $\Bbb R,$ iff every $x\in\Bbb R$ is a limit point of $G.$

With these two definitions, I made countless number of futile attempts and ultimately, I gave up. To be honest, I know that there are former posts relating to this particular topic such as :

Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$.

But the point is, I noticed the user there asks in the title, to prove that $S$ is dense in $\Bbb R.$ However, in the post the user wants to prove:

Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.

It seems that proving this fact, is equivalent to proving that $S$ is dense in $\Bbb R.$ This implies, that the definition of a dense set is:

A set $G$ is dense in $\Bbb R$ iff for every $\epsilon > 0$, the intersection of $G$ and $(0, \epsilon)$ is nonempty.

I don't understand if this is an alternative definition of dense sets, but all in all, I seem to have no idea about what is veing done in the post. For example, "Why is showing "for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty" is sufficient to prove that $S$ is dense in $\Bbb R$ ?"

Any help with this problem will be appreciated. I want to know, what are the strategies to solve these sort of questions. This would be very much helpful. To be noted, that I am familiar with "Basic Topology of $\Bbb R$".

Answer 1

The fact that for every $\epsilon>0$ the intersection of $G$ with $(0,\epsilon)$ is nonempty is not sufficient to prove that $G$ is dense in $\mathbb R$. It is merely sufficient to prove that $0$ is the limit point of $G$.

But - our particular set $G$ is not a random set here: it is closed for addition and taking additive inverse, i.e. if $x,y\in G$ then $x+y\in G$ and also if $x\in G$ then $−x\in G$. You can use those facts to prove that all the other points in $\mathbb R$ are limit points of G.

So, take a point $x\in\mathbb R$, $x\ne 0$. We will tackle the cases $x>0$ and $x<0$ separately.

For the start, let's assume $x>0$. Let $\epsilon>0$. We will prove that there is an element of $G$ in $(x, x+\epsilon)$.

Namely, take $g\in G\cap(0,\epsilon)$, which must exist as per the linked question. Now, use Archimedean property of $\mathbb R$ to conclude that there is $n\in\mathbb N$ such that $ng>x$. Take the smallest such $n$. Now notice:

• $ng\in G$ because $G$ is closed for addition, and $ng=\underbrace{g+g+\ldots+g}_{n\text{ times}}$.
• $ng>x$ (as chosen)
• $ng<x+\epsilon$. This is because, if we suppose the opposite, i.e., that $ng\ge\ x+\epsilon$, then, knowing that $-g>-\epsilon$ (as $g<\epsilon$) and adding those two inequalities, we get that $(n-1)g>x$, which is the contradiction with $n$ being the smallest.

Side note: one can prove that, in fact, $n=\lfloor x/g\rfloor+1$. The Archimedean property of $\mathbb R$ is nothing else than saying that "floor" function is well-defined, so if you can already use the "floor" function, there is no reason not to.

So, $ng\in G\cap(x,x+\epsilon)$. As this is true for every $\epsilon>0$, $x$ is the limit point of $G$.

The case $x<0$ is now easy: we have $-x>0$ and, for every $\epsilon>0$ there is an element of $g\in G\cap(-x, -x+\epsilon)$ (which we have just proven). Then, as $G$ is closed for additive inverse, $-g\in G\cap(x-\epsilon,x)$. So, again, $x$ is a limit point of $G$.

Answer 2

A simple counter-example for the equivalence is $G=(0,1)$.

The two properties are equivalent for an additive subgroup $G$ of $\mathbb R$. Your set $S$ is an additive subgroup.

Proof when $G$ is a subgroup:

Let $U$ be any open set. Suppose there is a number $x >0$ in $U$. There exists $r>0$ such that $(x-r,x+r) \subseteq U$. Let $0<\epsilon <r$. There is a point $y$ common to $G$ and $(0, \epsilon)$. The length of the interval $(\frac {x-r} y, \frac {x+r} y)$ is $\frac {2r} y$ which exceeds $1$ since $y<\epsilon <r$. Hence, this interval contains an integer $n$. Now check that $ny \in G\cap (x-r,x+r) \subseteq G \cap U$. This $G \cap U$ is non-empty. A similar argument wroks when $U$ contains a negative number.