Whether a certain set of lines through the origin is dense in R^2 Let $S=\bigcup_{m,n\in\mathbb{Z}}T_{m,n}$ in $\mathbb{R}^2$, where $T_{m,n}$ is the straight line passing through the origin and the point $(m,n)$. I need to check whether this set is dense in $\mathbb{R}^2$.
My intuition says, for a given $ \epsilon>0$ and a point $(x,y)\in\mathbb{R}^2$ we can find a rational $\frac{m'}{n'}$ such that $|\frac{y}{x}-\frac{m'}{n'}|<\epsilon$, then we have to show $T_{n',m'}$ being $\epsilon$-close to $(x,y)$.
How do I show this? Thanks for the help.
 A: Let $(x,y)$ be a point in the plane. As you've stated, by density of the rationals in $\mathbb{R}$, for each $\epsilon > 0$ we have $(p,q)$ such that $d_2((x,y),(p,q)) < \epsilon$. Now, it suffices to see that $(p,q)$ belongs to some of these lines, i.e. that there exist $m,n \in \mathbb{Z}$ such that 
$$
L(x) = \frac{n}{m}(x-m) + n
$$
satisfies $L(p) = q$. We want, then, $m, n$ such that
$$
q = \frac{n}{m}(p-m) + n 
$$
Or equivalently, multiplying by $m$, 
$$
mq = np - nm + mn = np
$$
So, $\frac{m}{n} = \frac{p}{q}$ if we assume $m, n, q, p \neq 0$. We can justify this last assumption: whitout loss of generality, $p$ and $q$ can be taken non-zero because we can adjust the second coordinate of $(p,q)$ to $(p + \delta,q + \delta)$ by $\delta$, a small rational number while still being at less than $\epsilon$ of $(x,y)$. Now, this will imply $\frac{p}{q} \neq 0$, and therefore there exist nonzero integers $n, m$ (which we can even take coprime) such that $\frac{m}{n} = \frac{p}{q}$. 
A: Late to the thread, but I wanted to provide a picture proof.  Let $Q=(x_0,y_0)$ be any point in $\Bbb R^2$.  For any radius $\delta > 0$, consider the open ball $B_{\delta}(x_0,y_0)$ centered at Q. Consider the two rays $T_1$ and $T_2$ from the origin that are tangent to the ball.  
As you go farther from the origin, a bigger and bigger ball can fit between these two tangent lines.  Far enough away, many points $(m,n)$ (denoted by the red dots) would fall within the large ball.  Let $P$ be any one of them. Then the ray $L$ connecting the point $P$ to the origin has to pass through the ball $B_{\delta}(x_0,y_0)$ by construction.  We have therefore found an element of S that intersects with the ball.  Since the $(x_0,y_0)$ as well as $\delta$ were arbitrary, we conclude that $S$ is dense in $\Bbb R^2$.

A: What you are doing is approximating the slope of the line passing between $(0,0)$ and $(x,y)$ with a rational slope. Now, if $x$ and $y$ are small, this is fine. But you can imagine that if we move $(x,y)$ further and further away from the origin and keeping $y/x$ constant, this approximation gets worse and worse. But also, knowing only the slope doesn't help you to actually find a point of $T_{m,n}$ that is close to $(x,y)$.
One thing we might try is to approximate $x$ by a rational and $y$ by a rational. This works, because I claim that $S$ contains every point with rational coordinates. One way to see this is to parameterize $T_{m,n}$ as $\{t(m,n) : t \in \mathbf{R}\}$. Then
$$ S = \{ t(m,n) : t \in \mathbf{R}, m, n \in \mathbf{Z}\} $$
If I have a pair of rational coordinates $\left( \frac{p_1}{q_1}, \frac{p_2}{q_2} \right)$, then I can put them over a common denominator and pull out that common denominator into the parameter $t$ as follows:
$$ \left( \frac{p_1}{q_1}, \frac{p_2}{q_2} \right) = \left( \frac{p_1q_2}{q_1q_2}, \frac{p_2q_1}{q_1q_2} \right) = \frac{1}{q_1q_2}(p_1q_2, p_2q_1) $$
So we see that $\left( \frac{p_1}{q_1}, \frac{p_2}{q_2} \right) \in S$ by taking $t = (q_1q_2)^{-1}$, $m = p_1q_2$ and $n = p_2q_1$.
Now that we've shown that $S$ contains $\mathbf{Q}^2$, we are done, because $\mathbf{Q}^2$ is dense in $\mathbf{R}^2$.
