# Closure of the Rational Plane

I came across this interesting problem in Conway's Complex Analysis book.

Let $A = \{a + bi ~|~ a,b \in \mathbb{Q} \}$. Show that $\overline{A} = \mathbb{C}$.

All we are given about the closure of a set is the bare definition---namely, that $\overline{A} = \bigcap_{A \subseteq C \subseteq \mathbb{C}} C$, where $C$ is a closed set. In other words, the smallest closed set containing. So, here is my attempt:

Suppose the contrary, that $K$ is the smallest closed set containing $A$; i.e., $\overline{A} = K \subset \mathbb{C}$. Then there exists a $z \in \mathbb{C}$ such that $z \notin K$. This means $z \in \mathbb{C} - K$, which is an open set. Therefore, $V_\epsilon (z) \subseteq \mathbb{C} - K$ for some $\epsilon > 0$. Consider the element $z + \frac{\epsilon}{2}(1+i)$. By the density of the rationals, there exist $p$ and $q$ such that

$x < p < x + \frac{\epsilon}{2}$ and $y < q < y + \frac{\epsilon}{2}$,

or

$0 < p - x < \frac{\epsilon}{2}$ and $0 < q - y < \frac{\epsilon}{2}$. Hence, $p + iq \in V_\epsilon (z)$ because

$|z -(p+iq)|^2 = |(x-p) +i(y-q)^2| = (x-p)^2 + (y-q)^2 < \frac{1}{2} \epsilon^2$

Therefore, $|z -(p+iq)| < \frac{1}{\sqrt{2}} \epsilon < \epsilon$. But $p+iq$ is by definition in $A$. Therefore, this is a contradiction.

Does this seem right?

Choose $z \in \mathbb{C}$ and let $z_n = {1 \over n} \lfloor n\operatorname{re} z \rfloor + i {1 \over n} \lfloor n\operatorname{im} z \rfloor$. Then $z_n \in A$ and $z_n \to z$.
So, if $C$ is closed and $A \subset C$, then $C$ must contain every $z$ (since it contains its limit points). That is, $\mathbb{C} \subset C$, hence $C = \mathbb{C}$ and so $\overline{A} = \mathbb{C}$.