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


$0 < p - x < \frac{\epsilon}{2}$ 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?


It is probably a little easier to prove this directly.

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 $\overline{A} = \mathbb{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.