Is this set neither open nor closed? Let $B=\{x+iy\in \mathbb{C} | x,y\in \mathbb{Q}\}$. I think that this set is neither closed nor open. Am I right?
 A: You are right, to see this, use the density of the rationals in the reals. Let $z\in B$. Then consider any disk of radius $\epsilon$ around $z$. It will have some $w$ with at least one irrational real or imaginary part. So $z$ is not in the interior. Then consider $z\in B^c$. Use the same steps to see there will be a point with both rational real and imaginary parts inside any disk. This shows that $B$ is not open, and $B^c$ is not open. So $B$ is neither open or closed.
A: First, you need to mention the topology with respect to which open and closed make sense. The usual (metric) topology on $\mathbb C$ is the natural choice here, so let us consider that.

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*$B$ is not closed.

It suffices to show that $B$ is not sequentially closed. Consider $x,y\notin \Bbb Q$. As $\Bbb Q$ is dense in $\Bbb R$, there exist sequences $\{x_n\}_{n=1}^\infty, \{y_n\}_{n=1}^\infty \subset \Bbb Q$ such that $x_n \xrightarrow{n\to\infty} x$ and $y_n \xrightarrow{n\to\infty} y$. It is clear that $x_n + iy_n \in B$ for all $n$, and $x_n +iy_n\xrightarrow{n\to\infty} x+iy \notin B$. So, $B$ is not closed.


*$B$ is not open.

To see this, I urge you to show that for $0\in B$, and all $r > 0$, $\mathcal B(0; r) \cap B^c \ne \varnothing$. Here, $\mathcal B(0; r)$ denotes the ball of radius $r$ around $0$ in $\mathbb C$, defined by the usual metric.
