Decide if $\mathbb{R} \subset \mathbb{C}$ is open or closed. The solution states that the ball of radius $\epsilon >0$ around a real number $x$ always contains the non-real number $x+i\epsilon/2$. 
I don't understand the answer, for every number $x \in \mathbb{R}$ there is an open ball, right? For every $x \in \mathbb{R}$ there is an $r>0$ such that I can form an open ball $B_r(x)\subset \mathbb{R}$.
 A: To decide if $\mathbb{R}$ is open in $\mathbb{C}$, you must use the topology of $\mathbb{C}$, not of $\mathbb{R}$.  That is, you must take $B_r(x)\subset \mathbb{C}$.
A: I think you're confusing open balls in $\mathbb{R}$ and $\mathbb{C}$. If you want to prove that $\mathbb{R}$ is open in $\mathbb{C}$, you have to prove that for any $x \in \mathbb{R}$, there exists $\epsilon > 0$ such that $B_x(\epsilon) \subseteq \mathbb{R}$, where $B_x(\epsilon)$ is a ball in $\mathbb{C}$, meaning that $B_x(\epsilon) = \{ z \mid z \in \mathbb{C}, |z-x| < \epsilon\}$.
Since for any $x \in \mathbb{R}$ and $\epsilon > 0$, $x + i\frac{\epsilon}{2} \in B_x(\epsilon)$ but $x + i\frac{\epsilon}{2} \notin \mathbb{R}$, you can't find any such ball.
A: You need to be careful what the definition of the sets is precisely. 
You study a subset of the complex numbers thus your topological objects are those for the complex numbers. 
Thus in this context $B_r (x) = \{z \in \mathbb{C} \colon |z-x| < r\}$.  So it is the set of all complex numbers at a distance less than $r$ from $x$. 
Would you be working in a context where the ambient set are the real numbers then $B_r (x) = \{z \in \mathbb{R} \colon |z-x| < r\}$ it is the set of all real numbers at a distance less than $r$ from $x$. 
Put differently, the notation $B_r (x)$ is slightly imprecise in that it does not make explicit the set relative to which the ball around $x$ is formed, and this information is only implicit. For the context of your exercise you need to complex version. 
