The set of reals is a closed subset of the complex numbers

I was working on this quiz review problem and was wondering if anyone could give me an opinion on my work.

There's a couple of ways to prove this, one would be to show that $\mathbb{C} \backslash \mathbb{R}$ is open, but I decided to show if $z \not \in \mathbb{R}$, then it is not a boundary point of $\mathbb{R}$.

Suppose that $z \in \mathbb{C} \backslash \mathbb{R}$. Then there are $x,y\in \mathbb{R}$ with $y\neq 0$ such that $z=x+iy$.

Then, $D_{|y|/2}(z)$ is a proper subset of $\mathbb{C}\backslash \mathbb{R}$, and so $z$ is not a boundary point of $\mathbb{R}$.

What do you guys think?

@complexcharacter Looks good. You've chosen $r=|y|/2$. Just out of curiosity, what is the largest possible $r$?

Well, $|y|$ would work as well, I believe. Any larger would not.

Wouldn't the open disc of radius | y | and center z interesect the real axis at the point x?

It wouldn't, as it is open and does not actually contain $x$.

I was wondering if a similar proof could be used by starting with $\mathbb{C}\cap\mathbb{R}\supset\mathbb{R}$ and $x,y\in\mathbb{R}$ with $y\equiv0\ni z=x+iy$. Then the $D_r(z)$ represented in polar form $re^{\theta i}$ would contain points $r,\theta \in \mathbb{R}$ would,for all $r>0$ contain points along $\theta=2k\pi$ $\forall$ $k\in\mathbb{N}$ that were elemets of $\mathbb{R}\cap\mathbb{C}$ and points that are not for all other $\theta$, making all the points $z\in\mathbb{R}\cap\mathbb{C}$ and thus all the points in $z\in\mathbb{R}$ boundry points. Do I co too far in the beginning by assuming $\mathbb{R}\cap\mathbb{C}\supset\mathbb{R}$?

Why is this done in terms of $\mathbb{C}\backslash \mathbb{R}$?

If you recall we proved in class that $E$ is closed iff $E^c$ is open. $\mathbb{C} \setminus\mathbb{R}$ is $E^c$ if $E=\mathbb{C}$.

Oh! Thank-you so much!