Is $|z-4|\geq |z|$ an open or closed region?

Question

After solving $|x+i y-4|\geq |x+iy|$ I found that $|z-4|\geq |z|$ if $x\leq 2$. Does this mean that $|z-4|\geq |z|$ is closed?

Answer 1

Yes. An alternative way to formalize this, is to not that $f\colon z\mapsto |z-4|-|z|$ is continuous (because subtraction and absolute value are continuous) and hence our region $f^{-1}([0,\infty))$ is closed because it is the pre-image of a closed set under continuous function.

Answer 2

We have $$ \lvert z - 4 \rvert \ge \lvert z \rvert \iff \\ \lvert z - 4 \rvert^2 \ge \lvert z \rvert^2 \iff \\ (x-4)^2 + y^2 \ge x^2 + y^2 \iff \\ -8x + 16 \ge 0 \iff \\ x \le 2 $$ so your solution set agrees with mine.

The question is how to characterize a complex set as open or closed.

According to this definition it would not be open, because e.g. for $z=2$ we could not fit in a disc with positive radius around it and stay within the solution set.

Now for the other property:

$S$ would be closed if its complement is open. The complement would be $$ S^C = \{ z = x + i y \mid (x,y) \in \mathbb{R}^2, x>2 \} $$ and this one indeed seems open, as one can fit in positive radius discs around every point of $S^C$.

So the set $S$ is closed and not open.