$\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ open or closed

Question

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Proving that a complex set in open/closed/neither and bounded/not bounded

I think $\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ is closed. But I have no idea how to show it since you have to take an element of the set (which lies on the axes) and take a neighbourhood around that (not all of the ball is within the set so it's not open).

But when you do the complement, not all of everything outside of the axes includes the ball (as some of the ball is on the axes). This would make it neither open nor closed but I'm sure it's closed! Can someone help me please?

It's definitely not bounded because no closed ball can cover all of the axes as they go on to infinity and beyond. Right?

Answer 1

The set $E=\{z\in C:|z|=|Re(z)|+|Im(z)|\}$ is precisely the union of the horizontal and vertical axes on the complex plane. This is closed but not open.

To see that it is closed, show that the complement is open: for any $z$ in the complement of $E$, let $r=\min(|Re(z)|,|Im(z)|)$. Then the ball with centre $z$ and radius $r$ lies in the complement of $E$.

To see that it is not open, consider the balls centred at $z=0$ with arbitrary radius. Any such ball certainly contains a point not lying on either of the axes.

Answer 2

EDIT

Looks like the question has now changed.

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I assume you are interested in the following set $X$. $$X = \{\left \vert \text{Real}(z) \right \vert + \left \vert \text{Imag}(z) \right \vert : z \in \mathbb{C}\}$$

HINT

Note that $X$ contains only non-negative real numbers. It is easy to show that (Why?) $$X = \mathbb{R}^+_0 = \mathbb{R}^+ \cup \{0\}$$

Now look at the complement of $\mathbb{R}^+_0$ in $\mathbb{C}$ i.e. $X^c = \mathbb{C} \backslash \mathbb{R}^+_0$. Can you show that it is open?

HINT

Consider $z \in X^c$ and $z$ is not purely real. Consider an open ball centered at $z$ and radius $$r = \dfrac{\left \vert \text{Imag(z)} \right \vert}2$$

For $z \in X^c$ and is purely real, then $z$ is negative, For this $z$, consider an open ball centered at $z$ and radius $$r = \dfrac{\left \vert \text{z} \right \vert}2$$

Answer 3

If you mean the set $\{z\in\mathbb C : |z|=|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\}$ you might want to consider the continuous function $g(z)=|z|-|\mathrm{Re}(z)|-|\mathrm{Im}(z)|$ and remember how continuous functions behave with closed sets.

Answer 4

The function $|z|-|\operatorname{im}(z)|-|\operatorname{re}(z)|$ is continuous. The set in question is the inverse-image of the closed set $\{0\}$ under this function. The inverse-image of a closed set under a continuous function is closed.