Is $\{z\in\mathbb C\mid|\text{Re }z|+|\text{Im }z|\le1\}$ open or closed?

Question

I am trying to figure out if the set $\{z\in\mathbb C\mid|\text{Re }z|+|\text{Im }z|\le1\}$ is open or closed or maybe none of that. I hope someone could provide a hint to solve this. Can this set be visualized? Thanks

Answer 1

It's a closed set. Actually if you identify $\mathbb{C}$ with $\mathbb{R}^2$, your set is just the closed unit ball of the 1-norm.

Answer 2

Function $z\mapsto\left|\text{Re }z\right|+\left|\text{Im }z\right|$ is continuous and the set you mention is the preimage of the closed set $\left[0,1\right]$. This implies that the set is closed.

Answer 3

The set is closed as preimage of a closed set under a continuous map.

Your set is the interior of the tetragon (including the boundary) in the complex plane spanned by the points $\pm i$, $\pm 1$ (see picture).

You can also think of it as the closed unit ball in the $1$-norm.

Answer 4

An idea: write a complex number as an ordered pair in the real plane:

$$\Bbb C\ni z=a+bi\longrightarrow (a,b)\in\Bbb R^2\;$$

You're interested in the set

$$\{(a,b)\in\Bbb R^2\;:\;\;|a|+|b|\le 1\}$$

Which is the closed square of side $\;\sqrt2\;$ and, of course, diagonals of length $\;\sqrt2\sqrt2=2\;$ and vertices on $\;(1,0)\,,\,(0,1)\,,\,(-1,0)\,,\,(0,-1)\;$

Clearly, this set is closed.

Answer 5

Real part, imaginary part, addition, absolute value are continuous. Hence your set is the inverse image under a continuous function of a closed subset of $\mathbb R$, hence closed.