Is there a partition of an open square into closed segments (not reduced to a point)?

Question

Let $C$ be an open square (for example $]0, 1[ \times ]0, 1[$)of the plane $\mathbf R^2$. Is there a partition of $C$ into closed segments (not reduced to a point) ?

Answer 1

Begin by covering the plane with exclusively horizontal and vertical closed line segments. Start by covering the closed unit square:

• Horizontal line segments from $(0,y)$ to $(1,y)$ for $y\in[0,1]$.

Now, whenever you have the closed square $[0,A]^2$ covered, double the size of the covered region:

• Vertical line segments from $(x,0)$ to $(x,A)$ for $x\in (A,2A]$;
• Horizontal line segments from $(0,y)$ to $(2A,y)$ for $y \in (A,2A]$.

Repeat this to cover the plane. Now you can cover the open square as well, simply by mapping $(x,y)\rightarrow \left(f(x), f(y)\right)$, where $f$ nicely takes $\mathbb{R}$ to $(0,1)$; for instance, $f(z)=1/(1+e^{-z})$ works.