Does there exist a function such that the preimage of $x ^ { 2 } + y ^ { 2 } \leq 1$ is the closed interval $[-1,1]?$

Question

Does there exist a continuous function $f : \mathbb { R } \rightarrow \mathbb { R } ^ { 2 }$ such that the preimage of the closed unit disk $x ^ { 2 } + y ^ { 2 } \leq 1$ is the closed interval $[ - 1,1 ] ?$ the open interval $( - 1,1 ) ?$

To be honest, I don't really know how to go about this problem.

Answer 1

Space filling curves certainly work but that's quite excessive.

For a function $f:\Bbb R\to\Bbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:=\{(x,y)\in\Bbb R^2:x^2+y^2\le 1 \}$ is merely the set $$\begin{align} f^{-1}(D) &= \{ x\in\Bbb R: (f_1(x),f_2(x))\in D \} \\ &= \{ x\in\Bbb R: f_1(x)^2+f_2(x)^2\le 1 \}. \end{align}$$

By letting $f(x)=(x,0)$, we have $$ f^{-1}(D)= \{ x\in\Bbb R: x^2\le 1 \} = [-1,1]. $$

On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.

Answer 2

f:R -> R×R, x -> (x,0) is continuous.
Let D be the closed unit disk.
The preimage of D by f is
f$^{-1}$(D) = { x : f(x) in D } = [-1,1].