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

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.

• en.wikipedia.org/wiki/Space-filling_curve – Will Jagy Dec 20 '18 at 1:39
• I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve. – MathematicsStudent1122 Dec 20 '18 at 1:39
• the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill – Rolf Hoyer Dec 20 '18 at 1:41
• @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that. – Mohammed Shahid Dec 20 '18 at 2:14

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 $$f^{-1}(D)=(-1,1)$$.

• Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set. – The_Sympathizer Dec 20 '18 at 6:13

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].

• Good example, but you really should format it better. – zhw. Dec 20 '18 at 17:25