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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.

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  • $\begingroup$ en.wikipedia.org/wiki/Space-filling_curve $\endgroup$ – Will Jagy Dec 20 '18 at 1:39
  • $\begingroup$ I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve. $\endgroup$ – MathematicsStudent1122 Dec 20 '18 at 1:39
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    $\begingroup$ 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 $\endgroup$ – Rolf Hoyer Dec 20 '18 at 1:41
  • $\begingroup$ @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. $\endgroup$ – Mohammed Shahid Dec 20 '18 at 2:14
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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 $f^{-1}(D)=(-1,1)$.

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  • $\begingroup$ 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. $\endgroup$ – The_Sympathizer Dec 20 '18 at 6:13
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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].

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

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