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Find map $f : \mathbb{R} \rightarrow \mathbb{R}$ such that inverse image $f^{-1}(${$x$}$)$ has two elements for any $x\in \mathbb{R}$.

I'm really struggling to find an example so any hints or examples would be much appreciated.

Asaf Karagila
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Vojtie
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  • What restrictions are on $R$? The statement is not true for a one element set. – Jay Jan 27 '21 at 17:48
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    I would start by doing this for the half-open interval $[0,1)$. This would naturally extend to the whole real line. – hardmath Jan 27 '21 at 17:49
  • I edited your post to clean up spelling and math notation, but then rolled it back after I became concerned I might be changing the meaning of your post. Have a look at the edit history to see what I thought might be corrections. – hardmath Jan 27 '21 at 17:57
  • @Jay what do you mean by restrictions on $R$? $\mathbb {R}$ - set of all real numbers is the domain for the map. – Vojtie Jan 27 '21 at 17:59
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    Hint: Both $(-\infty,0)$ and $[0,\infty)$ have the same cardinality as $\mathbb R.$ – zhw. Jan 27 '21 at 18:09
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    If you're looking for a continuous $f$, there isn't any. – Robert Israel Jan 27 '21 at 18:17
  • For reasons unknown to me my computer is ot displaying either the domain or the range of the function as blackboard math. – Jay Jan 28 '21 at 13:30

1 Answers1

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For every $n\in\Bbb Z$, map the interval $[n,n+1)$ to itself twice over, by:

$$f(x) = \begin{cases} n+2(x-n) & \text{if } x<n+\frac12 \\ n+2(x-n-\frac12) & \text{if }x\ge n+\frac12 \end{cases} $$

TonyK
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