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Dec 1, 2019 at 12:18 history edited bof CC BY-SA 4.0
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Dec 1, 2019 at 12:10 history edited bof CC BY-SA 4.0
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Dec 1, 2019 at 12:06 comment added bof Sounds right. Hmm. What if we define $f(x)=x$ for $x\notin\bigcup_nA_n$?
Dec 1, 2019 at 11:26 comment added Martin Sleziak If we choose the function $f_n$ in a manner similar to Devil's staircase, we get that $f^{-1}(x)$ is countable for each $x\ne0$, right? (For each $x$ we get at most two preimages inside $A_n$.) I am asking this since Vladimir Kanovei mentioned in a comment that they would also be interested in an example where the fibers are countable dense sets.
Dec 1, 2019 at 11:17 history answered bof CC BY-SA 4.0