3

Is there a continuous map $f\colon[0,1]\to[0,1]$ from the unit interval to itself such that the preimages $f^{-1}(t)$ are infinite for every point $0\le t\le 1$?

Rylee Lyman
  • 3,930
  • 5
    Take the projection to the first coordinate, of the Peano curve from $[0,1]$ filling the square $[0,1]^2$. For each $t\in[0,1]$ there is a pre-image $x_r\in[0,1]$ for each point $(t,r)\in[0,1]^2$, where $x_r$ is a time at which the Peano curve passes through $(t,r)$. – MoonLightSyzygy Jan 17 '20 at 22:50
  • 2
    @MoonLightSyzygy This is worth to be an offcial answer. – Paul Frost Jan 17 '20 at 23:30
  • @PaulFrost Stick to closing questions perfectly good questions. I will stick to doing math and posting answers wherever I want. I don't need you to tell me how worthy my answers are. It is there such that the answer gets to the person who asked, independently of the human sacrifices that sometimes, and sometimes not, people like you like to try to enforce in this website. – MoonLightSyzygy Jan 17 '20 at 23:58
  • 3
    @MoonLightSyzygy I appreciate your friendly response. – Paul Frost Jan 18 '20 at 00:17
  • 3
    @MoonLightSyzygy this is worth to be an official answer. – Ty Jensen Jan 18 '20 at 00:39

1 Answers1

1

Yes such functions do exist. For an example consider a space filling curve $[0,1]\to[0,1]^2$ and compose with the projection on the first coordinate.

Note that in this case the fibers all have cardinality $\mathfrak c$. It is also possible, though harder to have countable fibers. However for $\aleph_0<\kappa<\mathfrak c$ there is no continuous map $[0,1]\to[0,1]$ with fibers of cardinality $\kappa$.

Alessandro Codenotti
  • 12,285
  • I was going to post only the first part as a community wiki after seeing the comments, but I realized I had something more which could be interesting to say and went for a normal answer – Alessandro Codenotti Jan 18 '20 at 09:52
  • The final comment is a consequence of the fact that the fibres are Borel sets (or compact Hausdorff), of course. Not even a single fibre can ever have such a size. – Henno Brandsma Jan 18 '20 at 12:24