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I am trying to prove the existence of continuous onto function from (0,1) to [0,1]. Well, it was easy, since $\sin(\dfrac1{x(x-1)})$ does what I need.

BUT... a slight tweak to the definition of continuity grants that the preimage of a closed set is closed. Well, [0,1] is closed, but (0,1) is NOT!

What is wrong with it?

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    $\begingroup$ $(0,1)$ is closed in $(0,1)$: any topological space is a closed subset of itself. $\endgroup$
    – Pedro ♦
    Sep 16, 2016 at 4:33
  • $\begingroup$ And $(0,1/4]$ is likewise closed in $(0,1)$. Just sayin' ... $\endgroup$
    – Ted Shifrin
    Sep 16, 2016 at 4:40
  • $\begingroup$ And $(0,1)$ is open and closed. $\endgroup$
    – copper.hat
    Sep 16, 2016 at 4:49
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    $\begingroup$ Just wondering, is there any reason that a simple piecewise function wouldn't work? Take $$f(x) = \begin{cases}0 & x \in (0, 1/4) \\ 2(x - 1/4) & x \in [1/4, 3/4] \\ 1 & x \in (3/4, 1) \end{cases}$$ because I never would have thought of $\sin(1/(x(x - 1)))$ $\endgroup$
    – JMK
    Sep 16, 2016 at 4:53
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    $\begingroup$ Or how about $sin^2(2\pi x)$? $\endgroup$
    – David K
    Sep 16, 2016 at 4:55

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