# Exceptional values of real-valued functions on [0,1]

Given a continuous real-valued function $f$ from $[0,1]$ to itself with $f(0)=0$ and $f(1)=1$ such that $f^{-1}(c)$ is finite for all $c$ in $[0,1]$, let $E(f)$ be the set of $c$ in $[0,1]$ such that $|f^{-1}(c)|$ has (positive) even cardinality. Call a subset of $[0,1]$ an exceptional set if it is of the form $E(f)$ for some $f$ satisfying the stated conditions. How big can an exceptional set be? I intend "big" to be vague; you can interpret it in the sense of cardinality, measure, category, or Hausdorff dimension. It is not hard to construct an exceptional set that is countably infinite (I believe this is an exercise in Spivak's calculus book).

• Do you want that $f(x)$ is between $f(0)$ and $f(1)$ for all $x$? Otherwise, you can make $E(f)$ pretty much as large as you want: just take $f(x) = x-x^2+ \varepsilon x$ for $\varepsilon$ small. – Marco Golla Jun 7 '15 at 15:16
• Oops, you're right! I'll fix the question. Thanks. – James Propp Jun 7 '15 at 15:49

$E(f)$ is at most countable. An elementary argument based on the intermediate value theorem shows that any $c\in E(f)$ must be the value of a (strict) local extremum of $f$, at one of the points where this value is taken. However, if $c$ is a local maximum, say, then we can find $a,b\in\mathbb Q$ which let us recover $c$ as $c=\max_{a\le x\le b}f(x)$, so there are only countably many such values $c$.
• I'd like to know more about the elementary argument based on the IVT that Christian Remling refers to. (I see how to proceed if $f$ is piecewise monotone, but does this follow from my hypotheses?) – James Propp Jun 7 '15 at 18:56
• @JamesPropp: Say $0<x_1<\ldots < x_{2n}<1$ is the list of points where $f=c (\not= 0,1)$. Then $f(t)<c$ for $t<x_1$ (if not, then I'm getting more points with $f=c$), so $f>c$ at some points $>x_1$ (and close to $x_1$) at least if $x_1$ is not a maximum. Then in fact $f>c$ on all of $(x_1,x_2)$ (to avoid getting extra points with $f=c$). Continue in this style; if you want to avoid extrema, $f-c$ has to change sign at each $x_j$, but then you arrive with the wrong sign at $x_{2n}$, because the number of points is even. – Christian Remling Jun 7 '15 at 19:17