I took a discrete dynamical systems class last semester, and there is still something bothering me. We had a homework question asking us to prove the following (the "covering interval theorem"):

\begin{gather*} F: \mathbb{R} \rightarrow \mathbb{R} \; , \; F \in C^{\geq 1} \; , \; I \subset \mathbb{R} \\ F(I) \supset I \implies \exists x \in I \; | \; F(x) = x \end{gather*} or in other words, for a continuous real-valued function $F$, show that if there is some interval $I$ where the image of $I$ under $F$ covers $I$, that is $I$ is a subset of $F(I)$, then there exists a fixed point.

I found this problem very unsettling because it seems so obvious, but my professor insisted that there were multiple cases and how it was not trivial, but he never revealed the solution.

My idea was that since $F$ is continuous and $F(I) \supset I$, then

\begin{gather} \exists J \subset I \; | \; F(J) = I \end{gather}

and low and behold you can use the intermediate value theorem on that fact:

\begin{gather*} J = [a,b] \; , \; I = [c,d]\\ J \subset I \implies c \leq a < b \leq d \\ F(J) = I \implies c \leq F(a) \leq d \; \land \; c \leq F(b) \leq d\\ G(x)\equiv F(x) - x \in C^{\geq 1} \implies G(a) \geq 0 \; \land \; G(b) \leq 0 \end{gather*} so that completes the proof since if $G = 0$ at $a$ or $b$, then those are the fixed points, and if the strict inequalities hold, by the IVT we have $G = 0$ somewhere on the interval.

So my question is, why is this not a complete proof, or is it? Also does anyone know how to prove the statement about the existence of $J$?

Sorry if this is similar to other questions, one came up that was quite similar, but I'm looking for more explanation here.



1 Answer 1


You are right, but in fact even your argument can be made a bit shorter:

Write $I=[c,d]$. Since $F(I) \supset I$, there exist $a,b \in I$ such that $F(a) = c \le a$ and $F(b)=d\ge b$. Since $F(a) − a \le 0$ and $F(b) − b \ge 0$, the continuous function $F(x) − x$ has a zero in $I$.


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