Given a function $g:[0,1] \to \mathbb{R}$ satisfying the property that for every non empty dissection of the trivial $[0,1]$ to subsets $A,B$ we have either $\exists x \in A; g(x) \in B$ or $\exists x \in B; g(x) \in A$ and we have furthermore $g(x)>x$ for $x \in [0,1]$. Prove that there exist infinite $x \in [0,1]$ with $g(x)=1$. Proposed by Ali Zamani
Problem
Source: Iran TST1 Day2 P4
Tags: functional equation, Iranian TST, algebra
24.02.2020 03:09
Suppose, for contradiction, that there were only finitely many $x \in [0, 1]$ with $g(x) = 1.$ Call these numbers $x_1, x_2, \cdots, x_t,$ with $x_1$ being the largest. Call a real number $t \in [0, 1]$ tasty if and only if there exists a nonnegative integer $c$ so that $g^{(c)}(t) \in (x_1, 1).$ Observe that for tasty reals $t$ there is not nonnegative integer $c$ so that $g^{(c)}(t) = 1,$ as this would contradict the maximality of $x_1.$ Observe that the set $X$ of tasty reals is both nonempty and not $[0, 1].$ We claim that setting $A = X$ and $B = [0, 1] / X$ produces a contradiction. We need only to show the following lemma. Lemma. If $x \in [0, 1]$ is so that $g(x) \in [0, 1]$, then $x$ and $g(x)$ are either both in $A$ or both in $B.$ Proof. We need only show that $x \in A$ if and only if $g(x) \in A.$ If $g(x) \in A$, then we have that $g(x)$ is tasty. But since $g^{(i)}(g(x)) = g^{(i+1)}(x)$ for all nonnegative integers $i$, it's clear that $x$ would be tasty too. Now suppose that $x \in A.$ Suppose, for contradiction, that $g(x)$ was not tasty. Then, because $\{g^{(i)}(g(x))| i \in \mathbb{Z}_{\ge 0}\} = \{g^{(i)}(x) | i \in \mathbb{N}\}$ has no members in $(x_1, 1)$ but $\{g^{(i)}(x)| i \in \mathbb{Z}_{\ge 0}\}$ does, it's clear that $x \in (x_1, 1).$ But then $g(x) > x$ implies that $g(x) \in (x_1, 1]$ (since we're assuming $g(x) \in [0, 1]$). As $x> x_1$, we have that $g(x) \neq 1$, and hence $g(x) \in (x_1, 1)$ and $g(x)$ is actually tasty, contradiction. So the lemma is proven. $\blacksquare$ The above lemma shows that this partition of $[0, 1]$ into subsets $A$ and $B$ doesn't satisfy the property of the problem, and so we've reached a contradiction. $\square$
24.02.2020 18:31
Assume there are finitely many(possibly none) $x \in [0,1]$ with $g(x)=1$. The goal is designing partition $A \cup B=[0,1]$ with $x \in A \implies g(x) \notin B$ and vice versa. Starting from $1$, assume $A$ contains $1$. Then $A$ should also contain the preimage of $1$. $A$ should also contain the preimage of preimage of $1$. Repeating this, $A$ should at least contain every $x$ with $g^n (x)=1$ for some non-negative integer $n$. And stop. Let $B = [0,1] \cap A^c$. Straight from the definition of $A$, $g(x)\in A \implies x \in A$ and $1\ne x \in A \implies g(x) \in A$. If $A, B$ is not empty, then $1$ can be the only element satisfying the condition. But since $g(1)>1$, $g(1)$ cannot be the element of $B$. Next, since preimage of $1$ is finite, there exists $0<b$ such that every $b<x<1$ is not in the preimage of $1$. It can be easily seen that $B$ contains the open interval $(b,1)$. Also $A$ contains at least $1$. Therefore, $A$ and $B$ is both non-empty and contradiction came. Therefore, there are infinitely many $x$ with $g(x)=1$. Remark) Think of graph consists of vertices representing real numbers in $[0,1]$ and edge between $(x, g(x))$ if $0 \leq g(x) \leq 1$. To obey the condition, the graph should be consists of single connected component and both of the proof are finding some connected component that does not cover all the vertices.
15.03.2020 23:50
Take $A= \{ a | g^n(a)=1 , $ for some $ n\ge 0\} $ (here we define $g^0(x)=x, g^1(x)=g(x), g^2(x)= g(g(x)),...$ etc.) Clearly $A$ is nonempty cause $1\in A$ Suppose there are only finitely many $y$ such that $g(y)=1$ and let $m<1$ be the maximal one. If $x\in A$ then either $g^k(x)=1$ for $k>0$ so $x\le g^{k-1}(x) \le m$ or $x=1$. Then $[0,1]/A$ is nonempty for it contains $(m,1)$. This dissection however leads to a contradiction beacause -If $x\in A$ and $g^k(x)=1$ , $g^{k-1}(g(x))=1$ thus $g(x)\in A$ (or $x=1$). -If $x\in [0,1]/A$ then $ g^k(x) \neq 1 $ for any $k$ and the same holds for $g^k(g(x))$.
16.02.2022 14:52
Suppose not, and say there are finitely many preimages of $1$, and suppose $M$ is the largest, so no number in the range $(M,1)$ has image $1$. Define $S$ to be the set of reals such that on repeated applications of $g$ on it, it will eventually reach $1$. Clearly $S$ can't cover all of $[0,1)$ since nothing in the range $(M,1)$ can go to $1$ and $g$ only increases a number. Consider the partition $A = S \cup \{1\}$ and $B$ be the remaining numbers, which are both nonempty by the above line. Its easy to see that there do not exist $a \in A$ such that $f(a) \in B$ or $b \in B$ such that $f(b) \in A$, a contradiction to the problem statement, so we're done. $\blacksquare$