KAIST POW 2012-11 : Dividing a circle

Dividing a circle

Let $f$ be a continuous function from $[0,1]$ to a circle. Prove that there exists two closed intervals $I_1 , I_2 \subseteq [0,1]$ such that $I_1 \cap I_2$ has at most one point, $f(I_1)$ and $f(I_2)$ are semicircles, and $f(I_1)\cup f(I_2)$ is a circle.

Proof. Let the circle $C$. I will directly construct such intervals. Note that continuous function $f$ preserves connectedness and compactness. Since $[0,1]$ and $C$ is bounded, every closed subspace of them is compact.

Step1. We find minimal interval whose image is the circle.

We consider $A=\{x\in [0,1] : f([0,x])=C\}$. It is nonempty since $1\in A$. We will show that $d=\operatorname{inf}A\in A$. Assume the contrary that $f([0,d])\neq C$. Let $R\in C-f([0,d])$ By definition of $A$, $R\in f([d, d+(1-d)/n])$ for all $n\in \mathbb{N}$. Let $x_n\in [d, d+(1-d)/n]$ and $f(x_n)=R$. Then by continuity of $f$, $f(d)=f(\lim_{n\to \infty} x_n)=\lim_{n\to \infty} f(x_n)=R$. This contradicts to $R\in C-f([0,d])$. By same argument on $B=\{x\in [x,d] : f([x,d])=C\}$ with $0\in B$, we can take $a=\operatorname{sup}B\in B$. Let $f(a)=Q$ and $P$ be the opposite point of $Q$.

Claim. $f(a),f(d)\notin f((a,d))$, $f(a)=f(d)=Q$.


Assume the contrary that $f(a)\in f((a,d)), \exists r\in (a,d)$ such that $f(r)=f(a)$. We will use the fact that any $x>a$ cannot satisfy $f([x,d])=C$ to verify this assumption is impossible.
$f([a,r])\neq C$. Since $f([a,r])$ is closed, it contains boundary points. If $f([a,r])$ is singleton, $f([r,d])=C$. So $f([a,r])$ is not singleton, so contains 2 boundary points. Let $b_1< b_2\in [a,r]$ such that $ \{f(b_1), f(b_2)\}=\operatorname{bdy}f([a,r])$. Then $f([a,r])=f([b_1,b_2])$. So $f([b_1,d])=C$.  That is, $a=b_1$. So $f(a)(=f(r))$ and $f(b_2)$ are distinct boundary points, we deduce that $f([a, b_2])=f([b_2, r])$. So $f([b_2, d])=C$. That is, $a=b_2$. Then $f([a,r])=f([b_1,b_2])$ is singleton, and this is contradiction. That is, there is no such $r$. Finally, $Q\notin f((a,d))$.
 By similar argument with minimality of $d$ in $A$, $f(d)\notin f((a,d))$. Since $(a,d)$ is connected, $f((a,d))$ is connected so $C-f((a,d))$ is at most singleton. Thus, we have $f(a)=f(d)=Q$ and $Q\notin f((a,d))$.


Step2. We cut the path to get two semicircles.

Let $D =f^{-1}(P)\cap [a,d]$. This is nonempty because $f([a,d])=C$. Since $f$ is continuous and $C-\{P\}$ is open, $f^{-1}(C-\{P\}) $ is open. That is, $D=[a,d]-f^{-1}(C-\{P\})$ is closed. So $b=\operatorname{inf} D, c=\operatorname{sup} D \in D$. Then $S_1 = f([a,b]), S_2=f([c,d])$ are semicircles containing $P$ and $Q$ because $f((a,b)), f((c,d))$ do not contain $P$ and $Q$.
$[a,b]\cap [c,d]$ has at most one point because $b=\operatorname{inf} D\leq \operatorname{sup}D=c$. If $S_1=S_2, f([a,c])=f([a,d])=C$. This contradicts to minimality of $d$. So $S_1\neq S_2$, which means $S_1\cup S_2=C$.