Let $f:\mathbb R\to [1,3]$ be a continuous function such that
(i) $f^{-1}(1)$ and $f^{-1}(3)$ are singletons.
(ii) $f^{-1}(x)$ contains exactly two points for all $x\in (1,2)\cup (2,3)$.
How to find the possible cardinality of $f^{-1}(2)$?
Since $f$ is continuous, one can use intermediate value theorem. But after that I could not fix the problem. Any suggestions please!
On
All the possibilities are $1$, $2^{\aleph_0}$. Examples are given by others. It is helpful to draw a picture when you read the following proof.
Firstly we can suppose $f^{-1}(1)<f^{-1}(3)$. Otherwise we can consider $f(-x)$. So we will not lose the possibilities of cardinality of $f^{-1}(2)$ by doing this.
Let $a=f^{-1}(1)$, $b=f^{-1}(3)$. We will do the following steps.
(1) There is at least one element $x\in (a,b)$ such that $f(x)=2$.
(2) $f$ is increasing on $(a,b)$. $f$ is decreasing on $(b,+\infty)$. $f$ is decreasing on $(-\infty,a)$.
(3) If there are two distinct elements $x_1,x_2\in(a,b)$ such that $f(x_1)=f(x_2)=2$, Then Card$(f^{-1}(2))=2^{\aleph_0}$.
(4) It is not possible that $\exists x\in(-\infty,a)$, such that $f(x)=2$. Similar for $(b,+\infty)$.
(5) Conclude that Card$f^{-1}(2)$ is $1$ or $2^{\aleph_0}$.
Proof of (2): If $f$ is not increasing, we can get $c,d$ satisfying $a<c<d<b$ such that $f(c)>f(d)$. Then by intermediate value theorem, $f$ will hit any value in $(f(d),f(c))\backslash\{2\}$ at least three times. Contradiction! It is similar for the remaining two.
(3) Because $f$ is increasing on $(a,b)$.
(4) If $\exists x\in(-\infty,a)$ such that $f(x)=2$. Let $y=\lim_{x\rightarrow-\infty}f(x)$. Then $y\ne 3$ (Otherwise we can not arrang $x$ in $(b,+\infty)$) and $y\notin f^{-1}((-\infty,a))$. So there exists $x_2\in(b,+\infty)$ such that $f(b)=y$. Then we hit $f(b+1)$ at least three times. Contradiction!
This function has $|f^{-1}(2)|=1$, while this function has $|f^{-1}(2)|=c$. It's not clear whether any other possibilites exist.