I was trying to solve the following problem : Let $f : (0,1) \to \Bbb R$ be a continuous map such that $\lim_{n \to \infty} f(\frac{1}{2n}) = 1$ and $\lim_{n \to \infty} f(\frac{1}{2n+1}) = 2$ where $n \in \Bbb N$. Then does $\exists \lambda \in (1,2) \text{ such that } \{c \in (0,1) | f(c) = \lambda\}$ is finite?
My attempt :
Combining : (i) $f$ is continuous on $(0,1)$ , (ii) $0$ is the limit point of the set $\{\frac{1}{2}, \frac{1}{3} , \frac{1}{4} , \frac{1}{5}, \ldots\} $ , (iii) $f$ has IVP , intuitively it seems to me that $\nexists \lambda \in (1,2) \text{ such that } \{c \in (0,1) | f(c) = \lambda\}$ is finite . However, I'm unable to come up with a rigorous proof.
Is my thinking correct or wrong ? Please help me by providing an answer.
“Please help me by providing an answer” is not really how things work around here. But you have a conjecture worth exploring:
To unravel the negations and quantifiers, this is equivalent to: For all $\lambda\in(1,2)$, the set $\{c \in (0,1) \mid f(c) = \lambda\}$ is infinite. Focusing in on zero, it's sufficient to show: For all $\lambda \in (1,2)$ and $\epsilon > 0$, there exists $c$ such that $0 < c < \epsilon$ and $f(c) = \lambda$.
And I think you can prove that with your observations (i)–(iii).