Let $\{a_n\}$ be a sequence converging to $A$ in $\mathbb{R}$. Let $f:[a_i,A]\to\mathbb{R}\,,i\in\mathbb{N}$ be a continuous fucntion such that the restriction $f|_{[a_i,a_{i+1}]}:[a_i,a_{i+1}]\to\mathbb{R}$ is strictly monotone. Set $B=\{M\in\mathbb{R}|\exists \text{infinitely many}\ x\in[a_i,A]:f(x)=M\}$
Then, what is the cardinality of $B$? Typically, is it one, null or infinity, or some other finite number?
I think the answer should be one. This is because the restricted function is given monotone. Any hints? Thanks beforehand.
I will assume that by 'monotone' you really mean strictly monotone. Otherwise it is easy to construct examples in which $B$ contains many points. For example, by putting $f(x)=x_1$ for $x\in[a_1+(a_2-a_1)/3,a_1+2(a_2-a_1)/3]$ and $f(x)=x_2\neq x_1$ for $x\in[a_2+(a_3-a_2)/3,a_2+2(a_3-a_2)/3]$.
If $x_1,x_2$ were any two elements of $B$, then $f$ takes them as values at most once in each interval $[a_i,a_{i+1}]$. Therefore, $f^{-1}(x_i)$ is discrete for each $i=1,2$. Since $a_i\to A$, then $f^{-1}(x_i)$ are sets that accumulate at $A$. Taking a sequence $y_n\in f^{-1}(x_1)$ that tends to $A$, we get that $x_i=f(y_n)\to A$, as $n\to\infty$, for $i=1,2$. Therefore, $x_1=A=x_2$.