Proving Existence of Divergent Sequence Using Connectedness of $\mathbb{R}$ I am currently attempting to prove the following statement:
"Let $f:\mathbb{R}\to\mathbb{R}$ be continuous. Suppose there exists sequences $(a_{n})$ and $(b_{n})$ diverging to $\infty$ such that $f(a_{n})\to a$ and $f(b_{n})\to b$. Prove that there exists a sequence $(c_{j})$ such that $c_{j}\to\infty$ and $f(c_{j}) = c$ for all $j$, if $a<c<b$."
So far I have reasoned heuristically that since $\mathbb{R}$ is connected and the continuous image of a connected set is connected, $\lim f(a_{n})$ and $\lim f(b_{n})$ are in a connected subset of the image of $f$. Then, using the Intermediate Value Theorem it seems like there should be the possibility of deducing that there is some $(c_{j}) \in [(a_{n}),(b_{n})]$ with $\lim f(c_{j}) = c$, but I am not sure if this is a sound way of reasoning to the result. In particular, $(c_{j}) \in [(a_{n}),(b_{n})]$ doesn't make much sense to me as written, but I am having trouble conceiving of a different way of expressing what is going on. The difficulty I am presented with in my head is how one can have $\infty \in [\infty,\infty]$ while avoiding possibly circular reasoning? I thought that we could possibly think of a sequence of intervals instead, but I am unsure if this is appropriate for the given problem.
Overall, I feel as though I am having the most difficulty with picturing the geometry of the problem. It was recommended to me to think of $\sin x$ when trying to picture what such a function would look like concretely, but at the moment I am having trouble understanding how this would relate to the given function. For instance, I image $\sin x$ to be an oscillating function which does not converge, but I cannot think of how $\lim f(\sin x)$ could be constant valued given that the sequence itself does not converge.
Can someone please advise if this is an appropriate example for the above problem, and if my reasoning stated above is close to being on the right track? If not, can someone please provide feedback on where my reasoning is going wrong. This is a homework problem, so please do not give a full proof, I am trying to understand for myself but feel that my intuition is lacking. Any tips and feedback is greatly appreciated.
 A: First, le us see a particular case.
Consider $f(x)=\sin x$ as you have mentioned. Consider $a_n=n\pi$ and $b_n=\frac{\pi}{2}+2n\pi$. Then $a_n\to\infty , \; b_n\to\infty , \; f(a_n)=0$ and $f(b_n)=1$, $\forall n\in \mathbb{N}$.
Let $c_n=\frac{\pi}{4}+2n\pi$. Then $c_n\to\infty$ and $0<f(c_n)=\frac{\sqrt{2}}{2}<1$.
Let see now the general case.
Suppose $a_n,\; b_n$ and $f(x)$ are as given. WLOG let $a_n<b_n,\; \forall n\in \mathbb{N}$. Since $f(x)$ is continuous, we notice that $f([a_n,b_n])= [a,b]$ and this holds $\forall n\in\mathbb{N}$. Hence, we can construct a sequence $(c_n)$ such that $c_n\in(a_n,b_n)$ and $f(c_n)=c$, $\forall n\in\mathbb{N}$, where $c\in(a,b)$. Since $a_n\to\infty,\; b_n\to\infty$, from the construction of $c_n$ we also have $c_n\to \infty$.
A: The implicit assumption is that $a < b$. Then consider $c \in (a,b)$. Let $r = \min(c-a, b-c)$. There exists $N$ such that $f(a_n) \in (a-r,a+r)$ and $f(b_n) \in (b-r,b+r)$ for $n \ge N$. This implies $a_n \ne b_n$ and $c \in (f(a_n),f(b_n))$ for $n \ge N$. The IVT tells us that there exists $c_n$ between $a_n$ and $b_n$ such that $f(c_n) = c$. Since both $a_n \to \infty$ and $b_n \to \infty$ we see that $c_n \to \infty$.
