$\DeclareMathOperator{\im}{Im}\newcommand{\pme}{(-\epsilon,\epsilon)}\newcommand{\pmec}{[-\epsilon,\epsilon]}$This is oddly specific, but in an attempt to analyze a special case of this unanswered question, this came about.
I am interested in whether a continuous function with the following properties exists:
- $f\colon[0,\infty)\to\mathbb R$
- $f$ converges zo zero, i.e. $\forall \epsilon>0\exists x\in [0,\infty): f[x,\infty)\subseteq (-\epsilon,\epsilon)$
- $f$ has „countably infinite multiplicity in a neighborhood of 0“, by which I mean that for for every $y$ in the image and some neighborhood of 0, $f^{-1}(y)\simeq \mathbb N$
Intuition
My intuition says „no“, because the only function I can imagine that has this multiplicity property is e.g. $\sin x$, but that „diverges“ too heavily. If we fix that by „pressing it down“, like in $x^{-1}\sin x$, the only element of the image with countably infinite multiplicity is $y=0$.
Note that it is vital to include $0$ in the domain, because else we could use glue the left part of the topologists sine curve $\sin (x^{-1})$ to the right part of $\sin x/x$, and get a desired function.
To be perfectly honest, the convergence to the right and the inclusion of the $0$ to the left are more or less the same conditions: By using a homeomorphism $[0,\infty)\simeq [0,1)$, the existence of such a function is (by the convergence condition) equivalent to the ability to add a value at $1$ to extend it to a function $[0,1]\to\mathbb R$.
Weird, complicated attempt 1
The first observation is that for every $y\in \im f\setminus\{0\}$, convergence guarantees $f^{-1}$ to be bounded, and thus compact. By being countably infinite, it has to have one accumulation point. Let's call $x\in \mathbb R$ “horizontal accumulation point” if it is an accumulation point of $f^{-1}(f(x))$ and refer to the set of these as $\mathcal H(f)$. Note that we have to have uncountably many of them, because $\im f$ must be connected and cannot be just 0 (else $f^{-1}(0)$ would be uncountable).
Looking at how they lie in the graph of $f$, by considering some still uncountable subsets who are in a bounded subset in the graph (e.g. restricting attention to some $y>\epsilon$), we can conclude that there must be uncountably many accumulation points in the graph of $f\vert_{\mathcal H(f)}$.
I then hoped to arrive at some contradiction like that countability would require them to be isolated, but the only thing I came up with was the following IMO quite interesting example:
(You may also just look at https://www.desmos.com/calculator/kq0cq1irmc)
Construct $f\colon [0,1]\to \mathbb R$ by partitioning $[0,1]=\bigcup_{n=1}^\infty [2^{-n},2^{-n+1})$ on which we consider $f_{2^{-n},2^{-n}}$ defined as follows:
\begin{align*} &f_{C,w}\colon [C,C+w]\to\mathbb R,\\ &f_{C,w}(x):= \begin{cases} C+\left(x-C\right)\ \sin^2\left(\frac{2\pi w}{x-C}\right), &x\in [C,C+w/2] \\ -\frac{4C}{w}\cdot\left(x-C-\frac{w}{2}\right)+C, &x\in [C+w/2,C+3w/4] \\ \frac{4\left(C+w\right)}{w}\cdot\left(x-C-\frac{3w}{4}\right)\ , &x\in [C+3w/4,C+w] \\ \end{cases} \end{align*}
It is a well-defined, continuous function taking the values
\begin{align*} f_{C,w}(C) = f_{C,w}(C+w/2) &= C\\ f_{C,w}(C+3w/4) &= 0\\ f_{C,w}(C+w) &= C+w. \end{align*}
Notably, a subset of $f_{C,w}^{-1}(C)$ is $\{C\}\cup \left\{C+\frac{2w}{n} \mid n\in\mathbb N_{\geq 4}\right\}$, making $C$ a horizontal limit point.
Although this is not countably infinite in every horizontal fibre, it is an example of a function with a non-isolated horizontal accumulation point (0) where the value occurs countable infinitely often.
Weird, complicated attempt 2
I tried to weaken what is required by convergence to see what happens „before“ everything lands in the $\epsilon$-Strip. Convergence in this context means the nonemptiness of
$$ \mathcal C_\epsilon := \left\{\hat x\mid f((\hat x,\infty))\subseteq \pme\right\}, $$
which is just the region where the $\epsilon$-convergence-criterion holds. It is obviously upper directed. For all values excluded by this strip, namely $\im f \setminus \pme$, the countable cardinality must have its cause somewhere in $[0, \mathcal C_\epsilon]$. This suggests considering
$$ \mathcal C_\epsilon^n := \left\{ \hat x\bigm| \forall y\in \mathbb R\setminus \pme:\: \left| f^{-1}(y)\cap (\hat x, \infty) \right| \leq n \right\} $$
This just selects all points $\hat x$ for which $f\vert_{(\hat x,\infty)}$ maps everything into $\pme$ with the exception that each image point not in said interval may be reached at most $n$ times. We can easily observe $\mathcal C_\epsilon = \mathcal C_\epsilon^0$, and that $\mathcal C_\epsilon^n$ grows when making n larger.
However, I'm not sure what to look at after this, and especially, I'm not sure how the demand for $0$ to be in the domain may come into play.
Is there any (comparatively) simple, straightforward way to prove this?
