Vtyjkyuk

$DeclareMathOperatorimImnewcommandpme(-epsilon,epsilon)newcommandpmec[-epsilon,epsilon]$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:

• $fcolon[0,infty)tomathbb R$


• $f$ converges zo zero, i.e. $forall epsilon>0exists xin
  [0,infty): f[x,infty)subseteq (-epsilon,epsilon)$


• $f$ 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
$x^-1sin x$, 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 $[0,1]tomathbb R$.

Weird, complicated attempt 1

The first observation is that for every $yin im fsetminus0$,
convergence guarantees $f^-1$ to be bounded, and thus compact. By
being countably infinite, it has to have one accumulation point.
Let's call $xin mathbb R$ “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 $fvert_mathcal H(f)$.

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 $fcolon [0,1]to mathbb R$ 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:

beginalign*
&f_C,wcolon [C,C+w]tomathbb R,\
&f_C,w(x):=
begincases
C+left(x-Cright) sin^2left(frac2pi wx-Cright),
&xin [C,C+w/2] \
-frac4Cwcdotleft(x-C-fracw2right)+C,
&xin [C+w/2,C+3w/4] \
frac4left(C+wright)wcdotleft(x-C-frac3w4right) ,
&xin [C+3w/4,C+w] \
endcases
endalign*

It is a well-defined, continuous function taking the values

beginalign*
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.
endalign*

Notably, a subset of $f_C,w^-1(C)$ is
$Ccup leftC+frac2wn mid ninmathbb N_geq 4right$,
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

$$
mathcal C_epsilon := lefthat xmid f((hat x,infty))subseteq
pmeright,
$$

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

$$
mathcal C_epsilon^n := left f^-1(y)cap (hat x, infty) right
$$

This just selects all points $hat x$ for which
$fvert_(hat x,infty)$ 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?