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$\underline{\text{Motivation}:-}$

I started thinking about this question a few days ago while checking on my old questions. I thought about a similar question to this one, but instead of roots this time I want to know about singularities.

$\underline{\text{Details}:-}$

Let, $\mathcal{S}(f)$ be the set of all singularities of $f$ where $f$ is a real valued function.

For example,

$\mathcal{S}(\frac{1}{x})=\{0\}$

$\mathcal{S}(\Gamma(x))=\mathbb{Z}_{<1}$

And so on.

I was wondering whether there exists a real function for which $\mathcal{S}(f)$ is dense.

Maybe a function which have singularities at the rationals.

An easy answer to my question would be $\frac{1}{\sin(\frac{1}{x})}$, since $\sin(\frac{1}{x})$ has dense, disconnected roots.

$\underline{\text{My questions}:-}$

We can write a function $f$ with dense singularities as $\frac{1}{g}$, where $g$ is a function with dense roots, so does there exist a function $g$ which has the following properties-

●The roots of $g$ are dense and uncountable in an interval $[a,b]$.

●$g$ has no singularities in $[a,b]$.

●$g$ is continuous in $[a,b]$.

If there does exist such a function, I would like to have an example and the graph of $g$ and $f$ both.

(I just can't imagine how cool the graph of such a function would look like!)

Rounak Sarkar
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  • how is this different than $1\over f(x)$ where $f(x)$ is your dense root function? – user619894 Jul 19 '21 at 14:00
  • @user619894. I predicted that someone would comment this. But I unfortunately don't know about functions whose roots are dense but disconnected – Rounak Sarkar Jul 19 '21 at 14:19
  • $sin({1\over x})$ near zero? – user619894 Jul 19 '21 at 14:43
  • @user619894. How about a function whose roots are dense, disconnected and uncountable? – Rounak Sarkar Jul 19 '21 at 15:04
  • Do you require $f$ to be continuous in any way? – G. Chiusole Jul 19 '21 at 15:17
  • @G. Chiusole. I legit don't know how to apply the concept of continuousness to $f$. If there does exist a function $f$ such that $\mathcal{S}(f)$ is dense, disconnected and uncountable, then it would behave very weirdly. – Rounak Sarkar Jul 19 '21 at 16:11
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    https://en.m.wikipedia.org/wiki/Dirichlet_function – user619894 Jul 19 '21 at 17:15
  • Let me just change the question a little bit. – Rounak Sarkar Jul 20 '21 at 07:30
  • Your continuity requirement prevents many types of densely occurring behavior, but moderately relaxing continuity a bit (relative to the kinds of pathological behavior people mean when the full Axiom of Choice must be invoked), one can have a function $f$ such that given any open interval $(a,b),$ every real number is an output value for some number in this interval. This implies, among other things, that for each real number $r$ (e.g. $r=0$ gives the zeros of $f),$ the function equals $r$ for a dense set of real number inputs; (continued) – Dave L. Renfro Jul 22 '21 at 06:03
  • also, the graph of $f$ is dense in the coordinate plane. Such functions are now called everywhere surjective, although earlier than about 15 or 20 years ago there was no standard term for them and in the literature they were either unnamed or given various names by different authors (e.g. strongly Darboux function is one such name). – Dave L. Renfro Jul 22 '21 at 06:07
  • With continuity, you can't have roots dense in any open interval without being the zero function there. However, this answer has nice graphs of Volterra's function, which has uncountably many roots in $[0,1]$ (not only this, but in infinitely many different places in $[0,1])$ and has no singularities of the type you want to avoid (because all values of the function are between $-1$ and $1)$ and is differentiable at each point of $[0,1]$ (a much stronger requirement than being continuous). – Dave L. Renfro Jul 23 '21 at 08:09

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