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Consider the following simple example of a function $f: \mathbb{R}\to\mathbb{R}$ which is open and discontinuous at all points. If $x\in\mathbb{R}$ is represented as something.$x_1x_2x_3\dots$ in the binary system, then set $$f(x)=\lim_{n\to\infty}\frac{x_1+\cdots+x_n}{n}$$ if the limit exists and belongs to $(0,1)$, and set $f(x)$ to (say) $\frac{1}{2}$ otherwise.

Is this example known (I suppose it is), and what's the reference for it?

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2 Answers 2

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I saw it as an exercise (E1.4.2) in the book "Analysis Now" by "Gert. K. Pedersen" with slight difference; $f$ is defined by $$f:\mathbb R\rightarrow\mathbb R,\quad x\mapsto\limsup\frac1n{\sum_{k=1}^nx_n}$$ where $x-\lfloor x\rfloor=0.x_1x_2\dots$ is the binary expansion of the fractional part of $x$.

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  • $\begingroup$ $f$ should be valued in $[0,1]$ if you want it to be open. To have it open while valued in $\mathbf{R}$, change the values $0$, $1$ to $1/2$. $\endgroup$
    – YCor
    May 29, 2018 at 16:33
  • $\begingroup$ For what it's worth, this is often called the Cesàro-Vietoris function (see my various comments here also). $\endgroup$ May 29, 2018 at 18:30
  • $\begingroup$ @YCor Of course to get an open map, it is hinted in the mentioned exercise to consider $h\circ f$, where $h$ is any surjective continuous functions from $[0,1]$ on $\mathbb R$. $\endgroup$ May 29, 2018 at 19:24
  • $\begingroup$ @DaveLRenfro, see your "here" for the Wacław Sierpiński's example. $\endgroup$
    – Wlod AA
    May 30, 2018 at 6:52
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The following open question was popular in some places:

Q:   does there exist $\ x\in(0;1)\ $ such that $\,\ s(x)=x,\,\ $ where

$$ s(x)\ :=\ \lim_{n=\infty} \frac{\sum_{k=1}^n x_k}n $$

and $\ x_n\ $ are binary digits $\ 0\ $ or $\ 1,\ $ and

$$ x\ =\ \sum_{n=1}^\infty\frac{x_n}{2^n} $$

Thus, in 1959/60 I've formulated and proved a more general theorem, and it was recognized in the spring of 1961, namely:

Theorem   For every function $\ f: (0;1)\rightarrow (0;1)\ $ which is a pointwise limit of a sequence of continuous functions, and for every non-empty $\ (a;b)\subseteq (0;1),\ $ the equation

$$ s(x)\ =\ f(x) $$

has $\ 2^{\aleph_0} $ of different solutions $\ x\in (a;b).$

Remark   there is something of a paradox due to the juxtaposition of the equation $\ s(x)=f(x)\ $ and the classical Jacques Hadamard equation $\ s(x)=\frac 12,\ $ which holds for almost all $\ x\in[0;1].$

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