I wonder whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic? for every two real numbers $\alpha,\beta,\alpha\lt\beta$, $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$

I can't say such a function does not exist, neither can I construct a example

Thanks a lot!


The Conway base 13 function is one such function. From Wikipedia:

$f$ takes as its value every real number somewhere within every open interval $(a,b)$.

The construction of the function is a little bit complicated. Refer to the wiki page for details.

  • $\begingroup$ I am really grateful for your prompt help! $\endgroup$ – Clin May 11 '14 at 22:41
  • $\begingroup$ @Clin Happy to help. $\endgroup$ – Ayman Hourieh May 11 '14 at 22:41

Let $K$ be the Cantor set. For any $a, b\in \mathbb R,$ $a\ne 0,$ I'll also call $aK+b$ a Cantor set. Since $K$ has the cardinality of $\mathbb R,$ so do these other Cantor sets.

Note that if $U$ is any nonempty open subset of $\mathbb R,$ there exists a Cantor set inside of $U.$

Let $I_1,I_2, \dots$ be the open intervals in $\mathbb R$ with rational end points.

Claim: There are pairwise disjoint Cantor sets $K_1,K_2, \dots$ such that for each $n,$ $K_n \subset I_n.$

The claim is a nice exercise. Let's assume the result. For each $n$ there exists a bijection $f_n:K_n \to \mathbb R.$ Define $f:\mathbb R \to \mathbb R$ as follows. For each $n,$ set $f= f_n$ on $K_n.$ Set $f=0$ everywhere else.

Then $f$ has the desired property. Proof: Suppose $I$ is a nonempty open interval. Then $I_n\subset I$ for some $n.$ Since $K_n \subset I_n,$ we have $\mathbb R = f(K_n) \subset f(I_n) \subset f(I),$ and we're done.


My answer to this question is precisely such a function, constructed in an easy way:

There is in fact a rather easy example of a function $\mathbb R \to \mathbb R$ such that the image of every open set is $\mathbb R$: Let $(x_i)_{i\in\mathbb Z_+}$ be the binary decimal expansion of $x$, so that each $x_i\in\{0,1\}$. Let then $$f(x)=\sum_{k=1}^∞\frac{(−1)^{x_k}}k\quad \textrm{if the series converges}$$ $$f(x)=0\quad\textrm{otherwise.}$$ Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $f((a,b))=\mathbb R$ for any real $a,b$.

The harmonic series can be substituted with any other unbounded series where the summand goes to zero.


I find it easier to define one from $(0,1)$ to $(0,1)$. You can use your favorite bijection to stretch each axis to $\Bbb R$. Express $x \in (0,1)$ in base $3$, choosing the expansion ending in all $2$'s for numbers which would terminate. If $x$ has an infinite number of $2$'s in the expansion, set $f(x)=x$ and ignore it in the rest. If $x$ has a finite number of $2$'s in the expansion, multiply it by $3^k$ so that the last two is just to the left of the ternary point. The expansion is now all $0$'s and $1$'s. Read it as a binary number and return that value. To show it takes all values in an interval, let somebody give you a value $y$ it should take. Express $y$ in binary. Find an interval $(\frac {3m+2}{3^k},\frac {3m+3}{3^k})$ that is within the given interval for some naturals $m,k$. Express $\frac {3m+2}{3^k}$ in ternary and append the binary of $y$ on the right. That will give you an $x$ such that $f(x)=y$ This needs a slight patch because some numbers are mapped to $1$, which is outside the range, so map them to $0.5$ and we are done.


A function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for each $a<b$, $f[(a,b)]= \mathbb{R}$ is known as a strongly Darboux function. It can be constructed easily by transfinite induction (or more accurately, by the Recursion Theomrem).

You may consult Section 7.2 of the book "Set theory for the working mathematician" written by Krzysztof Ciesielski.


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