How continuous are functions that map dense sets to dense sets?

Question

Let $f:[0,1] \to [0,1]$ be a function with the property that for every dense $D \subset [0,1]$, $f(D)$ is dense in $f([0,1])$. We can note that $f$ need not be continuous. For instance, consider $f(x)=\left|\sin \ \left(\frac{1}{x-1/2}\right)\right|$ with $f(1/2)=0$.

Is there a nowhere continuous example? If not, how large must the continuity set of such functions be?

As a user with 10k+ rep, why did you post a problem-statement question? — Ѕᴀᴀᴅ, May 13 '20 at 06:26
@Saad You're not expected to provide an essay for every question. Here, I provided a non-continuous example, which is sufficient context. — MathematicsStudent1122, May 13 '20 at 06:31
@Saad I've seen you make this comment several times in recent days on essentially reasonable posts. Consider practicing what you preach (see this and this, which have as much if not less context than my question above does). — MathematicsStudent1122, May 13 '20 at 06:34
Take $f(x)$ to be sampled from uniform random iid variables. — Jair Taylor, May 13 '20 at 07:13
On second thought, my idea above doesn't work. With probability $1$ you can construct a sequence that maps into $[0,1/2)$. — Jair Taylor, May 13 '20 at 08:01

bof · Accepted Answer · 2020-05-13T20:26:59.057

Here is a nowhere continuous example.

Let $U_1,U_2,U_3,\dots$ enumerate all rational intervals contained in $[0,1]$.

For each $n$ choose two distinct points $u_n,v_n\in U_n$.

Let $I_n=\left(\frac1{n+1},\frac1n\right]$ for $n=1,2,3,\dots$.

Define $f:[0,1]\to[0,1]$ as follows:

$$f(x)=u_n\text{ if }x\in I_n\cap\mathbb Q;$$ $$f(x)=v_n\text{ if }x\in I_n\setminus\mathbb Q;$$ $$f(0)=0.$$

Plainly $f$ is nowhere continuous. If $D$ is a dense subset of $[0,1]$, then for each $n$ we have $D\cap I_n\ne\emptyset$, so $f(D)\cap U_n\ne\emptyset$.

How continuous are functions that map dense sets to dense sets?

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