Is there an example of continuous map $f:\mathbb R\to \mathbb R^2$ such that there exists $a\in \mathbb R$ such that for all $]y,z[\subset \mathbb R$ , $f([y,z])$ has non empty interior and such that for any $x>a$, $f(]a,x])$ is a neighborhood of $f(a)$.

I think that if $f$ sends a n intervalle on a convexe there exists such an $a$. The existence of such a fonction is an open problem discussed heure : Continuous maps which send intervals of $\mathbb{R}$ to convex subsets of $\mathbb{R}^2$)

(We can ask the same question without asking that for all $]y,z[\subset \mathbb R$ , $f([y,z])$ has non empty interior (is it strictly weaker?) , but my motivation is the problem I just spoke about, where this property holds)

  • I forget to add "up to a fine translation" in the "motivation" part, and there were a typo with forgetting to ask $x>0$ in the question , but is it a good reason to down vote? or did I missed something else? – jcdornano Sep 13 at 19:29
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    note that the Peano curve has this property wrto the square, that is for any $x>0$ the set $f(]0,x[)$ is a nbd of $f(0)=(0,0)$ in $[0,1]^2$. – Pietro Majer Sep 13 at 20:26
  • that's right, Pietro Majer, do you think we can build this way an example wrto the plane? – jcdornano Sep 13 at 20:33
  • I am going to edit the question and remplace $0$ by $a$ without "win" of generality, but, it should be better to communicate and avoid sentences like "up to translation". – jcdornano Sep 13 at 20:37
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    For the Peano curve, it follows from the self-similarity that $f\big(\big]{n\over9^k},{n+1\over9^k} \big[\big)$ is a square of edge ${1\over3^k}$ centered at $f\big({2n+1\over2\cdot 9^k}\big)$, so it fulfills your first request. So maybe another more "curly" self-similar curve could do. – Pietro Majer Sep 14 at 10:23

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