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Does there exist a surjective continuous function $f:[0,1]\to [0,1]^2$ which maps every convex set to a convex set?

Such a function could be considered an especially "regular" sort of space-filling curve. There are of course many well-known examples of continuous surjections $[0,1]\to[0,1]^2$ such as the Peano curve but none of them seem to map convex sets to convex sets.

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    $\begingroup$ Since the convex subsets of $\mathbb R$ are intervals, $f$ should thus map every interval $I \subseteq [0, 1]$ to a convex subset of $[0, 1]^2$. $\endgroup$
    – md2perpe
    Jul 19 '17 at 11:01
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    $\begingroup$ I'm guessing no $\endgroup$ Aug 2 '17 at 14:58
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    $\begingroup$ This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/… $\endgroup$
    – charmd
    Oct 16 '17 at 15:01
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    $\begingroup$ @CharlesMadeline Difficult problem or not, it is a very poor question on math.se. Please see what we expect of a good question: How to aske a good question on math.se. $\endgroup$
    – amWhy
    Jan 7 '19 at 0:32
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    $\begingroup$ I personally welcome your desire to help the question and to be explicit, I don't really have any problem with this question staying open even though it is a PSQ. But it seems that it is not the community consensus. $\endgroup$
    – user170039
    Jan 7 '19 at 4:26
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Partial answer: Here is a discontinuous surjective function $f:[0,1]\rightarrow[0,1]^2$ that maps convex sets to convex sets. We can make sure every interval maps to the entire (convex) set $[0,1]^2$.

Let's say that two real numbers are in the same equivalence class if their unique decimal expansion (with no infinite tail of 9s) differs only in a finite number of digits. For example $0=0.0000...$, $1=1.0000...$, $1/2=0.50000...$ are all in the same equivalence class.

Let $A$ be the set of all equivalence classes. The cardinality of $A$ is the same as that of the reals, which is the same as that of $[0,1]^2$. So there is a surjective function $g:A\rightarrow [0,1]^2$. For each $x\in[0,1]$ let $a(x) \in A$ denote its equivalence class. Define $f:[0,1]\rightarrow[0,1]^2$ by $$ f(x) = g(a(x))$$ Let $I\subseteq[0,1]$ be an interval (containing more than one point). Then $I$ contains points from all equivalence classes in $A$, so $f(I)=[0,1]^2$.

All convex subsets $C \subseteq [0,1]$ that contain at least two distinct points must contain the interval between those points, so $f(C)=[0,1]^2$. And of course all single-point sets map to single-point sets.


Note: I originally overlooked the continuity requirement, as zhw notes below. So I have edited to emphasize that.

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    $\begingroup$ What about the continuity of $f?$ $\endgroup$
    – zhw.
    Dec 31 '18 at 19:39
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    $\begingroup$ I wouldn't delete it; it's a partial solution and now you've made that clear. $\endgroup$
    – zhw.
    Jan 1 '19 at 17:03
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    $\begingroup$ @zhw. : Okay I will keep my answer but I will offer a bounty so that I will (hopefully) not distract attention from this question by giving a (partial) answer. $\endgroup$
    – Michael
    Jan 1 '19 at 21:10
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    $\begingroup$ Please read my answer to your meta post, centering around this question, @Michael. You've offered roughly $\frac 1{24}$th of your total rep to bring attention to a very low quality question, which deserves to be closed, not remain, and further, not fully answered. The asker's account has been deleted. I'm afraid your sincere generosity has been misdirected. Also, this site discourages the answering of questions lacking any form of context, whatsoever. For more information about forms of context an asker can provide, please see your meta post, and my answer to it. $\endgroup$
    – amWhy
    Jan 6 '19 at 23:43
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    $\begingroup$ @amWhy This is a wonderful question. Very intriguing concept, and written straight to the point with no fluff. I want to know the answer too. $\endgroup$
    – Nick Alger
    Jan 7 '19 at 8:32

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