Does there exist a space filling curve which sends every convex set to a convex set ?












30












$begingroup$


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






general-topology metric-spaces continuity convex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 2




    $begingroup$
    I'm guessing no
    $endgroup$
    – Akiva Weinberger
    Aug 2 '17 at 14:58






  • 9




    $begingroup$
    This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
    $endgroup$
    – Charles Madeline
    Oct 16 '17 at 15:01






  • 5




    $begingroup$
    @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 at 0:32






  • 2




    $begingroup$
    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.
    $endgroup$
    – user 170039
    Jan 7 at 4:26
















30












$begingroup$


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






general-topology metric-spaces continuity convex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 2




    $begingroup$
    I'm guessing no
    $endgroup$
    – Akiva Weinberger
    Aug 2 '17 at 14:58






  • 9




    $begingroup$
    This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
    $endgroup$
    – Charles Madeline
    Oct 16 '17 at 15:01






  • 5




    $begingroup$
    @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 at 0:32






  • 2




    $begingroup$
    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.
    $endgroup$
    – user 170039
    Jan 7 at 4:26














30












30








30


19



$begingroup$


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






general-topology metric-spaces continuity convex-analysis







share|cite|improve this question











$endgroup$




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






general-topology metric-spaces continuity convex-analysis




general-topology metric-spaces continuity convex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 20 '17 at 1:21









Alex Ravsky

42.4k32383




42.4k32383










asked Jul 18 '17 at 18:21







user456828















  • 1




    $begingroup$
    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






  • 2




    $begingroup$
    I'm guessing no
    $endgroup$
    – Akiva Weinberger
    Aug 2 '17 at 14:58






  • 9




    $begingroup$
    This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
    $endgroup$
    – Charles Madeline
    Oct 16 '17 at 15:01






  • 5




    $begingroup$
    @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 at 0:32






  • 2




    $begingroup$
    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.
    $endgroup$
    – user 170039
    Jan 7 at 4:26














  • 1




    $begingroup$
    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






  • 2




    $begingroup$
    I'm guessing no
    $endgroup$
    – Akiva Weinberger
    Aug 2 '17 at 14:58






  • 9




    $begingroup$
    This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
    $endgroup$
    – Charles Madeline
    Oct 16 '17 at 15:01






  • 5




    $begingroup$
    @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 at 0:32






  • 2




    $begingroup$
    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.
    $endgroup$
    – user 170039
    Jan 7 at 4:26








1




1




$begingroup$
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




$begingroup$
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




2




2




$begingroup$
I'm guessing no
$endgroup$
– Akiva Weinberger
Aug 2 '17 at 14:58




$begingroup$
I'm guessing no
$endgroup$
– Akiva Weinberger
Aug 2 '17 at 14:58




9




9




$begingroup$
This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
$endgroup$
– Charles Madeline
Oct 16 '17 at 15:01




$begingroup$
This is a difficult problem which was briefly discussed here : mathoverflow.net/questions/200535/…
$endgroup$
– Charles Madeline
Oct 16 '17 at 15:01




5




5




$begingroup$
@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 at 0:32




$begingroup$
@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 at 0:32




2




2




$begingroup$
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.
$endgroup$
– user 170039
Jan 7 at 4:26




$begingroup$
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.
$endgroup$
– user 170039
Jan 7 at 4:26










1 Answer
1






active

oldest

votes


















7












$begingroup$

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 $g:Arightarrow [0,1]^2$. For each $xin[0,1]$ let $a(x) in A$ denote its equivalence class. Define $f:[0,1]rightarrow[0,1]^2$ by
$$ f(x) = g(a(x))$$
Let $Isubseteq[0,1]$ 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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    What about the continuity of $f?$
    $endgroup$
    – zhw.
    Dec 31 '18 at 19:39






  • 6




    $begingroup$
    I wouldn't delete it; it's a partial solution and now you've made that clear.
    $endgroup$
    – zhw.
    Jan 1 at 17:03








  • 1




    $begingroup$
    @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 at 21:10






  • 4




    $begingroup$
    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 at 23:43








  • 7




    $begingroup$
    @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 at 8:32











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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 $g:Arightarrow [0,1]^2$. For each $xin[0,1]$ let $a(x) in A$ denote its equivalence class. Define $f:[0,1]rightarrow[0,1]^2$ by
$$ f(x) = g(a(x))$$
Let $Isubseteq[0,1]$ 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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    What about the continuity of $f?$
    $endgroup$
    – zhw.
    Dec 31 '18 at 19:39






  • 6




    $begingroup$
    I wouldn't delete it; it's a partial solution and now you've made that clear.
    $endgroup$
    – zhw.
    Jan 1 at 17:03








  • 1




    $begingroup$
    @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 at 21:10






  • 4




    $begingroup$
    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 at 23:43








  • 7




    $begingroup$
    @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 at 8:32
















7












$begingroup$

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 $g:Arightarrow [0,1]^2$. For each $xin[0,1]$ let $a(x) in A$ denote its equivalence class. Define $f:[0,1]rightarrow[0,1]^2$ by
$$ f(x) = g(a(x))$$
Let $Isubseteq[0,1]$ 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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    What about the continuity of $f?$
    $endgroup$
    – zhw.
    Dec 31 '18 at 19:39






  • 6




    $begingroup$
    I wouldn't delete it; it's a partial solution and now you've made that clear.
    $endgroup$
    – zhw.
    Jan 1 at 17:03








  • 1




    $begingroup$
    @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 at 21:10






  • 4




    $begingroup$
    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 at 23:43








  • 7




    $begingroup$
    @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 at 8:32














7












7








7





$begingroup$

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 $g:Arightarrow [0,1]^2$. For each $xin[0,1]$ let $a(x) in A$ denote its equivalence class. Define $f:[0,1]rightarrow[0,1]^2$ by
$$ f(x) = g(a(x))$$
Let $Isubseteq[0,1]$ 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.






share|cite|improve this answer











$endgroup$



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 $g:Arightarrow [0,1]^2$. For each $xin[0,1]$ let $a(x) in A$ denote its equivalence class. Define $f:[0,1]rightarrow[0,1]^2$ by
$$ f(x) = g(a(x))$$
Let $Isubseteq[0,1]$ 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 21:19

























answered Dec 31 '18 at 19:20









MichaelMichael

12.9k11429




12.9k11429








  • 3




    $begingroup$
    What about the continuity of $f?$
    $endgroup$
    – zhw.
    Dec 31 '18 at 19:39






  • 6




    $begingroup$
    I wouldn't delete it; it's a partial solution and now you've made that clear.
    $endgroup$
    – zhw.
    Jan 1 at 17:03








  • 1




    $begingroup$
    @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 at 21:10






  • 4




    $begingroup$
    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 at 23:43








  • 7




    $begingroup$
    @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 at 8:32














  • 3




    $begingroup$
    What about the continuity of $f?$
    $endgroup$
    – zhw.
    Dec 31 '18 at 19:39






  • 6




    $begingroup$
    I wouldn't delete it; it's a partial solution and now you've made that clear.
    $endgroup$
    – zhw.
    Jan 1 at 17:03








  • 1




    $begingroup$
    @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 at 21:10






  • 4




    $begingroup$
    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 at 23:43








  • 7




    $begingroup$
    @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 at 8:32








3




3




$begingroup$
What about the continuity of $f?$
$endgroup$
– zhw.
Dec 31 '18 at 19:39




$begingroup$
What about the continuity of $f?$
$endgroup$
– zhw.
Dec 31 '18 at 19:39




6




6




$begingroup$
I wouldn't delete it; it's a partial solution and now you've made that clear.
$endgroup$
– zhw.
Jan 1 at 17:03






$begingroup$
I wouldn't delete it; it's a partial solution and now you've made that clear.
$endgroup$
– zhw.
Jan 1 at 17:03






1




1




$begingroup$
@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 at 21:10




$begingroup$
@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 at 21:10




4




4




$begingroup$
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 at 23:43






$begingroup$
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 at 23:43






7




7




$begingroup$
@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 at 8:32




$begingroup$
@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 at 8:32


















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