Must compact bijections be continuous?












6












$begingroup$


We say a map $f:X to Y$ is compact if compact sets are mapped to compact sets.




If $f:X to Y$ is a compact bijection, must it be continuous?




The bijection condition is necessary. Otherwise consider an appropriately constructed piecewise constant function $mathbb{R} to mathbb{R}$ as a counterexample.



You might notice I never specified what exactly $X,Y$ are. Let us first consider simply the case where $X=mathbb{R}^n$, $Y=mathbb{R}^m$. If that holds, then work our way up to $X,Y$ being generic metric spaces. If that holds, then work our way up to $X,Y$ being topological spaces [EDIT: it's not true for general topological spaces, see the comments].










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$endgroup$








  • 3




    $begingroup$
    Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
    $endgroup$
    – Randall
    Dec 18 '18 at 0:41












  • $begingroup$
    Related questions: question 1 and question 2
    $endgroup$
    – jgon
    Dec 18 '18 at 0:44








  • 2




    $begingroup$
    @Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 0:46










  • $begingroup$
    Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
    $endgroup$
    – Eric Wofsey
    Dec 18 '18 at 4:02
















6












$begingroup$


We say a map $f:X to Y$ is compact if compact sets are mapped to compact sets.




If $f:X to Y$ is a compact bijection, must it be continuous?




The bijection condition is necessary. Otherwise consider an appropriately constructed piecewise constant function $mathbb{R} to mathbb{R}$ as a counterexample.



You might notice I never specified what exactly $X,Y$ are. Let us first consider simply the case where $X=mathbb{R}^n$, $Y=mathbb{R}^m$. If that holds, then work our way up to $X,Y$ being generic metric spaces. If that holds, then work our way up to $X,Y$ being topological spaces [EDIT: it's not true for general topological spaces, see the comments].










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
    $endgroup$
    – Randall
    Dec 18 '18 at 0:41












  • $begingroup$
    Related questions: question 1 and question 2
    $endgroup$
    – jgon
    Dec 18 '18 at 0:44








  • 2




    $begingroup$
    @Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 0:46










  • $begingroup$
    Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
    $endgroup$
    – Eric Wofsey
    Dec 18 '18 at 4:02














6












6








6


2



$begingroup$


We say a map $f:X to Y$ is compact if compact sets are mapped to compact sets.




If $f:X to Y$ is a compact bijection, must it be continuous?




The bijection condition is necessary. Otherwise consider an appropriately constructed piecewise constant function $mathbb{R} to mathbb{R}$ as a counterexample.



You might notice I never specified what exactly $X,Y$ are. Let us first consider simply the case where $X=mathbb{R}^n$, $Y=mathbb{R}^m$. If that holds, then work our way up to $X,Y$ being generic metric spaces. If that holds, then work our way up to $X,Y$ being topological spaces [EDIT: it's not true for general topological spaces, see the comments].










share|cite|improve this question











$endgroup$




We say a map $f:X to Y$ is compact if compact sets are mapped to compact sets.




If $f:X to Y$ is a compact bijection, must it be continuous?




The bijection condition is necessary. Otherwise consider an appropriately constructed piecewise constant function $mathbb{R} to mathbb{R}$ as a counterexample.



You might notice I never specified what exactly $X,Y$ are. Let us first consider simply the case where $X=mathbb{R}^n$, $Y=mathbb{R}^m$. If that holds, then work our way up to $X,Y$ being generic metric spaces. If that holds, then work our way up to $X,Y$ being topological spaces [EDIT: it's not true for general topological spaces, see the comments].







real-analysis general-topology continuity metric-spaces compactness






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 4:08









Eric Wofsey

190k14216348




190k14216348










asked Dec 18 '18 at 0:34









MathematicsStudent1122MathematicsStudent1122

8,98322668




8,98322668








  • 3




    $begingroup$
    Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
    $endgroup$
    – Randall
    Dec 18 '18 at 0:41












  • $begingroup$
    Related questions: question 1 and question 2
    $endgroup$
    – jgon
    Dec 18 '18 at 0:44








  • 2




    $begingroup$
    @Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 0:46










  • $begingroup$
    Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
    $endgroup$
    – Eric Wofsey
    Dec 18 '18 at 4:02














  • 3




    $begingroup$
    Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
    $endgroup$
    – Randall
    Dec 18 '18 at 0:41












  • $begingroup$
    Related questions: question 1 and question 2
    $endgroup$
    – jgon
    Dec 18 '18 at 0:44








  • 2




    $begingroup$
    @Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 0:46










  • $begingroup$
    Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
    $endgroup$
    – Eric Wofsey
    Dec 18 '18 at 4:02








3




3




$begingroup$
Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
$endgroup$
– Randall
Dec 18 '18 at 0:41






$begingroup$
Generally no. Finite spaces are necessarily compact so all these notions become vacuous. And it’s not true that all bijections between finite spaces are continuous.
$endgroup$
– Randall
Dec 18 '18 at 0:41














$begingroup$
Related questions: question 1 and question 2
$endgroup$
– jgon
Dec 18 '18 at 0:44






$begingroup$
Related questions: question 1 and question 2
$endgroup$
– jgon
Dec 18 '18 at 0:44






2




2




$begingroup$
@Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 0:46




$begingroup$
@Randall Thank you. This rules out the case for general topological spaces, although I believe for metric spaces, the only finite metric spaces are those with a discrete metric, and we certainly have continuity then.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 0:46












$begingroup$
Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
$endgroup$
– Eric Wofsey
Dec 18 '18 at 4:02




$begingroup$
Incidentally, I had a vague feeling that I had answered this question before, but I believe I was instead thinking of this tangentially related question which is another neat result along these lines.
$endgroup$
– Eric Wofsey
Dec 18 '18 at 4:02










1 Answer
1






active

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6












$begingroup$

Yes, this is true for any metric spaces or more generally for any compactly generated Hausdorff spaces. A space $X$ is compactly generated if a subset $Asubseteq X$ is closed iff $Acap K$ is closed in $K$ for all compact subspaces $Ksubseteq X$. Equivalently, $X$ is compactly generated iff every map from $X$ to another space which is continuous when restricted to every compact subset of $X$ is continuous. Note that any metric space is compactly generated, since closedness of sets is detected by convergent sequences and the closure of any convergent sequence is compact.



Here is the precise statement in what seems to be the maximal generality.




Let $X$ be a compactly generated Hausdorff space and let $Y$ be a Hausdorff space. Let $f:Xto Y$ be a compact injection. Then $f$ is continuous.




To show that $f$ is continuous, it suffices to show its restriction to any compact subset of $X$ is continuous. So, let $Ksubseteq X$ be compact. Then the restriction of $f$ to $K$ is a closed map, since $Y$ is Hausdorff. This means that the inverse of the restriction $g:f(K)to K$ is a continuous bijection. Since $f(K)$ is compact and $K$ is Hausdorff, this implies $g$ is a homeomorphism. Thus $f|_K=g^{-1}$ is continuous, as desired.





Here are some quick examples to show each of the hypotheses above are needed.





  • $X$ needs to be compactly generated: If $X$ is the 1-point Lindelöfification of an uncountable discrete space and $Y$ is the same set with the discrete topology, then $X$ and $Y$ are Hausdorff and the identity map $Xto Y$ is compact but not continuous.


  • $X$ needs to be Hausdorff: If $X$ is a 2-point indiscrete space and $Y$ is a 2-point discrete space, then $X$ is compactly generated, $Y$ is Hausdorff, and a bijection $Xto Y$ is compact but not continuous.


  • $Y$ needs to be Hausdorff: If $X=[0,1]$ with the usual topology and $Y=[0,1]$ with the topology that the only nontrivial open set is ${0}$, then $X$ is compactly generated and Hausdorff and the identity map $Xto Y$ is compact but not continuous.






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    $begingroup$

    Yes, this is true for any metric spaces or more generally for any compactly generated Hausdorff spaces. A space $X$ is compactly generated if a subset $Asubseteq X$ is closed iff $Acap K$ is closed in $K$ for all compact subspaces $Ksubseteq X$. Equivalently, $X$ is compactly generated iff every map from $X$ to another space which is continuous when restricted to every compact subset of $X$ is continuous. Note that any metric space is compactly generated, since closedness of sets is detected by convergent sequences and the closure of any convergent sequence is compact.



    Here is the precise statement in what seems to be the maximal generality.




    Let $X$ be a compactly generated Hausdorff space and let $Y$ be a Hausdorff space. Let $f:Xto Y$ be a compact injection. Then $f$ is continuous.




    To show that $f$ is continuous, it suffices to show its restriction to any compact subset of $X$ is continuous. So, let $Ksubseteq X$ be compact. Then the restriction of $f$ to $K$ is a closed map, since $Y$ is Hausdorff. This means that the inverse of the restriction $g:f(K)to K$ is a continuous bijection. Since $f(K)$ is compact and $K$ is Hausdorff, this implies $g$ is a homeomorphism. Thus $f|_K=g^{-1}$ is continuous, as desired.





    Here are some quick examples to show each of the hypotheses above are needed.





    • $X$ needs to be compactly generated: If $X$ is the 1-point Lindelöfification of an uncountable discrete space and $Y$ is the same set with the discrete topology, then $X$ and $Y$ are Hausdorff and the identity map $Xto Y$ is compact but not continuous.


    • $X$ needs to be Hausdorff: If $X$ is a 2-point indiscrete space and $Y$ is a 2-point discrete space, then $X$ is compactly generated, $Y$ is Hausdorff, and a bijection $Xto Y$ is compact but not continuous.


    • $Y$ needs to be Hausdorff: If $X=[0,1]$ with the usual topology and $Y=[0,1]$ with the topology that the only nontrivial open set is ${0}$, then $X$ is compactly generated and Hausdorff and the identity map $Xto Y$ is compact but not continuous.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Yes, this is true for any metric spaces or more generally for any compactly generated Hausdorff spaces. A space $X$ is compactly generated if a subset $Asubseteq X$ is closed iff $Acap K$ is closed in $K$ for all compact subspaces $Ksubseteq X$. Equivalently, $X$ is compactly generated iff every map from $X$ to another space which is continuous when restricted to every compact subset of $X$ is continuous. Note that any metric space is compactly generated, since closedness of sets is detected by convergent sequences and the closure of any convergent sequence is compact.



      Here is the precise statement in what seems to be the maximal generality.




      Let $X$ be a compactly generated Hausdorff space and let $Y$ be a Hausdorff space. Let $f:Xto Y$ be a compact injection. Then $f$ is continuous.




      To show that $f$ is continuous, it suffices to show its restriction to any compact subset of $X$ is continuous. So, let $Ksubseteq X$ be compact. Then the restriction of $f$ to $K$ is a closed map, since $Y$ is Hausdorff. This means that the inverse of the restriction $g:f(K)to K$ is a continuous bijection. Since $f(K)$ is compact and $K$ is Hausdorff, this implies $g$ is a homeomorphism. Thus $f|_K=g^{-1}$ is continuous, as desired.





      Here are some quick examples to show each of the hypotheses above are needed.





      • $X$ needs to be compactly generated: If $X$ is the 1-point Lindelöfification of an uncountable discrete space and $Y$ is the same set with the discrete topology, then $X$ and $Y$ are Hausdorff and the identity map $Xto Y$ is compact but not continuous.


      • $X$ needs to be Hausdorff: If $X$ is a 2-point indiscrete space and $Y$ is a 2-point discrete space, then $X$ is compactly generated, $Y$ is Hausdorff, and a bijection $Xto Y$ is compact but not continuous.


      • $Y$ needs to be Hausdorff: If $X=[0,1]$ with the usual topology and $Y=[0,1]$ with the topology that the only nontrivial open set is ${0}$, then $X$ is compactly generated and Hausdorff and the identity map $Xto Y$ is compact but not continuous.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Yes, this is true for any metric spaces or more generally for any compactly generated Hausdorff spaces. A space $X$ is compactly generated if a subset $Asubseteq X$ is closed iff $Acap K$ is closed in $K$ for all compact subspaces $Ksubseteq X$. Equivalently, $X$ is compactly generated iff every map from $X$ to another space which is continuous when restricted to every compact subset of $X$ is continuous. Note that any metric space is compactly generated, since closedness of sets is detected by convergent sequences and the closure of any convergent sequence is compact.



        Here is the precise statement in what seems to be the maximal generality.




        Let $X$ be a compactly generated Hausdorff space and let $Y$ be a Hausdorff space. Let $f:Xto Y$ be a compact injection. Then $f$ is continuous.




        To show that $f$ is continuous, it suffices to show its restriction to any compact subset of $X$ is continuous. So, let $Ksubseteq X$ be compact. Then the restriction of $f$ to $K$ is a closed map, since $Y$ is Hausdorff. This means that the inverse of the restriction $g:f(K)to K$ is a continuous bijection. Since $f(K)$ is compact and $K$ is Hausdorff, this implies $g$ is a homeomorphism. Thus $f|_K=g^{-1}$ is continuous, as desired.





        Here are some quick examples to show each of the hypotheses above are needed.





        • $X$ needs to be compactly generated: If $X$ is the 1-point Lindelöfification of an uncountable discrete space and $Y$ is the same set with the discrete topology, then $X$ and $Y$ are Hausdorff and the identity map $Xto Y$ is compact but not continuous.


        • $X$ needs to be Hausdorff: If $X$ is a 2-point indiscrete space and $Y$ is a 2-point discrete space, then $X$ is compactly generated, $Y$ is Hausdorff, and a bijection $Xto Y$ is compact but not continuous.


        • $Y$ needs to be Hausdorff: If $X=[0,1]$ with the usual topology and $Y=[0,1]$ with the topology that the only nontrivial open set is ${0}$, then $X$ is compactly generated and Hausdorff and the identity map $Xto Y$ is compact but not continuous.






        share|cite|improve this answer











        $endgroup$



        Yes, this is true for any metric spaces or more generally for any compactly generated Hausdorff spaces. A space $X$ is compactly generated if a subset $Asubseteq X$ is closed iff $Acap K$ is closed in $K$ for all compact subspaces $Ksubseteq X$. Equivalently, $X$ is compactly generated iff every map from $X$ to another space which is continuous when restricted to every compact subset of $X$ is continuous. Note that any metric space is compactly generated, since closedness of sets is detected by convergent sequences and the closure of any convergent sequence is compact.



        Here is the precise statement in what seems to be the maximal generality.




        Let $X$ be a compactly generated Hausdorff space and let $Y$ be a Hausdorff space. Let $f:Xto Y$ be a compact injection. Then $f$ is continuous.




        To show that $f$ is continuous, it suffices to show its restriction to any compact subset of $X$ is continuous. So, let $Ksubseteq X$ be compact. Then the restriction of $f$ to $K$ is a closed map, since $Y$ is Hausdorff. This means that the inverse of the restriction $g:f(K)to K$ is a continuous bijection. Since $f(K)$ is compact and $K$ is Hausdorff, this implies $g$ is a homeomorphism. Thus $f|_K=g^{-1}$ is continuous, as desired.





        Here are some quick examples to show each of the hypotheses above are needed.





        • $X$ needs to be compactly generated: If $X$ is the 1-point Lindelöfification of an uncountable discrete space and $Y$ is the same set with the discrete topology, then $X$ and $Y$ are Hausdorff and the identity map $Xto Y$ is compact but not continuous.


        • $X$ needs to be Hausdorff: If $X$ is a 2-point indiscrete space and $Y$ is a 2-point discrete space, then $X$ is compactly generated, $Y$ is Hausdorff, and a bijection $Xto Y$ is compact but not continuous.


        • $Y$ needs to be Hausdorff: If $X=[0,1]$ with the usual topology and $Y=[0,1]$ with the topology that the only nontrivial open set is ${0}$, then $X$ is compactly generated and Hausdorff and the identity map $Xto Y$ is compact but not continuous.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 4:14

























        answered Dec 18 '18 at 3:56









        Eric WofseyEric Wofsey

        190k14216348




        190k14216348






























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