Is every continious image of a non-compact space is non-compact?











up vote
2
down vote

favorite












Is every continuous image of a non-compact space is non-compact?



I was thinking about constant function. I think it will be false.



Am I right?






general-topology compactness







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  • 3




    If the function is constant, then the image is a singleton, which is definetely compact.
    – José Carlos Santos
    Nov 22 at 13:10










  • okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
    – jasmine
    Nov 22 at 13:17






  • 1




    Yes, the statement is false, for the reason that you gave.
    – José Carlos Santos
    Nov 22 at 13:23















up vote
2
down vote

favorite












Is every continuous image of a non-compact space is non-compact?



I was thinking about constant function. I think it will be false.



Am I right?






general-topology compactness







share|cite|improve this question




















  • 3




    If the function is constant, then the image is a singleton, which is definetely compact.
    – José Carlos Santos
    Nov 22 at 13:10










  • okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
    – jasmine
    Nov 22 at 13:17






  • 1




    Yes, the statement is false, for the reason that you gave.
    – José Carlos Santos
    Nov 22 at 13:23













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Is every continuous image of a non-compact space is non-compact?



I was thinking about constant function. I think it will be false.



Am I right?






general-topology compactness







share|cite|improve this question















Is every continuous image of a non-compact space is non-compact?



I was thinking about constant function. I think it will be false.



Am I right?






general-topology compactness




general-topology compactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 13:29









Martin Sleziak

44.6k7115269




44.6k7115269










asked Nov 22 at 13:07









jasmine

1,505416




1,505416








  • 3




    If the function is constant, then the image is a singleton, which is definetely compact.
    – José Carlos Santos
    Nov 22 at 13:10










  • okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
    – jasmine
    Nov 22 at 13:17






  • 1




    Yes, the statement is false, for the reason that you gave.
    – José Carlos Santos
    Nov 22 at 13:23














  • 3




    If the function is constant, then the image is a singleton, which is definetely compact.
    – José Carlos Santos
    Nov 22 at 13:10










  • okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
    – jasmine
    Nov 22 at 13:17






  • 1




    Yes, the statement is false, for the reason that you gave.
    – José Carlos Santos
    Nov 22 at 13:23








3




3




If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 at 13:10




If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 at 13:10












okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 at 13:17




okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 at 13:17




1




1




Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 at 13:23




Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 at 13:23










1 Answer
1






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accepted










Let $f: [0, infty) to mathbb R$ be defined by



$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.



Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.






share|cite|improve this answer





















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    up vote
    2
    down vote



    accepted










    Let $f: [0, infty) to mathbb R$ be defined by



    $f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.



    Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Let $f: [0, infty) to mathbb R$ be defined by



      $f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.



      Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $f: [0, infty) to mathbb R$ be defined by



        $f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.



        Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.






        share|cite|improve this answer












        Let $f: [0, infty) to mathbb R$ be defined by



        $f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.



        Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 13:20









        Fred

        43.6k1644




        43.6k1644






























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