$( mathscr{X}, d_x)$ and $( mathscr{Y}, d_y)$ metric spaces ..Proof Verification












0












$begingroup$


Let
$( mathscr{X}, d_x)$ and $( mathscr{Y}, d_y)$ be two metric spaces and



$ D subset mathscr{X} $ a compact set.



To show is, that if



$f: D rightarrow mathscr{Y} $is continous and injective, then



$ f^{-1} :f(D) rightarrow D $



is also continous.



My Idea:



Since $D$ is compact, it is also closed.
$$ (f^{-1})^{-1} (D)= f(D) $$
And since $f$ is continous and $D$ compact, is also $f(D)$ compact.
As Subset of the metric space $mathscr{Y} $ is $f(D)$ also closed.
So the inverse image of a closed image is also closed, therefore is $f^{-1}$ is also continous.



Does that makes sense to you? :)
but where does the injectivity comes in?
Any help is appreciated !






real-analysis general-topology proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by “completed”? Do you mean “closed”?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:21










  • $begingroup$
    yes, I edited it
    $endgroup$
    – constant94
    Dec 5 '18 at 18:39










  • $begingroup$
    Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:40












  • $begingroup$
    @JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
    $endgroup$
    – Federico
    Dec 5 '18 at 18:41












  • $begingroup$
    @Federico, how do I do that?
    $endgroup$
    – constant94
    Dec 5 '18 at 18:43
















0












$begingroup$


Let
$( mathscr{X}, d_x)$ and $( mathscr{Y}, d_y)$ be two metric spaces and



$ D subset mathscr{X} $ a compact set.



To show is, that if



$f: D rightarrow mathscr{Y} $is continous and injective, then



$ f^{-1} :f(D) rightarrow D $



is also continous.



My Idea:



Since $D$ is compact, it is also closed.
$$ (f^{-1})^{-1} (D)= f(D) $$
And since $f$ is continous and $D$ compact, is also $f(D)$ compact.
As Subset of the metric space $mathscr{Y} $ is $f(D)$ also closed.
So the inverse image of a closed image is also closed, therefore is $f^{-1}$ is also continous.



Does that makes sense to you? :)
but where does the injectivity comes in?
Any help is appreciated !






real-analysis general-topology proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by “completed”? Do you mean “closed”?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:21










  • $begingroup$
    yes, I edited it
    $endgroup$
    – constant94
    Dec 5 '18 at 18:39










  • $begingroup$
    Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:40












  • $begingroup$
    @JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
    $endgroup$
    – Federico
    Dec 5 '18 at 18:41












  • $begingroup$
    @Federico, how do I do that?
    $endgroup$
    – constant94
    Dec 5 '18 at 18:43














0












0








0





$begingroup$


Let
$( mathscr{X}, d_x)$ and $( mathscr{Y}, d_y)$ be two metric spaces and



$ D subset mathscr{X} $ a compact set.



To show is, that if



$f: D rightarrow mathscr{Y} $is continous and injective, then



$ f^{-1} :f(D) rightarrow D $



is also continous.



My Idea:



Since $D$ is compact, it is also closed.
$$ (f^{-1})^{-1} (D)= f(D) $$
And since $f$ is continous and $D$ compact, is also $f(D)$ compact.
As Subset of the metric space $mathscr{Y} $ is $f(D)$ also closed.
So the inverse image of a closed image is also closed, therefore is $f^{-1}$ is also continous.



Does that makes sense to you? :)
but where does the injectivity comes in?
Any help is appreciated !






real-analysis general-topology proof-verification







share|cite|improve this question











$endgroup$




Let
$( mathscr{X}, d_x)$ and $( mathscr{Y}, d_y)$ be two metric spaces and



$ D subset mathscr{X} $ a compact set.



To show is, that if



$f: D rightarrow mathscr{Y} $is continous and injective, then



$ f^{-1} :f(D) rightarrow D $



is also continous.



My Idea:



Since $D$ is compact, it is also closed.
$$ (f^{-1})^{-1} (D)= f(D) $$
And since $f$ is continous and $D$ compact, is also $f(D)$ compact.
As Subset of the metric space $mathscr{Y} $ is $f(D)$ also closed.
So the inverse image of a closed image is also closed, therefore is $f^{-1}$ is also continous.



Does that makes sense to you? :)
but where does the injectivity comes in?
Any help is appreciated !






real-analysis general-topology proof-verification




real-analysis general-topology proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 18:38







constant94

















asked Dec 5 '18 at 18:15









constant94constant94

6310




6310








  • 1




    $begingroup$
    What do you mean by “completed”? Do you mean “closed”?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:21










  • $begingroup$
    yes, I edited it
    $endgroup$
    – constant94
    Dec 5 '18 at 18:39










  • $begingroup$
    Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:40












  • $begingroup$
    @JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
    $endgroup$
    – Federico
    Dec 5 '18 at 18:41












  • $begingroup$
    @Federico, how do I do that?
    $endgroup$
    – constant94
    Dec 5 '18 at 18:43














  • 1




    $begingroup$
    What do you mean by “completed”? Do you mean “closed”?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:21










  • $begingroup$
    yes, I edited it
    $endgroup$
    – constant94
    Dec 5 '18 at 18:39










  • $begingroup$
    Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 18:40












  • $begingroup$
    @JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
    $endgroup$
    – Federico
    Dec 5 '18 at 18:41












  • $begingroup$
    @Federico, how do I do that?
    $endgroup$
    – constant94
    Dec 5 '18 at 18:43








1




1




$begingroup$
What do you mean by “completed”? Do you mean “closed”?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 18:21




$begingroup$
What do you mean by “completed”? Do you mean “closed”?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 18:21












$begingroup$
yes, I edited it
$endgroup$
– constant94
Dec 5 '18 at 18:39




$begingroup$
yes, I edited it
$endgroup$
– constant94
Dec 5 '18 at 18:39












$begingroup$
Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 18:40






$begingroup$
Your proof is correct. If $f$ was not injective, then $f^{-1}$ would not exist.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 18:40














$begingroup$
@JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
$endgroup$
– Federico
Dec 5 '18 at 18:41






$begingroup$
@JoséCarlosSantos The proof is not correct at all. He needs to verify the condition for any closed subset of $D$, not just $D$ itself.
$endgroup$
– Federico
Dec 5 '18 at 18:41














$begingroup$
@Federico, how do I do that?
$endgroup$
– constant94
Dec 5 '18 at 18:43




$begingroup$
@Federico, how do I do that?
$endgroup$
– constant94
Dec 5 '18 at 18:43










1 Answer
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$begingroup$

Let $g=f^{-1}:f(D)to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $Esubset D$ closed.



Fix $Esubset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.



Notice that the proof holds for every $f:Xto Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.






share|cite|improve this answer









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    $begingroup$

    Let $g=f^{-1}:f(D)to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $Esubset D$ closed.



    Fix $Esubset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.



    Notice that the proof holds for every $f:Xto Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $g=f^{-1}:f(D)to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $Esubset D$ closed.



      Fix $Esubset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.



      Notice that the proof holds for every $f:Xto Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $g=f^{-1}:f(D)to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $Esubset D$ closed.



        Fix $Esubset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.



        Notice that the proof holds for every $f:Xto Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.






        share|cite|improve this answer









        $endgroup$



        Let $g=f^{-1}:f(D)to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $Esubset D$ closed.



        Fix $Esubset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.



        Notice that the proof holds for every $f:Xto Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 18:47









        FedericoFederico

        5,039514




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