Proving that function is continuous (Alexandroff extension, Hausdorff space)












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$begingroup$


Let $(Z,W)$ be a compact Hausdorff space and $tilde Xsubseteq Z$ an open subset of $Z$.



Furthermore let $h:(tilde X, W_{|tilde X}) to (X, mathcal T)$ be a homeomorphism.



$Y:=X cup {infty}$




How can I show that $f: Z to Y, f(x)=h(x)$ for $x in tilde X$ and $f(z)=infty$ for $z in Zsetminustilde X$ is continuous?




$h$ is a homeomorphism, so $f$ is continuous on $tilde X$. And $f$ is constant on $Zsetminus tilde X$ so $f$ is continuous on $Zsetminus tilde X$. But how do I show that $f$ is continuous on $Z$ as a whole?






real-analysis general-topology analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $(Z,W)$ be a compact Hausdorff space and $tilde Xsubseteq Z$ an open subset of $Z$.



    Furthermore let $h:(tilde X, W_{|tilde X}) to (X, mathcal T)$ be a homeomorphism.



    $Y:=X cup {infty}$




    How can I show that $f: Z to Y, f(x)=h(x)$ for $x in tilde X$ and $f(z)=infty$ for $z in Zsetminustilde X$ is continuous?




    $h$ is a homeomorphism, so $f$ is continuous on $tilde X$. And $f$ is constant on $Zsetminus tilde X$ so $f$ is continuous on $Zsetminus tilde X$. But how do I show that $f$ is continuous on $Z$ as a whole?






    real-analysis general-topology analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $(Z,W)$ be a compact Hausdorff space and $tilde Xsubseteq Z$ an open subset of $Z$.



      Furthermore let $h:(tilde X, W_{|tilde X}) to (X, mathcal T)$ be a homeomorphism.



      $Y:=X cup {infty}$




      How can I show that $f: Z to Y, f(x)=h(x)$ for $x in tilde X$ and $f(z)=infty$ for $z in Zsetminustilde X$ is continuous?




      $h$ is a homeomorphism, so $f$ is continuous on $tilde X$. And $f$ is constant on $Zsetminus tilde X$ so $f$ is continuous on $Zsetminus tilde X$. But how do I show that $f$ is continuous on $Z$ as a whole?






      real-analysis general-topology analysis







      share|cite|improve this question









      $endgroup$




      Let $(Z,W)$ be a compact Hausdorff space and $tilde Xsubseteq Z$ an open subset of $Z$.



      Furthermore let $h:(tilde X, W_{|tilde X}) to (X, mathcal T)$ be a homeomorphism.



      $Y:=X cup {infty}$




      How can I show that $f: Z to Y, f(x)=h(x)$ for $x in tilde X$ and $f(z)=infty$ for $z in Zsetminustilde X$ is continuous?




      $h$ is a homeomorphism, so $f$ is continuous on $tilde X$. And $f$ is constant on $Zsetminus tilde X$ so $f$ is continuous on $Zsetminus tilde X$. But how do I show that $f$ is continuous on $Z$ as a whole?






      real-analysis general-topology analysis




      real-analysis general-topology analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 10:14









      user626880user626880

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          1 Answer
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          $begingroup$

          You need to prove that for any open subset $U$ of $Y$ containing $infty$, $f^{-1}(U)$ is an open subset of $Z$.



          An open subset $U$ of $Y$ containing $infty$ can be written as a $Y backslash K$, where $K$ is a compact subset of $X$.



          So $f^{-1}(U) = Z backslash h^{-1}(K)$ and you can do the rest.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
            $endgroup$
            – user626880
            Dec 16 '18 at 16:42










          • $begingroup$
            No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 16:49










          • $begingroup$
            Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
            $endgroup$
            – user626880
            Dec 16 '18 at 16:56












          • $begingroup$
            I think it is right.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 17:00











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          1 Answer
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          active

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          active

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          1












          $begingroup$

          You need to prove that for any open subset $U$ of $Y$ containing $infty$, $f^{-1}(U)$ is an open subset of $Z$.



          An open subset $U$ of $Y$ containing $infty$ can be written as a $Y backslash K$, where $K$ is a compact subset of $X$.



          So $f^{-1}(U) = Z backslash h^{-1}(K)$ and you can do the rest.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
            $endgroup$
            – user626880
            Dec 16 '18 at 16:42










          • $begingroup$
            No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 16:49










          • $begingroup$
            Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
            $endgroup$
            – user626880
            Dec 16 '18 at 16:56












          • $begingroup$
            I think it is right.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 17:00
















          1












          $begingroup$

          You need to prove that for any open subset $U$ of $Y$ containing $infty$, $f^{-1}(U)$ is an open subset of $Z$.



          An open subset $U$ of $Y$ containing $infty$ can be written as a $Y backslash K$, where $K$ is a compact subset of $X$.



          So $f^{-1}(U) = Z backslash h^{-1}(K)$ and you can do the rest.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
            $endgroup$
            – user626880
            Dec 16 '18 at 16:42










          • $begingroup$
            No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 16:49










          • $begingroup$
            Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
            $endgroup$
            – user626880
            Dec 16 '18 at 16:56












          • $begingroup$
            I think it is right.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 17:00














          1












          1








          1





          $begingroup$

          You need to prove that for any open subset $U$ of $Y$ containing $infty$, $f^{-1}(U)$ is an open subset of $Z$.



          An open subset $U$ of $Y$ containing $infty$ can be written as a $Y backslash K$, where $K$ is a compact subset of $X$.



          So $f^{-1}(U) = Z backslash h^{-1}(K)$ and you can do the rest.






          share|cite|improve this answer









          $endgroup$



          You need to prove that for any open subset $U$ of $Y$ containing $infty$, $f^{-1}(U)$ is an open subset of $Z$.



          An open subset $U$ of $Y$ containing $infty$ can be written as a $Y backslash K$, where $K$ is a compact subset of $X$.



          So $f^{-1}(U) = Z backslash h^{-1}(K)$ and you can do the rest.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 11:24









          MindlackMindlack

          4,945211




          4,945211












          • $begingroup$
            Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
            $endgroup$
            – user626880
            Dec 16 '18 at 16:42










          • $begingroup$
            No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 16:49










          • $begingroup$
            Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
            $endgroup$
            – user626880
            Dec 16 '18 at 16:56












          • $begingroup$
            I think it is right.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 17:00


















          • $begingroup$
            Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
            $endgroup$
            – user626880
            Dec 16 '18 at 16:42










          • $begingroup$
            No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 16:49










          • $begingroup$
            Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
            $endgroup$
            – user626880
            Dec 16 '18 at 16:56












          • $begingroup$
            I think it is right.
            $endgroup$
            – Mindlack
            Dec 16 '18 at 17:00
















          $begingroup$
          Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
          $endgroup$
          – user626880
          Dec 16 '18 at 16:42




          $begingroup$
          Do you really mean compact or do you mean closed instead? Assuming you mean closed, we have that $Z setminus h^{-1}(K)$ is open in $Z$
          $endgroup$
          – user626880
          Dec 16 '18 at 16:42












          $begingroup$
          No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
          $endgroup$
          – Mindlack
          Dec 16 '18 at 16:49




          $begingroup$
          No, I mean compact. By definition of Alexandroff completion $X$ must be locally compact and Hausdorff.
          $endgroup$
          – Mindlack
          Dec 16 '18 at 16:49












          $begingroup$
          Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
          $endgroup$
          – user626880
          Dec 16 '18 at 16:56






          $begingroup$
          Okay thanks. So because $K$ is a compact subset of $X$ which is Hausdorff, it is also closed, therefore $h^{-1}(K)$ is closed as well and its complement, $Z setminus h^{-1}(K)$ is open and this implicates that $f$ is continuous. Is that correct?
          $endgroup$
          – user626880
          Dec 16 '18 at 16:56














          $begingroup$
          I think it is right.
          $endgroup$
          – Mindlack
          Dec 16 '18 at 17:00




          $begingroup$
          I think it is right.
          $endgroup$
          – Mindlack
          Dec 16 '18 at 17:00


















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