Pre-image of Hausdorff space is Hausdorff












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Let $X, Y$ be topological spaces, with Y a Hausdorff space. Prove that if there exists an injective and continuous function $f: X rightarrow Y$, then $X$ is Hausdorff.






Here's my idea:



Since $Y$ is Hausdorff, different points $x,y in Y$ have disjoint neighborhood $U subset tau_x,V subset tau_y$.



But $f$ is also continuous, so the pre-image of these sets is an open set in $X$, say $f^{-1}(U)=A$, $f^{-1}(V)=B$. Now, because $f$ is injective, I got that $A cap B = emptyset$, and $f^{-1}(x)$, $f^{-1}(y)$ have disjoint neighborhoods.






general-topology proof-verification separation-axioms







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$endgroup$

















    5












    $begingroup$



    Let $X, Y$ be topological spaces, with Y a Hausdorff space. Prove that if there exists an injective and continuous function $f: X rightarrow Y$, then $X$ is Hausdorff.






    Here's my idea:



    Since $Y$ is Hausdorff, different points $x,y in Y$ have disjoint neighborhood $U subset tau_x,V subset tau_y$.



    But $f$ is also continuous, so the pre-image of these sets is an open set in $X$, say $f^{-1}(U)=A$, $f^{-1}(V)=B$. Now, because $f$ is injective, I got that $A cap B = emptyset$, and $f^{-1}(x)$, $f^{-1}(y)$ have disjoint neighborhoods.






    general-topology proof-verification separation-axioms







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$



      Let $X, Y$ be topological spaces, with Y a Hausdorff space. Prove that if there exists an injective and continuous function $f: X rightarrow Y$, then $X$ is Hausdorff.






      Here's my idea:



      Since $Y$ is Hausdorff, different points $x,y in Y$ have disjoint neighborhood $U subset tau_x,V subset tau_y$.



      But $f$ is also continuous, so the pre-image of these sets is an open set in $X$, say $f^{-1}(U)=A$, $f^{-1}(V)=B$. Now, because $f$ is injective, I got that $A cap B = emptyset$, and $f^{-1}(x)$, $f^{-1}(y)$ have disjoint neighborhoods.






      general-topology proof-verification separation-axioms







      share|cite|improve this question











      $endgroup$





      Let $X, Y$ be topological spaces, with Y a Hausdorff space. Prove that if there exists an injective and continuous function $f: X rightarrow Y$, then $X$ is Hausdorff.






      Here's my idea:



      Since $Y$ is Hausdorff, different points $x,y in Y$ have disjoint neighborhood $U subset tau_x,V subset tau_y$.



      But $f$ is also continuous, so the pre-image of these sets is an open set in $X$, say $f^{-1}(U)=A$, $f^{-1}(V)=B$. Now, because $f$ is injective, I got that $A cap B = emptyset$, and $f^{-1}(x)$, $f^{-1}(y)$ have disjoint neighborhoods.






      general-topology proof-verification separation-axioms




      general-topology proof-verification separation-axioms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 17:09









      José Carlos Santos

      159k22126231




      159k22126231










      asked Aug 22 '17 at 21:50









      VoBVoB

      729413




      729413






















          3 Answers
          3






          active

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          6












          $begingroup$

          Let $x_1,x_2 in X$ such that $x_1 neq x_2$



          Then $f(x_1),f(x_2) in Y$ and $f(x_1) neq f(x_2)$ because $f$ is injective.



          Then there exist two disjoint open sets $U_1,U_2 in mathcal{T_Y}$ such that $f(x_1) in U_1$ and $f(x_2) in U_2$



          Thus $$x_1 in f^{-1}(U_1)$$ $$x_2 in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.



          Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 cap V_2 =f^{-1}(U_1) cap f^{-1}(U_2)=f^{-1}(U_1 cap U_2)=f^{-1}(emptyset)=emptyset$$



          Therefore $X$ is Hausdorf.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $Uintau_x$ and $Vintau_y$ (note that you wrote $Usubsettau_x$ and $Vsubsettau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?



            There are two ways of solving this:




            1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.

            2. You can assume without loss of generality that $U$ and $V$ are open sets.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I said that $U, V subset Y$ are open sets just because $f$ is continuous.
              $endgroup$
              – VoB
              Aug 22 '17 at 22:14










            • $begingroup$
              The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
              $endgroup$
              – José Carlos Santos
              Aug 22 '17 at 22:20





















            2












            $begingroup$

            Take distinct $x,y in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              3 Answers
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              6












              $begingroup$

              Let $x_1,x_2 in X$ such that $x_1 neq x_2$



              Then $f(x_1),f(x_2) in Y$ and $f(x_1) neq f(x_2)$ because $f$ is injective.



              Then there exist two disjoint open sets $U_1,U_2 in mathcal{T_Y}$ such that $f(x_1) in U_1$ and $f(x_2) in U_2$



              Thus $$x_1 in f^{-1}(U_1)$$ $$x_2 in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.



              Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 cap V_2 =f^{-1}(U_1) cap f^{-1}(U_2)=f^{-1}(U_1 cap U_2)=f^{-1}(emptyset)=emptyset$$



              Therefore $X$ is Hausdorf.






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                Let $x_1,x_2 in X$ such that $x_1 neq x_2$



                Then $f(x_1),f(x_2) in Y$ and $f(x_1) neq f(x_2)$ because $f$ is injective.



                Then there exist two disjoint open sets $U_1,U_2 in mathcal{T_Y}$ such that $f(x_1) in U_1$ and $f(x_2) in U_2$



                Thus $$x_1 in f^{-1}(U_1)$$ $$x_2 in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.



                Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 cap V_2 =f^{-1}(U_1) cap f^{-1}(U_2)=f^{-1}(U_1 cap U_2)=f^{-1}(emptyset)=emptyset$$



                Therefore $X$ is Hausdorf.






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Let $x_1,x_2 in X$ such that $x_1 neq x_2$



                  Then $f(x_1),f(x_2) in Y$ and $f(x_1) neq f(x_2)$ because $f$ is injective.



                  Then there exist two disjoint open sets $U_1,U_2 in mathcal{T_Y}$ such that $f(x_1) in U_1$ and $f(x_2) in U_2$



                  Thus $$x_1 in f^{-1}(U_1)$$ $$x_2 in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.



                  Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 cap V_2 =f^{-1}(U_1) cap f^{-1}(U_2)=f^{-1}(U_1 cap U_2)=f^{-1}(emptyset)=emptyset$$



                  Therefore $X$ is Hausdorf.






                  share|cite|improve this answer











                  $endgroup$



                  Let $x_1,x_2 in X$ such that $x_1 neq x_2$



                  Then $f(x_1),f(x_2) in Y$ and $f(x_1) neq f(x_2)$ because $f$ is injective.



                  Then there exist two disjoint open sets $U_1,U_2 in mathcal{T_Y}$ such that $f(x_1) in U_1$ and $f(x_2) in U_2$



                  Thus $$x_1 in f^{-1}(U_1)$$ $$x_2 in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.



                  Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 cap V_2 =f^{-1}(U_1) cap f^{-1}(U_2)=f^{-1}(U_1 cap U_2)=f^{-1}(emptyset)=emptyset$$



                  Therefore $X$ is Hausdorf.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 22 '17 at 23:07

























                  answered Aug 22 '17 at 22:17









                  Marios GretsasMarios Gretsas

                  8,48011437




                  8,48011437























                      3












                      $begingroup$

                      There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $Uintau_x$ and $Vintau_y$ (note that you wrote $Usubsettau_x$ and $Vsubsettau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?



                      There are two ways of solving this:




                      1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.

                      2. You can assume without loss of generality that $U$ and $V$ are open sets.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                        $endgroup$
                        – VoB
                        Aug 22 '17 at 22:14










                      • $begingroup$
                        The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                        $endgroup$
                        – José Carlos Santos
                        Aug 22 '17 at 22:20


















                      3












                      $begingroup$

                      There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $Uintau_x$ and $Vintau_y$ (note that you wrote $Usubsettau_x$ and $Vsubsettau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?



                      There are two ways of solving this:




                      1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.

                      2. You can assume without loss of generality that $U$ and $V$ are open sets.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                        $endgroup$
                        – VoB
                        Aug 22 '17 at 22:14










                      • $begingroup$
                        The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                        $endgroup$
                        – José Carlos Santos
                        Aug 22 '17 at 22:20
















                      3












                      3








                      3





                      $begingroup$

                      There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $Uintau_x$ and $Vintau_y$ (note that you wrote $Usubsettau_x$ and $Vsubsettau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?



                      There are two ways of solving this:




                      1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.

                      2. You can assume without loss of generality that $U$ and $V$ are open sets.






                      share|cite|improve this answer











                      $endgroup$



                      There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $Uintau_x$ and $Vintau_y$ (note that you wrote $Usubsettau_x$ and $Vsubsettau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?



                      There are two ways of solving this:




                      1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.

                      2. You can assume without loss of generality that $U$ and $V$ are open sets.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 14 '18 at 21:34

























                      answered Aug 22 '17 at 22:09









                      José Carlos SantosJosé Carlos Santos

                      159k22126231




                      159k22126231












                      • $begingroup$
                        I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                        $endgroup$
                        – VoB
                        Aug 22 '17 at 22:14










                      • $begingroup$
                        The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                        $endgroup$
                        – José Carlos Santos
                        Aug 22 '17 at 22:20




















                      • $begingroup$
                        I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                        $endgroup$
                        – VoB
                        Aug 22 '17 at 22:14










                      • $begingroup$
                        The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                        $endgroup$
                        – José Carlos Santos
                        Aug 22 '17 at 22:20


















                      $begingroup$
                      I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                      $endgroup$
                      – VoB
                      Aug 22 '17 at 22:14




                      $begingroup$
                      I said that $U, V subset Y$ are open sets just because $f$ is continuous.
                      $endgroup$
                      – VoB
                      Aug 22 '17 at 22:14












                      $begingroup$
                      The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                      $endgroup$
                      – José Carlos Santos
                      Aug 22 '17 at 22:20






                      $begingroup$
                      The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^star$ and $V^star$ of $f(x)$ and of $f(y)$ respectively such that $U^starcap V^star=emptyset$ and then say that there are open sets $Uintau_x$ and $Vintau_y$ such that $xin Usubset U^star$ and $yin Vsubset V^star$. Since $U^starcap V^star=emptyset$, $Ucap V=emptyset$.
                      $endgroup$
                      – José Carlos Santos
                      Aug 22 '17 at 22:20













                      2












                      $begingroup$

                      Take distinct $x,y in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Take distinct $x,y in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Take distinct $x,y in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.






                          share|cite|improve this answer









                          $endgroup$



                          Take distinct $x,y in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 22 '17 at 22:07









                          Zain PatelZain Patel

                          15.7k51949




                          15.7k51949






























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