Are subspaces of paracompact spaces normal?












3












$begingroup$


Are all subspaces of a paracompact space normal?



This is what I think about this question...
First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.



Am I right about this and what counter-example do we have?






general-topology examples-counterexamples separation-axioms paracompactness







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Are all subspaces of a paracompact space normal?



    This is what I think about this question...
    First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.



    Am I right about this and what counter-example do we have?






    general-topology examples-counterexamples separation-axioms paracompactness







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Are all subspaces of a paracompact space normal?



      This is what I think about this question...
      First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.



      Am I right about this and what counter-example do we have?






      general-topology examples-counterexamples separation-axioms paracompactness







      share|cite|improve this question











      $endgroup$




      Are all subspaces of a paracompact space normal?



      This is what I think about this question...
      First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.



      Am I right about this and what counter-example do we have?






      general-topology examples-counterexamples separation-axioms paracompactness




      general-topology examples-counterexamples separation-axioms paracompactness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 3:22









      Eric Wofsey

      184k13212338




      184k13212338










      asked Dec 4 '14 at 3:25









      BirkaryBirkary

      397




      397






















          1 Answer
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          $begingroup$

          Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace



          $$Z=(Xtimes Y)setminus{langle p,qrangle}$$



          is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.



          To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
            $endgroup$
            – Birkary
            Dec 4 '14 at 3:58










          • $begingroup$
            @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:06










          • $begingroup$
            thanks,this was very helpful. I will make sure to understand the answer you give.
            $endgroup$
            – Birkary
            Dec 4 '14 at 4:08










          • $begingroup$
            @Birkary: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:09










          • $begingroup$
            More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
            $endgroup$
            – DanielWainfleet
            Dec 9 '18 at 3:36











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          $begingroup$

          Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace



          $$Z=(Xtimes Y)setminus{langle p,qrangle}$$



          is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.



          To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
            $endgroup$
            – Birkary
            Dec 4 '14 at 3:58










          • $begingroup$
            @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:06










          • $begingroup$
            thanks,this was very helpful. I will make sure to understand the answer you give.
            $endgroup$
            – Birkary
            Dec 4 '14 at 4:08










          • $begingroup$
            @Birkary: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:09










          • $begingroup$
            More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
            $endgroup$
            – DanielWainfleet
            Dec 9 '18 at 3:36
















          2












          $begingroup$

          Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace



          $$Z=(Xtimes Y)setminus{langle p,qrangle}$$



          is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.



          To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
            $endgroup$
            – Birkary
            Dec 4 '14 at 3:58










          • $begingroup$
            @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:06










          • $begingroup$
            thanks,this was very helpful. I will make sure to understand the answer you give.
            $endgroup$
            – Birkary
            Dec 4 '14 at 4:08










          • $begingroup$
            @Birkary: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:09










          • $begingroup$
            More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
            $endgroup$
            – DanielWainfleet
            Dec 9 '18 at 3:36














          2












          2








          2





          $begingroup$

          Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace



          $$Z=(Xtimes Y)setminus{langle p,qrangle}$$



          is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.



          To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.






          share|cite|improve this answer











          $endgroup$



          Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace



          $$Z=(Xtimes Y)setminus{langle p,qrangle}$$



          is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.



          To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '14 at 3:41

























          answered Dec 4 '14 at 3:36









          Brian M. ScottBrian M. Scott

          456k38509909




          456k38509909












          • $begingroup$
            your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
            $endgroup$
            – Birkary
            Dec 4 '14 at 3:58










          • $begingroup$
            @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:06










          • $begingroup$
            thanks,this was very helpful. I will make sure to understand the answer you give.
            $endgroup$
            – Birkary
            Dec 4 '14 at 4:08










          • $begingroup$
            @Birkary: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:09










          • $begingroup$
            More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
            $endgroup$
            – DanielWainfleet
            Dec 9 '18 at 3:36


















          • $begingroup$
            your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
            $endgroup$
            – Birkary
            Dec 4 '14 at 3:58










          • $begingroup$
            @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:06










          • $begingroup$
            thanks,this was very helpful. I will make sure to understand the answer you give.
            $endgroup$
            – Birkary
            Dec 4 '14 at 4:08










          • $begingroup$
            @Birkary: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Dec 4 '14 at 4:09










          • $begingroup$
            More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
            $endgroup$
            – DanielWainfleet
            Dec 9 '18 at 3:36
















          $begingroup$
          your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
          $endgroup$
          – Birkary
          Dec 4 '14 at 3:58




          $begingroup$
          your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
          $endgroup$
          – Birkary
          Dec 4 '14 at 3:58












          $begingroup$
          @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
          $endgroup$
          – Brian M. Scott
          Dec 4 '14 at 4:06




          $begingroup$
          @Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
          $endgroup$
          – Brian M. Scott
          Dec 4 '14 at 4:06












          $begingroup$
          thanks,this was very helpful. I will make sure to understand the answer you give.
          $endgroup$
          – Birkary
          Dec 4 '14 at 4:08




          $begingroup$
          thanks,this was very helpful. I will make sure to understand the answer you give.
          $endgroup$
          – Birkary
          Dec 4 '14 at 4:08












          $begingroup$
          @Birkary: You’re welcome.
          $endgroup$
          – Brian M. Scott
          Dec 4 '14 at 4:09




          $begingroup$
          @Birkary: You’re welcome.
          $endgroup$
          – Brian M. Scott
          Dec 4 '14 at 4:09












          $begingroup$
          More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
          $endgroup$
          – DanielWainfleet
          Dec 9 '18 at 3:36




          $begingroup$
          More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
          $endgroup$
          – DanielWainfleet
          Dec 9 '18 at 3:36


















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