a relationship between local compactness and closure












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Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?






general-topology







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?






    general-topology







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?






      general-topology







      share|cite|improve this question









      $endgroup$




      Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?






      general-topology




      general-topology






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 6 at 15:08









      User12239User12239

      364216




      364216






















          2 Answers
          2






          active

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          2












          $begingroup$

          Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
            $endgroup$
            – User12239
            Apr 6 at 15:26










          • $begingroup$
            You can find a proof in 'Topology' by Munkres.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:28










          • $begingroup$
            Also see this.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:31






          • 1




            $begingroup$
            Thanks I’m looking them up
            $endgroup$
            – User12239
            Apr 6 at 15:33



















          2












          $begingroup$

          In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.



          Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin bar{S}$






          share|cite|improve this answer









          $endgroup$














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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
              $endgroup$
              – User12239
              Apr 6 at 15:26










            • $begingroup$
              You can find a proof in 'Topology' by Munkres.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:28










            • $begingroup$
              Also see this.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:31






            • 1




              $begingroup$
              Thanks I’m looking them up
              $endgroup$
              – User12239
              Apr 6 at 15:33
















            2












            $begingroup$

            Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
              $endgroup$
              – User12239
              Apr 6 at 15:26










            • $begingroup$
              You can find a proof in 'Topology' by Munkres.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:28










            • $begingroup$
              Also see this.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:31






            • 1




              $begingroup$
              Thanks I’m looking them up
              $endgroup$
              – User12239
              Apr 6 at 15:33














            2












            2








            2





            $begingroup$

            Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.






            share|cite|improve this answer











            $endgroup$



            Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 6 at 15:25

























            answered Apr 6 at 15:20









            Thomas ShelbyThomas Shelby

            4,7362727




            4,7362727












            • $begingroup$
              Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
              $endgroup$
              – User12239
              Apr 6 at 15:26










            • $begingroup$
              You can find a proof in 'Topology' by Munkres.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:28










            • $begingroup$
              Also see this.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:31






            • 1




              $begingroup$
              Thanks I’m looking them up
              $endgroup$
              – User12239
              Apr 6 at 15:33


















            • $begingroup$
              Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
              $endgroup$
              – User12239
              Apr 6 at 15:26










            • $begingroup$
              You can find a proof in 'Topology' by Munkres.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:28










            • $begingroup$
              Also see this.
              $endgroup$
              – Thomas Shelby
              Apr 6 at 15:31






            • 1




              $begingroup$
              Thanks I’m looking them up
              $endgroup$
              – User12239
              Apr 6 at 15:33
















            $begingroup$
            Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
            $endgroup$
            – User12239
            Apr 6 at 15:26




            $begingroup$
            Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
            $endgroup$
            – User12239
            Apr 6 at 15:26












            $begingroup$
            You can find a proof in 'Topology' by Munkres.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:28




            $begingroup$
            You can find a proof in 'Topology' by Munkres.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:28












            $begingroup$
            Also see this.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:31




            $begingroup$
            Also see this.
            $endgroup$
            – Thomas Shelby
            Apr 6 at 15:31




            1




            1




            $begingroup$
            Thanks I’m looking them up
            $endgroup$
            – User12239
            Apr 6 at 15:33




            $begingroup$
            Thanks I’m looking them up
            $endgroup$
            – User12239
            Apr 6 at 15:33











            2












            $begingroup$

            In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.



            Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin bar{S}$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.



              Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin bar{S}$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.



                Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin bar{S}$






                share|cite|improve this answer









                $endgroup$



                In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.



                Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin bar{S}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 6 at 15:18









                MaksimMaksim

                1,01719




                1,01719






























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