Finite subcover with special property












2












$begingroup$



Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






general-topology metric-spaces compactness







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




    Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



    $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



    How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






    general-topology metric-spaces compactness







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




      Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



      $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



      How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






      general-topology metric-spaces compactness







      share|cite|improve this question











      $endgroup$





      Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




      Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



      $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



      How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






      general-topology metric-spaces compactness




      general-topology metric-spaces compactness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 22:28









      Kenny Wong

      18.6k21439




      18.6k21439










      asked Dec 26 '18 at 22:20









      Lucas CorrêaLucas Corrêa

      1,6181321




      1,6181321






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:34










          • $begingroup$
            Is this a good idea?
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:35






          • 1




            $begingroup$
            @LucasCorrêa Yes, that's the right idea.
            $endgroup$
            – Kenny Wong
            Dec 26 '18 at 22:35











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:34










          • $begingroup$
            Is this a good idea?
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:35






          • 1




            $begingroup$
            @LucasCorrêa Yes, that's the right idea.
            $endgroup$
            – Kenny Wong
            Dec 26 '18 at 22:35
















          4












          $begingroup$

          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:34










          • $begingroup$
            Is this a good idea?
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:35






          • 1




            $begingroup$
            @LucasCorrêa Yes, that's the right idea.
            $endgroup$
            – Kenny Wong
            Dec 26 '18 at 22:35














          4












          4








          4





          $begingroup$

          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer









          $endgroup$



          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 22:26









          Kenny WongKenny Wong

          18.6k21439




          18.6k21439












          • $begingroup$
            Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:34










          • $begingroup$
            Is this a good idea?
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:35






          • 1




            $begingroup$
            @LucasCorrêa Yes, that's the right idea.
            $endgroup$
            – Kenny Wong
            Dec 26 '18 at 22:35


















          • $begingroup$
            Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:34










          • $begingroup$
            Is this a good idea?
            $endgroup$
            – Lucas Corrêa
            Dec 26 '18 at 22:35






          • 1




            $begingroup$
            @LucasCorrêa Yes, that's the right idea.
            $endgroup$
            – Kenny Wong
            Dec 26 '18 at 22:35
















          $begingroup$
          Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
          $endgroup$
          – Lucas Corrêa
          Dec 26 '18 at 22:34




          $begingroup$
          Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
          $endgroup$
          – Lucas Corrêa
          Dec 26 '18 at 22:34












          $begingroup$
          Is this a good idea?
          $endgroup$
          – Lucas Corrêa
          Dec 26 '18 at 22:35




          $begingroup$
          Is this a good idea?
          $endgroup$
          – Lucas Corrêa
          Dec 26 '18 at 22:35




          1




          1




          $begingroup$
          @LucasCorrêa Yes, that's the right idea.
          $endgroup$
          – Kenny Wong
          Dec 26 '18 at 22:35




          $begingroup$
          @LucasCorrêa Yes, that's the right idea.
          $endgroup$
          – Kenny Wong
          Dec 26 '18 at 22:35


















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