Finite subcover with special property












2















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





























    2















    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



























      2












      2








      2








      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
















      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 6 hours ago









      Kenny Wong

      17k21135




      17k21135










      asked 6 hours ago









      Lucas Corrêa

      1,3451321




      1,3451321






















          1 Answer
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          4














          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





















          • 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.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4














          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





















          • 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.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago
















          4














          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





















          • 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.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago














          4












          4








          4






          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












          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 6 hours ago









          Kenny Wong

          17k21135




          17k21135












          • 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.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago


















          • 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.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago
















          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.
          – Lucas Corrêa
          6 hours ago




          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.
          – Lucas Corrêa
          6 hours ago












          Is this a good idea?
          – Lucas Corrêa
          6 hours ago




          Is this a good idea?
          – Lucas Corrêa
          6 hours ago




          1




          1




          @LucasCorrêa Yes, that's the right idea.
          – Kenny Wong
          6 hours ago




          @LucasCorrêa Yes, that's the right idea.
          – Kenny Wong
          6 hours ago


















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