Length of countably many intervals (2)












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I already posted this question along with a related one but only the other one got answered, so I'm reposting this.




The following is a proof from my textbook. I have 2 questions, which are in bold.



Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.



Proof:



Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.



Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .



Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .



But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.



Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.



And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?






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    0














    I already posted this question along with a related one but only the other one got answered, so I'm reposting this.




    The following is a proof from my textbook. I have 2 questions, which are in bold.



    Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.



    Proof:



    Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.



    Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .



    Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .



    But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.



    Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.



    And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?






    real-analysis







    share|cite|improve this question

























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      I already posted this question along with a related one but only the other one got answered, so I'm reposting this.




      The following is a proof from my textbook. I have 2 questions, which are in bold.



      Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.



      Proof:



      Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.



      Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .



      Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .



      But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.



      Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.



      And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?






      real-analysis







      share|cite|improve this question













      I already posted this question along with a related one but only the other one got answered, so I'm reposting this.




      The following is a proof from my textbook. I have 2 questions, which are in bold.



      Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.



      Proof:



      Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.



      Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .



      Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .



      But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.



      Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.



      And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?






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      asked Nov 29 '18 at 14:15









      ThomasThomas

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          Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.



          This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.



          You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $






          share|cite|improve this answer





















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            Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.



            This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.



            You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $






            share|cite|improve this answer


























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              Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.



              This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.



              You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $






              share|cite|improve this answer
























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                Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.



                This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.



                You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $






                share|cite|improve this answer












                Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.



                This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.



                You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 14:26









                Sorin TircSorin Tirc

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