Expanding or shrinking an interval slightly












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Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. Suppose the $I_n$ are pairwise disjoint. My textbook says "We have $bigcup_{n=1}^infty I_nsubset bigcup_{k=1}^infty J_k$. 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."



My question: What does that sentence mean? How can you expand each $J_k$ slightly and end up with it being open? How can you shrink $I_n$ slightly and end up with it being closed? My point is that you can expand an open interval slightly so that it becomes closed or shrink a closed interval slightly so that it becomes open. The textbook seems to have it reversed.






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    Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. Suppose the $I_n$ are pairwise disjoint. My textbook says "We have $bigcup_{n=1}^infty I_nsubset bigcup_{k=1}^infty J_k$. 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."



    My question: What does that sentence mean? How can you expand each $J_k$ slightly and end up with it being open? How can you shrink $I_n$ slightly and end up with it being closed? My point is that you can expand an open interval slightly so that it becomes closed or shrink a closed interval slightly so that it becomes open. The textbook seems to have it reversed.






    real-analysis







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      Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. Suppose the $I_n$ are pairwise disjoint. My textbook says "We have $bigcup_{n=1}^infty I_nsubset bigcup_{k=1}^infty J_k$. 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."



      My question: What does that sentence mean? How can you expand each $J_k$ slightly and end up with it being open? How can you shrink $I_n$ slightly and end up with it being closed? My point is that you can expand an open interval slightly so that it becomes closed or shrink a closed interval slightly so that it becomes open. The textbook seems to have it reversed.






      real-analysis







      share|cite|improve this question















      Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. Suppose the $I_n$ are pairwise disjoint. My textbook says "We have $bigcup_{n=1}^infty I_nsubset bigcup_{k=1}^infty J_k$. 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."



      My question: What does that sentence mean? How can you expand each $J_k$ slightly and end up with it being open? How can you shrink $I_n$ slightly and end up with it being closed? My point is that you can expand an open interval slightly so that it becomes closed or shrink a closed interval slightly so that it becomes open. The textbook seems to have it reversed.






      real-analysis




      real-analysis






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      edited Nov 29 at 0:36









      Gerry Myerson

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      asked Nov 29 at 0:19









      Thomas

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          $(a,b)$ becomes closed if you shrink it to $[a+epsilon, b-epsilon]$ and $[a,b]$ becomes open if you expand it to $(a-epsilon, b+epsilon)$. You can do a similar thing starting with any interval (half closed intervals, for example).






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            $(a,b)$ becomes closed if you shrink it to $[a+epsilon, b-epsilon]$ and $[a,b]$ becomes open if you expand it to $(a-epsilon, b+epsilon)$. You can do a similar thing starting with any interval (half closed intervals, for example).






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              $(a,b)$ becomes closed if you shrink it to $[a+epsilon, b-epsilon]$ and $[a,b]$ becomes open if you expand it to $(a-epsilon, b+epsilon)$. You can do a similar thing starting with any interval (half closed intervals, for example).






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                $(a,b)$ becomes closed if you shrink it to $[a+epsilon, b-epsilon]$ and $[a,b]$ becomes open if you expand it to $(a-epsilon, b+epsilon)$. You can do a similar thing starting with any interval (half closed intervals, for example).






                share|cite|improve this answer












                $(a,b)$ becomes closed if you shrink it to $[a+epsilon, b-epsilon]$ and $[a,b]$ becomes open if you expand it to $(a-epsilon, b+epsilon)$. You can do a similar thing starting with any interval (half closed intervals, for example).







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                answered Nov 29 at 0:24









                Kavi Rama Murthy

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