Expanding or shrinking an interval slightly
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
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
asked Nov 29 at 0:19
Thomas
701415
add a comment |
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
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
asked Nov 29 at 0:19
Thomas
701415
add a comment |
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
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
asked Nov 29 at 0:19
Thomas
701415
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
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
asked Nov 29 at 0:19
Thomas
701415
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
asked Nov 29 at 0:19
Thomas
701415
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
edited Nov 29 at 0:36
Gerry Myerson
146k8147298
146k8147298
asked Nov 29 at 0:19
Thomas
701415
asked Nov 29 at 0:19
Thomas
701415
asked Nov 29 at 0:19
Thomas
701415
701415
add a comment |
add a comment |
1 Answer
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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).
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
add a comment |
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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).
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
add a comment |
$(a,b)$ 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).
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
add a comment |
$(a,b)$ 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).
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
$(a,b)$ 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).
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:24
Kavi Rama Murthy
48.7k31854
48.7k31854
add a comment |
add a comment |
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