Why is the intersection empty for open nested intervals?
$begingroup$
Is this correct?
The length of, for example $(0, 1/n)$ as n appoaches infinity, doesn't reach 0 because it's not in the set. But, for the closed set it does because it is in the set?
Apologies for the duplicate question, but I'm having trouble with understanding the answers of others. Below is a link to a a similar question.
Why doesn't the nested interval theorem hold for open intervals?
real-analysis proof-explanation intuition
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
$endgroup$
add a comment |
$begingroup$
Is this correct?
The length of, for example $(0, 1/n)$ as n appoaches infinity, doesn't reach 0 because it's not in the set. But, for the closed set it does because it is in the set?
Apologies for the duplicate question, but I'm having trouble with understanding the answers of others. Below is a link to a a similar question.
Why doesn't the nested interval theorem hold for open intervals?
real-analysis proof-explanation intuition
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
$endgroup$
$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30
add a comment |
$begingroup$
Is this correct?
The length of, for example $(0, 1/n)$ as n appoaches infinity, doesn't reach 0 because it's not in the set. But, for the closed set it does because it is in the set?
Apologies for the duplicate question, but I'm having trouble with understanding the answers of others. Below is a link to a a similar question.
Why doesn't the nested interval theorem hold for open intervals?
real-analysis proof-explanation intuition
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
$endgroup$
Is this correct?
The length of, for example $(0, 1/n)$ as n appoaches infinity, doesn't reach 0 because it's not in the set. But, for the closed set it does because it is in the set?
Apologies for the duplicate question, but I'm having trouble with understanding the answers of others. Below is a link to a a similar question.
Why doesn't the nested interval theorem hold for open intervals?
real-analysis proof-explanation intuition
real-analysis proof-explanation intuition
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
edited Dec 26 '18 at 8:36
Asaf Karagila♦
304k32432763
304k32432763
asked Dec 26 '18 at 1:25
user606466
$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30
add a comment |
$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30
$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30
$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30
add a comment |
2 Answers
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$begingroup$
It is not that the intersection of a sequence of open nested intervals is always empty, just that it can be empty.
The quintessential example is the one you have given; define the sequence $(I_n)_{ninmathbb{N}}$ such that $I_n=(0,1/n)$. Question: If I pick a real number $x$, does it belong to the set $cap_{n=1}^k I_n$? Only if $0<x<1/k$. Since the reals have the property that there are no non-zero infinitesimals, every real number $x$ has some natural number $k_x$ such that $x>1/k_x$.
Thus, eventually, after a finite number of intersections, $x$ will not belong to our set. Thus, if we take the entire intersection, no real number $x$ gets in our set, and we are empty.
I would ignore the lengths of the intervals for this particular portion of the problem; we only wish to demonstrate that the theorem for nested, closed, and bounded intervals cannot be generalized to nested, open, and bounded intervals.
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
It does not have to do with length. Think about any positive number $x$. By the archimedean property there is some natural $n$ s.t. $1/n<x$, in particular,$xnotin (0,1/n)$ and thus is not in the intersection. Because the only number that could be in the intersection is $0$ but $0notin (0,1)$, this intersection is empty.
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is not that the intersection of a sequence of open nested intervals is always empty, just that it can be empty.
The quintessential example is the one you have given; define the sequence $(I_n)_{ninmathbb{N}}$ such that $I_n=(0,1/n)$. Question: If I pick a real number $x$, does it belong to the set $cap_{n=1}^k I_n$? Only if $0<x<1/k$. Since the reals have the property that there are no non-zero infinitesimals, every real number $x$ has some natural number $k_x$ such that $x>1/k_x$.
Thus, eventually, after a finite number of intersections, $x$ will not belong to our set. Thus, if we take the entire intersection, no real number $x$ gets in our set, and we are empty.
I would ignore the lengths of the intervals for this particular portion of the problem; we only wish to demonstrate that the theorem for nested, closed, and bounded intervals cannot be generalized to nested, open, and bounded intervals.
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
It is not that the intersection of a sequence of open nested intervals is always empty, just that it can be empty.
The quintessential example is the one you have given; define the sequence $(I_n)_{ninmathbb{N}}$ such that $I_n=(0,1/n)$. Question: If I pick a real number $x$, does it belong to the set $cap_{n=1}^k I_n$? Only if $0<x<1/k$. Since the reals have the property that there are no non-zero infinitesimals, every real number $x$ has some natural number $k_x$ such that $x>1/k_x$.
Thus, eventually, after a finite number of intersections, $x$ will not belong to our set. Thus, if we take the entire intersection, no real number $x$ gets in our set, and we are empty.
I would ignore the lengths of the intervals for this particular portion of the problem; we only wish to demonstrate that the theorem for nested, closed, and bounded intervals cannot be generalized to nested, open, and bounded intervals.
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
It is not that the intersection of a sequence of open nested intervals is always empty, just that it can be empty.
The quintessential example is the one you have given; define the sequence $(I_n)_{ninmathbb{N}}$ such that $I_n=(0,1/n)$. Question: If I pick a real number $x$, does it belong to the set $cap_{n=1}^k I_n$? Only if $0<x<1/k$. Since the reals have the property that there are no non-zero infinitesimals, every real number $x$ has some natural number $k_x$ such that $x>1/k_x$.
Thus, eventually, after a finite number of intersections, $x$ will not belong to our set. Thus, if we take the entire intersection, no real number $x$ gets in our set, and we are empty.
I would ignore the lengths of the intervals for this particular portion of the problem; we only wish to demonstrate that the theorem for nested, closed, and bounded intervals cannot be generalized to nested, open, and bounded intervals.
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
$endgroup$
It is not that the intersection of a sequence of open nested intervals is always empty, just that it can be empty.
The quintessential example is the one you have given; define the sequence $(I_n)_{ninmathbb{N}}$ such that $I_n=(0,1/n)$. Question: If I pick a real number $x$, does it belong to the set $cap_{n=1}^k I_n$? Only if $0<x<1/k$. Since the reals have the property that there are no non-zero infinitesimals, every real number $x$ has some natural number $k_x$ such that $x>1/k_x$.
Thus, eventually, after a finite number of intersections, $x$ will not belong to our set. Thus, if we take the entire intersection, no real number $x$ gets in our set, and we are empty.
I would ignore the lengths of the intervals for this particular portion of the problem; we only wish to demonstrate that the theorem for nested, closed, and bounded intervals cannot be generalized to nested, open, and bounded intervals.
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
answered Dec 26 '18 at 1:37
ImNotTheGuyImNotTheGuy
38516
38516
add a comment |
add a comment |
$begingroup$
It does not have to do with length. Think about any positive number $x$. By the archimedean property there is some natural $n$ s.t. $1/n<x$, in particular,$xnotin (0,1/n)$ and thus is not in the intersection. Because the only number that could be in the intersection is $0$ but $0notin (0,1)$, this intersection is empty.
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
$endgroup$
add a comment |
$begingroup$
It does not have to do with length. Think about any positive number $x$. By the archimedean property there is some natural $n$ s.t. $1/n<x$, in particular,$xnotin (0,1/n)$ and thus is not in the intersection. Because the only number that could be in the intersection is $0$ but $0notin (0,1)$, this intersection is empty.
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
$endgroup$
add a comment |
$begingroup$
It does not have to do with length. Think about any positive number $x$. By the archimedean property there is some natural $n$ s.t. $1/n<x$, in particular,$xnotin (0,1/n)$ and thus is not in the intersection. Because the only number that could be in the intersection is $0$ but $0notin (0,1)$, this intersection is empty.
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
$endgroup$
It does not have to do with length. Think about any positive number $x$. By the archimedean property there is some natural $n$ s.t. $1/n<x$, in particular,$xnotin (0,1/n)$ and thus is not in the intersection. Because the only number that could be in the intersection is $0$ but $0notin (0,1)$, this intersection is empty.
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
answered Dec 26 '18 at 1:40
Guacho PerezGuacho Perez
3,92911132
3,92911132
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. As this is a duplicate question, please provide a link to it so that anybody considering answering it can see what what was already asked and stated on the other question, thus avoiding things like repeating the same answer here. Thanks.
$endgroup$
– John Omielan
Dec 26 '18 at 1:30