Nested Interval Property and the intersection of infinite sequences

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Given sequences such that $A_n = n,n+1,cdots $, then it can be shown that $bigcaplimits_n=1^infty A_n = emptyset$

Now, according to nested interval property if $mathbfI_n = [a_b,b_n] = x in Re: a_nleq xleq b_n$, then $bigcaplimits_n=1^infty mathbfI_n neq emptyset$

The above two statements looks very similar, but the results are just opposite. From what I can see is, $A_n$ is a countable infinite set whereas $mathbfI_n$ is an uncountable infinite set. Is that the only difference, if it is true ?. Or is there anything more than that which relates two statements above ?

asked Jul 31 at 20:28

Shew

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up vote
3
down vote

favorite

Given sequences such that $A_n = n,n+1,cdots $, then it can be shown that $bigcaplimits_n=1^infty A_n = emptyset$

Now, according to nested interval property if $mathbfI_n = [a_b,b_n] = x in Re: a_nleq xleq b_n$, then $bigcaplimits_n=1^infty mathbfI_n neq emptyset$

asked Jul 31 at 20:28

Shew

451411

add a commentÂ |Â

up vote
3
down vote

favorite

Given sequences such that $A_n = n,n+1,cdots $, then it can be shown that $bigcaplimits_n=1^infty A_n = emptyset$

Now, according to nested interval property if $mathbfI_n = [a_b,b_n] = x in Re: a_nleq xleq b_n$, then $bigcaplimits_n=1^infty mathbfI_n neq emptyset$

asked Jul 31 at 20:28

Shew

451411

Given sequences such that $A_n = n,n+1,cdots $, then it can be shown that $bigcaplimits_n=1^infty A_n = emptyset$

Now, according to nested interval property if $mathbfI_n = [a_b,b_n] = x in Re: a_nleq xleq b_n$, then $bigcaplimits_n=1^infty mathbfI_n neq emptyset$

asked Jul 31 at 20:28

Shew

451411

asked Jul 31 at 20:28

Shew

451411

asked Jul 31 at 20:28

Shew

451411

asked Jul 31 at 20:28

Shew

451411

add a commentÂ |Â

3 Answers
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First note that there are countably many $A_n$ and $I_n$ so the problem is not the countability or non-countability of the intersection. In other words the set containing all the $A_n$ and the set containing all the $I_n$ are both countable being indexed by the countable set $mathbbN$.

Secondly it is true that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable. This is a different statement from the above, for here we are talking about what the sets $I_n$ and $A_n$ contain, not what the sets that contain them contain.

However, the fact that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable is not the reason for the discrepancy.

In fact it is possible to have a countable collection of countable sets $J_n = x in mathbbQ : 1 leq x leq 1 + frac1n$ such that the intersection is non-empty, in this case equal to the set containing the number 1.

Conversely it is possible to have a collection of uncountable sets $B_n = x in mathbbR : x geq n$ such that the intersection is empty.

Thus the crucial difference between your sets $A_n$ and $I_n$ is that the $I_n$ are $bounded$ and nested whilst the $A_n$ are $unbounded$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

answered Jul 31 at 21:20

YeatsL

815

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

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up vote
6
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There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1supseteq K_2supseteq K_3dots$ be a nested sequence of nonepmty compact sets in a topological space, $X$. Then $bigcap_n K_n$ is nonempty.

Proof: If $bigcap_n K_n=varnothing$ , then $Xsetminus K_n_nge 2$ is an open cover of $K_1$ with no finite subcover. $square$

However, the assumption of compactness is necessary. In a general topological space, a nested sequence of sets $A_n$ can have null intersection. Your sets $A_n$ are a typical counter-example. Note that the $A_n$ are not compact.

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

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up vote
2
down vote

First of all note that both $A_n$ and $I_n$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

However, the fact that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable is not the reason for the discrepancy.

Conversely it is possible to have a collection of uncountable sets $B_n = x in mathbbR : x geq n$ such that the intersection is empty.

Thus the crucial difference between your sets $A_n$ and $I_n$ is that the $I_n$ are $bounded$ and nested whilst the $A_n$ are $unbounded$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

answered Jul 31 at 21:20

YeatsL

815

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

add a commentÂ |Â

up vote
1
down vote

accepted

However, the fact that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable is not the reason for the discrepancy.

Conversely it is possible to have a collection of uncountable sets $B_n = x in mathbbR : x geq n$ such that the intersection is empty.

Thus the crucial difference between your sets $A_n$ and $I_n$ is that the $I_n$ are $bounded$ and nested whilst the $A_n$ are $unbounded$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

answered Jul 31 at 21:20

YeatsL

815

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

add a commentÂ |Â

up vote
1
down vote

accepted

However, the fact that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable is not the reason for the discrepancy.

Conversely it is possible to have a collection of uncountable sets $B_n = x in mathbbR : x geq n$ such that the intersection is empty.

Thus the crucial difference between your sets $A_n$ and $I_n$ is that the $I_n$ are $bounded$ and nested whilst the $A_n$ are $unbounded$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

answered Jul 31 at 21:20

YeatsL

815

However, the fact that the sets $I_n$ are uncountable whilst the sets $A_n$ are countable is not the reason for the discrepancy.

Conversely it is possible to have a collection of uncountable sets $B_n = x in mathbbR : x geq n$ such that the intersection is empty.

Thus the crucial difference between your sets $A_n$ and $I_n$ is that the $I_n$ are $bounded$ and nested whilst the $A_n$ are $unbounded$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

answered Jul 31 at 21:20

YeatsL

815

answered Jul 31 at 21:20

YeatsL

815

answered Jul 31 at 21:20

YeatsL

815

answered Jul 31 at 21:20

YeatsL

815

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

add a commentÂ |Â

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection.
â€“Â Mike Earnest
Aug 1 at 3:03

@Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_n$. The answer about compactness (of which closed and bounded is an example) gets my vote.
â€“Â YeatsL
Aug 1 at 8:59

add a commentÂ |Â

up vote
6
down vote

There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1supseteq K_2supseteq K_3dots$ be a nested sequence of nonepmty compact sets in a topological space, $X$. Then $bigcap_n K_n$ is nonempty.

Proof: If $bigcap_n K_n=varnothing$ , then $Xsetminus K_n_nge 2$ is an open cover of $K_1$ with no finite subcover. $square$

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

add a commentÂ |Â

up vote
6
down vote

There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1supseteq K_2supseteq K_3dots$ be a nested sequence of nonepmty compact sets in a topological space, $X$. Then $bigcap_n K_n$ is nonempty.

Proof: If $bigcap_n K_n=varnothing$ , then $Xsetminus K_n_nge 2$ is an open cover of $K_1$ with no finite subcover. $square$

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

add a commentÂ |Â

up vote
6
down vote

There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1supseteq K_2supseteq K_3dots$ be a nested sequence of nonepmty compact sets in a topological space, $X$. Then $bigcap_n K_n$ is nonempty.

Proof: If $bigcap_n K_n=varnothing$ , then $Xsetminus K_n_nge 2$ is an open cover of $K_1$ with no finite subcover. $square$

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1supseteq K_2supseteq K_3dots$ be a nested sequence of nonepmty compact sets in a topological space, $X$. Then $bigcap_n K_n$ is nonempty.

Proof: If $bigcap_n K_n=varnothing$ , then $Xsetminus K_n_nge 2$ is an open cover of $K_1$ with no finite subcover. $square$

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

edited Jul 31 at 21:05

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

answered Jul 31 at 20:58

Mike Earnest

14.7k11644

add a commentÂ |Â

up vote
2
down vote

First of all note that both $A_n$ and $I_n$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

add a commentÂ |Â

up vote
2
down vote

First of all note that both $A_n$ and $I_n$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

add a commentÂ |Â

up vote
2
down vote

First of all note that both $A_n$ and $I_n$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

First of all note that both $A_n$ and $I_n$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

edited Jul 31 at 20:40

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

answered Jul 31 at 20:34

Mohammad Riazi-Kermani

27.3k41851

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

add a commentÂ |Â

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers.
â€“Â Shew
Jul 31 at 20:38

I mean the collection of $I_n$ not each set.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:42

Please check the edited version.
â€“Â Mohammad Riazi-Kermani
Jul 31 at 20:44

add a commentÂ |Â

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