Infinite descending subsequence of closed intervals of $[0, 1]$
$begingroup$
The question i'm asking arises from lemma $12$ of Laver's proof for the consistency of Borel's conjecture. It has some set theoretic assumptions but i doubt they are needed in the proof. And since i'm not really good with analysis i would be really glad if someone could point me in the right direction.
Let $I_n$, $n lt omega$, be closed intervals in $[0, 1]$ such that for all $n$, there exists some $0 le m le n-1$ such that $I_n = [m/n, (m+1)/n]$. Then there exists an infinite descending subsequence of the above intervals.
My semi-idea: Let $I_n = [a_n , b_n]$. Then the $a_n$'s have either an increasing infinite subsequence or a decreasing infinite subsequence. WLOG assume the $a_n$'s themselves are increasing. Since $[0, 1]$ is bounded, the $a_n$'s converge to a real $u$. By the same argument we may assume the $b_n$'s are decreasing and converge to the same real $u$ because the length of $I_n$'s is decreasing. Now here i wanted to somehow consider an $epsilon gt 0$ and find intervals included in eachother but i'm failing to go further. Any help would be appreciated.
real-analysis set-theory
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
$endgroup$
|
show 4 more comments
$begingroup$
The question i'm asking arises from lemma $12$ of Laver's proof for the consistency of Borel's conjecture. It has some set theoretic assumptions but i doubt they are needed in the proof. And since i'm not really good with analysis i would be really glad if someone could point me in the right direction.
Let $I_n$, $n lt omega$, be closed intervals in $[0, 1]$ such that for all $n$, there exists some $0 le m le n-1$ such that $I_n = [m/n, (m+1)/n]$. Then there exists an infinite descending subsequence of the above intervals.
My semi-idea: Let $I_n = [a_n , b_n]$. Then the $a_n$'s have either an increasing infinite subsequence or a decreasing infinite subsequence. WLOG assume the $a_n$'s themselves are increasing. Since $[0, 1]$ is bounded, the $a_n$'s converge to a real $u$. By the same argument we may assume the $b_n$'s are decreasing and converge to the same real $u$ because the length of $I_n$'s is decreasing. Now here i wanted to somehow consider an $epsilon gt 0$ and find intervals included in eachother but i'm failing to go further. Any help would be appreciated.
real-analysis set-theory
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
$endgroup$
$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
1
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
1
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02
|
show 4 more comments
$begingroup$
The question i'm asking arises from lemma $12$ of Laver's proof for the consistency of Borel's conjecture. It has some set theoretic assumptions but i doubt they are needed in the proof. And since i'm not really good with analysis i would be really glad if someone could point me in the right direction.
Let $I_n$, $n lt omega$, be closed intervals in $[0, 1]$ such that for all $n$, there exists some $0 le m le n-1$ such that $I_n = [m/n, (m+1)/n]$. Then there exists an infinite descending subsequence of the above intervals.
My semi-idea: Let $I_n = [a_n , b_n]$. Then the $a_n$'s have either an increasing infinite subsequence or a decreasing infinite subsequence. WLOG assume the $a_n$'s themselves are increasing. Since $[0, 1]$ is bounded, the $a_n$'s converge to a real $u$. By the same argument we may assume the $b_n$'s are decreasing and converge to the same real $u$ because the length of $I_n$'s is decreasing. Now here i wanted to somehow consider an $epsilon gt 0$ and find intervals included in eachother but i'm failing to go further. Any help would be appreciated.
real-analysis set-theory
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
$endgroup$
The question i'm asking arises from lemma $12$ of Laver's proof for the consistency of Borel's conjecture. It has some set theoretic assumptions but i doubt they are needed in the proof. And since i'm not really good with analysis i would be really glad if someone could point me in the right direction.
Let $I_n$, $n lt omega$, be closed intervals in $[0, 1]$ such that for all $n$, there exists some $0 le m le n-1$ such that $I_n = [m/n, (m+1)/n]$. Then there exists an infinite descending subsequence of the above intervals.
My semi-idea: Let $I_n = [a_n , b_n]$. Then the $a_n$'s have either an increasing infinite subsequence or a decreasing infinite subsequence. WLOG assume the $a_n$'s themselves are increasing. Since $[0, 1]$ is bounded, the $a_n$'s converge to a real $u$. By the same argument we may assume the $b_n$'s are decreasing and converge to the same real $u$ because the length of $I_n$'s is decreasing. Now here i wanted to somehow consider an $epsilon gt 0$ and find intervals included in eachother but i'm failing to go further. Any help would be appreciated.
real-analysis set-theory
real-analysis set-theory
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
asked Dec 25 '18 at 12:08
Shervin SorouriShervin Sorouri
519211
519211
$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
1
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
1
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02
|
show 4 more comments
$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
1
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
1
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02
$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
1
1
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
1
1
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02
|
show 4 more comments
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$begingroup$
If, as you claim, you can (by passing to a subsequence) get the $a_n$'s to increase and the $b_n$'s to decrease, then your intervals are nested. If $m<n$ then $a_mleq a_n<b_nleq b_m$, and so $I_nsubseteq I_m$. I think the real problem is your "WLOG". What if, for example, the $a_n$'s are decreasing?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:24
1
$begingroup$
Are you sure that "descending sequence of intervals" is intended to mean that the intervals are nested? If so, isn't the sequence with $I_n=[frac1n,frac2n]$ a counterexample?
$endgroup$
– Andreas Blass
Dec 25 '18 at 18:29
$begingroup$
This does seem like a counter-example, maybe the set theoretic assumptions are needed after all. Because the same argument that there is a descending subsequence is used in every text i've seen on borel's conjecture without any proof.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:16
$begingroup$
Althouth to be fair Laver actually doesn't say nested but he says a subsequence of the intervals which converges to a single real $u$. But in Freidman's notes he explicitly says descending.
$endgroup$
– Shervin Sorouri
Dec 25 '18 at 19:18
1
$begingroup$
Getting the subsequence to converge to a single real is much easier. By passing to a subsequence, arrange that the $a_n$'s converge to some real $u$. Then observe that the $b_n$'s also converge to the same $u$, because the distance between $a_n$ and $b_n$ converges to $0$. (The distance was $1/n$ for the original sequence; it'll be the same or smaller for the subsequence.)
$endgroup$
– Andreas Blass
Dec 25 '18 at 20:02