Question involving Nested Intervals Theorem
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Let $[a_n,b_n]_ninmathbbN$ be a sequence of closed bounded intervals in $mathbbR$ such that $(forall ninmathbbN)([a_n+1,b_n+1]subset [a_n,b_n])$ e $lim_ntoinfty(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbbN)(Scap [a_n,b_n]neq emptyset )$.
We know from the Nested Intervals Theorem that exists $sigmainmathbbR$ such that $bigcap _n=1^infty [a_n,b_n]=sigma$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?
I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:
Assume that $sigmanotin S$, then $Scap left(bigcap _n=1^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$.
Given $xin S$, we have $xnotin bigcap _n=1^infty [a_n,b_n]Rightarrow (exists ninmathbbN)(xnotin [a_n,b_n])$.
Therefore,
$(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$
But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.
Can someone help me please?
real-analysis elementary-set-theory
asked Sep 10 at 21:38
rfloc
12518
add a comment |Â
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Let $[a_n,b_n]_ninmathbbN$ be a sequence of closed bounded intervals in $mathbbR$ such that $(forall ninmathbbN)([a_n+1,b_n+1]subset [a_n,b_n])$ e $lim_ntoinfty(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbbN)(Scap [a_n,b_n]neq emptyset )$.
We know from the Nested Intervals Theorem that exists $sigmainmathbbR$ such that $bigcap _n=1^infty [a_n,b_n]=sigma$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?
I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:
Assume that $sigmanotin S$, then $Scap left(bigcap _n=1^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$.
Given $xin S$, we have $xnotin bigcap _n=1^infty [a_n,b_n]Rightarrow (exists ninmathbbN)(xnotin [a_n,b_n])$.
Therefore,
$(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$
But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.
Can someone help me please?
real-analysis elementary-set-theory
asked Sep 10 at 21:38
rfloc
12518
Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $[a_n,b_n]_ninmathbbN$ be a sequence of closed bounded intervals in $mathbbR$ such that $(forall ninmathbbN)([a_n+1,b_n+1]subset [a_n,b_n])$ e $lim_ntoinfty(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbbN)(Scap [a_n,b_n]neq emptyset )$.
We know from the Nested Intervals Theorem that exists $sigmainmathbbR$ such that $bigcap _n=1^infty [a_n,b_n]=sigma$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?
I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:
Assume that $sigmanotin S$, then $Scap left(bigcap _n=1^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$.
Given $xin S$, we have $xnotin bigcap _n=1^infty [a_n,b_n]Rightarrow (exists ninmathbbN)(xnotin [a_n,b_n])$.
Therefore,
$(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$
But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.
Can someone help me please?
real-analysis elementary-set-theory
asked Sep 10 at 21:38
rfloc
12518
Let $[a_n,b_n]_ninmathbbN$ be a sequence of closed bounded intervals in $mathbbR$ such that $(forall ninmathbbN)([a_n+1,b_n+1]subset [a_n,b_n])$ e $lim_ntoinfty(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbbN)(Scap [a_n,b_n]neq emptyset )$.
We know from the Nested Intervals Theorem that exists $sigmainmathbbR$ such that $bigcap _n=1^infty [a_n,b_n]=sigma$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?
I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:
Assume that $sigmanotin S$, then $Scap left(bigcap _n=1^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$.
Given $xin S$, we have $xnotin bigcap _n=1^infty [a_n,b_n]Rightarrow (exists ninmathbbN)(xnotin [a_n,b_n])$.
Therefore,
$(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$
But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _n=1^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbbN)(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.
Can someone help me please?
real-analysis elementary-set-theory
real-analysis elementary-set-theory
asked Sep 10 at 21:38
rfloc
12518
asked Sep 10 at 21:38
rfloc
12518
edited Sep 10 at 21:57
asked Sep 10 at 21:38
rfloc
12518
asked Sep 10 at 21:38
rfloc
12518
asked Sep 10 at 21:38
rfloc
12518
12518
Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55
add a comment |Â
Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55
Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55
add a comment |Â
1 Answer
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oldest
votes
up vote
1
down vote
By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=a_nmid ninmathbb N$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn't have to belong to $S$.
answered Sep 10 at 21:48
SMM
2,03049
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=a_nmid ninmathbb N$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn't have to belong to $S$.
answered Sep 10 at 21:48
SMM
2,03049
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
add a comment |Â
up vote
1
down vote
By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=a_nmid ninmathbb N$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn't have to belong to $S$.
answered Sep 10 at 21:48
SMM
2,03049
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=a_nmid ninmathbb N$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn't have to belong to $S$.
answered Sep 10 at 21:48
SMM
2,03049
By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=a_nmid ninmathbb N$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn't have to belong to $S$.
answered Sep 10 at 21:48
SMM
2,03049
answered Sep 10 at 21:48
SMM
2,03049
answered Sep 10 at 21:48
SMM
2,03049
answered Sep 10 at 21:48
SMM
2,03049
2,03049
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
add a comment |Â
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
– rfloc
Sep 10 at 22:41
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
To the proposer: Suppose $a_n<a_n+1<b_n+1<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash sigma.$
– DanielWainfleet
Sep 11 at 6:53
add a comment |Â
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Is there a material difference between $a_n$ and $a^(n)$?
– Saucy O'Path
Sep 10 at 21:48
No, I just forgot to change the notation. I will correct.
– rfloc
Sep 10 at 21:55