Mixing
Proof that a strictly decreasing sequence of nested intervals boils down to a single point.
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The nested intervals theorem says the following.
If a sequence of intervals $langle I_nrangle$ is decreasing, then $bigcap_n=1^infty I_n$ is not empty.
However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $bigcap_n=1^infty I_n$ should be a single point.
What I tried:
Let $I_n = [ a_n, b_n]$. Then $langle a_nrangle$ is increasing, while $langle b_nrangle$ is decreasing. Since $langle a_nrangle$ is bounded, it converges to a point, say $alpha$. And, since $b_k$ is an upper bound $forall k$, thus $alpha le b_k$.
Even if $langle b_nrangle$ is decreasing and bounded, I don't know how I can say $lim_n to infty b_n = alpha$.
Is there anyone to help me?
analysis
edited Jul 28 at 14:10
Nosrati
19.2k41544
asked Jul 27 at 12:54
moreblue
1738
 |Â
show 5 more comments
up vote
6
down vote
favorite
The nested intervals theorem says the following.
If a sequence of intervals $langle I_nrangle$ is decreasing, then $bigcap_n=1^infty I_n$ is not empty.
However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $bigcap_n=1^infty I_n$ should be a single point.
What I tried:
Let $I_n = [ a_n, b_n]$. Then $langle a_nrangle$ is increasing, while $langle b_nrangle$ is decreasing. Since $langle a_nrangle$ is bounded, it converges to a point, say $alpha$. And, since $b_k$ is an upper bound $forall k$, thus $alpha le b_k$.
Even if $langle b_nrangle$ is decreasing and bounded, I don't know how I can say $lim_n to infty b_n = alpha$.
Is there anyone to help me?
analysis
edited Jul 28 at 14:10
Nosrati
19.2k41544
asked Jul 27 at 12:54
moreblue
1738
1
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
10
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
3
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11
 |Â
show 5 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
The nested intervals theorem says the following.
If a sequence of intervals $langle I_nrangle$ is decreasing, then $bigcap_n=1^infty I_n$ is not empty.
However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $bigcap_n=1^infty I_n$ should be a single point.
What I tried:
Let $I_n = [ a_n, b_n]$. Then $langle a_nrangle$ is increasing, while $langle b_nrangle$ is decreasing. Since $langle a_nrangle$ is bounded, it converges to a point, say $alpha$. And, since $b_k$ is an upper bound $forall k$, thus $alpha le b_k$.
Even if $langle b_nrangle$ is decreasing and bounded, I don't know how I can say $lim_n to infty b_n = alpha$.
Is there anyone to help me?
analysis
edited Jul 28 at 14:10
Nosrati
19.2k41544
asked Jul 27 at 12:54
moreblue
1738
The nested intervals theorem says the following.
If a sequence of intervals $langle I_nrangle$ is decreasing, then $bigcap_n=1^infty I_n$ is not empty.
However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $bigcap_n=1^infty I_n$ should be a single point.
What I tried:
Let $I_n = [ a_n, b_n]$. Then $langle a_nrangle$ is increasing, while $langle b_nrangle$ is decreasing. Since $langle a_nrangle$ is bounded, it converges to a point, say $alpha$. And, since $b_k$ is an upper bound $forall k$, thus $alpha le b_k$.
Even if $langle b_nrangle$ is decreasing and bounded, I don't know how I can say $lim_n to infty b_n = alpha$.
Is there anyone to help me?
analysis
edited Jul 28 at 14:10
Nosrati
19.2k41544
asked Jul 27 at 12:54
moreblue
1738
edited Jul 28 at 14:10
Nosrati
19.2k41544
edited Jul 28 at 14:10
Nosrati
19.2k41544
edited Jul 28 at 14:10
Nosrati
19.2k41544
19.2k41544
asked Jul 27 at 12:54
moreblue
1738
asked Jul 27 at 12:54
moreblue
1738
asked Jul 27 at 12:54
moreblue
1738
1738
1
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
10
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
3
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11
 |Â
show 5 more comments
1
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
10
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
3
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11
1
1
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
10
10
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
3
3
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
30
down vote
accepted
The statement is false. Take $a_n=-1-frac1n$ and $b_n=1+frac1n$. Then the sequence $(I_n)_ninmathbb N$ is strictly decreasing, but$$bigcap_ninmathbb N[a_n,b_n]=[-1,1].$$
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
30
down vote
accepted
The statement is false. Take $a_n=-1-frac1n$ and $b_n=1+frac1n$. Then the sequence $(I_n)_ninmathbb N$ is strictly decreasing, but$$bigcap_ninmathbb N[a_n,b_n]=[-1,1].$$
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
add a comment |Â
up vote
30
down vote
accepted
The statement is false. Take $a_n=-1-frac1n$ and $b_n=1+frac1n$. Then the sequence $(I_n)_ninmathbb N$ is strictly decreasing, but$$bigcap_ninmathbb N[a_n,b_n]=[-1,1].$$
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
add a comment |Â
up vote
30
down vote
accepted
up vote
30
down vote
accepted
The statement is false. Take $a_n=-1-frac1n$ and $b_n=1+frac1n$. Then the sequence $(I_n)_ninmathbb N$ is strictly decreasing, but$$bigcap_ninmathbb N[a_n,b_n]=[-1,1].$$
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
The statement is false. Take $a_n=-1-frac1n$ and $b_n=1+frac1n$. Then the sequence $(I_n)_ninmathbb N$ is strictly decreasing, but$$bigcap_ninmathbb N[a_n,b_n]=[-1,1].$$
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
edited Jul 27 at 20:01
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
answered Jul 27 at 13:12
José Carlos Santos
113k1696173
113k1696173
add a comment |Â
add a comment |Â
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1
If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-frac1n$ and $b_n=frac1n$, for instance. The intersection will be $0$ then.
– José Carlos Santos
Jul 27 at 12:59
10
The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point.
– Mark
Jul 27 at 13:01
@José Carlos Santos I modified the expression.
– moreblue
Jul 27 at 13:09
@Mark Is there any counter-example?
– moreblue
Jul 27 at 13:10
3
@moreblue Consider $I_n = left(0, 1+frac1nright)$.
– mechanodroid
Jul 27 at 13:11