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floor of a Cauchy sequence is Cauchy? [closed]
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I am required to show that if $(a_n)$ is a Cauchy sequence then so is $(operatornamefloor(a_n))$. Any hints on how we may go about showing this please do not provide complete solutions.
real-analysis cauchy-sequences
asked Jul 31 at 16:16
Atif Farooq
2,7352824
closed as off-topic by Andrés E. Caicedo, John Ma, Atif Farooq, Mostafa Ayaz, ΘΣΦGenSan Jul 31 at 22:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, ΘΣΦGenSan
add a comment |Â
up vote
-2
down vote
favorite
I am required to show that if $(a_n)$ is a Cauchy sequence then so is $(operatornamefloor(a_n))$. Any hints on how we may go about showing this please do not provide complete solutions.
real-analysis cauchy-sequences
asked Jul 31 at 16:16
Atif Farooq
2,7352824
closed as off-topic by Andrés E. Caicedo, John Ma, Atif Farooq, Mostafa Ayaz, ΘΣΦGenSan Jul 31 at 22:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, ΘΣΦGenSan
5
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
5
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
3
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I am required to show that if $(a_n)$ is a Cauchy sequence then so is $(operatornamefloor(a_n))$. Any hints on how we may go about showing this please do not provide complete solutions.
real-analysis cauchy-sequences
asked Jul 31 at 16:16
Atif Farooq
2,7352824
I am required to show that if $(a_n)$ is a Cauchy sequence then so is $(operatornamefloor(a_n))$. Any hints on how we may go about showing this please do not provide complete solutions.
real-analysis cauchy-sequences
asked Jul 31 at 16:16
Atif Farooq
2,7352824
asked Jul 31 at 16:16
Atif Farooq
2,7352824
asked Jul 31 at 16:16
Atif Farooq
2,7352824
asked Jul 31 at 16:16
Atif Farooq
2,7352824
2,7352824
closed as off-topic by Andrés E. Caicedo, John Ma, Atif Farooq, Mostafa Ayaz, ΘΣΦGenSan Jul 31 at 22:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, ΘΣΦGenSan
closed as off-topic by Andrés E. Caicedo, John Ma, Atif Farooq, Mostafa Ayaz, ΘΣΦGenSan Jul 31 at 22:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, ΘΣΦGenSan
5
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
5
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
3
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34
add a comment |Â
5
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
5
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
3
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34
5
5
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
5
5
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
3
3
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
Consider this particular example:
$$1+(-1)^n frac1n$$
The sequence converges to $1$, hence it is Cauchy.
Try to examine the floor function.
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
add a comment |Â
up vote
3
down vote
$lfloor a_nrfloor - lfloor a_mrfloor in mathbb Z$, so for any $0 < epsilon < 1$ the only way to have $|lfloor a_nrfloor - lfloor a_mrfloor| < epsilon$ is to have $|lfloor a_nrfloor - lfloor a_mrfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k in mathbb Z$ so that $k le a_n < k+1$ and $k le a_m < k+1$).
So for $lfloor a_nrfloor$ to be cauchy is for $a_n$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M in mathbb N$ and $k in mathbb Z$ where $n > M implies k le a_n < k+1$).
Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.
A simple example is $a_n=(-1)^n*frac 1n$. Clearly Cauchy and $a_n to 0$. But for even $n$, $a_n > 0$ and $lfloor a_n rfloor = 0$ but for odd $n$, $a_n < 0$ and $lfloor a_n rfloor = -1$.
So the statement is not true.
However if $a_n to L$ and $L not in mathbb Z$ then statement is true.
Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.
Formally:
If $L not in mathbb Z$ then $rfloor L lfloor < L < rfloor L lfloor + 1$ and if $D = min(L - rfloor L lfloor,rfloor L lfloor + 1 - L)$ then, as $a_n$ is cauchy, for any $epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < min(D, epsilon) le D$.
So if $a_n ge L$ we have $rfloor L lfloor < L le a_n$ and $a_n-L < rfloor L lfloor + 1 - L$ so $a_n < rfloor L lfloor + 1$.
And if $a_n < L$ then we have $L - a_n < L - rfloor L lfloor$ so $rfloor L lfloor< a_n <L < rfloor L lfloor + 1$.
Either way $rfloor L lfloor< a_n < rfloor L lfloor + 1$ so $rfloor a_n lfloor = rfloor L lfloor$ and $rfloor a_n lfloorto rfloor L lfloor$
answered Jul 31 at 17:04
fleablood
60.2k22575
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Consider this particular example:
$$1+(-1)^n frac1n$$
The sequence converges to $1$, hence it is Cauchy.
Try to examine the floor function.
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
add a comment |Â
up vote
9
down vote
accepted
Consider this particular example:
$$1+(-1)^n frac1n$$
The sequence converges to $1$, hence it is Cauchy.
Try to examine the floor function.
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Consider this particular example:
$$1+(-1)^n frac1n$$
The sequence converges to $1$, hence it is Cauchy.
Try to examine the floor function.
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
Consider this particular example:
$$1+(-1)^n frac1n$$
The sequence converges to $1$, hence it is Cauchy.
Try to examine the floor function.
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
answered Jul 31 at 16:21
Siong Thye Goh
76.7k134794
76.7k134794
add a comment |Â
add a comment |Â
up vote
3
down vote
$lfloor a_nrfloor - lfloor a_mrfloor in mathbb Z$, so for any $0 < epsilon < 1$ the only way to have $|lfloor a_nrfloor - lfloor a_mrfloor| < epsilon$ is to have $|lfloor a_nrfloor - lfloor a_mrfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k in mathbb Z$ so that $k le a_n < k+1$ and $k le a_m < k+1$).
So for $lfloor a_nrfloor$ to be cauchy is for $a_n$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M in mathbb N$ and $k in mathbb Z$ where $n > M implies k le a_n < k+1$).
Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.
A simple example is $a_n=(-1)^n*frac 1n$. Clearly Cauchy and $a_n to 0$. But for even $n$, $a_n > 0$ and $lfloor a_n rfloor = 0$ but for odd $n$, $a_n < 0$ and $lfloor a_n rfloor = -1$.
So the statement is not true.
However if $a_n to L$ and $L not in mathbb Z$ then statement is true.
Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.
Formally:
If $L not in mathbb Z$ then $rfloor L lfloor < L < rfloor L lfloor + 1$ and if $D = min(L - rfloor L lfloor,rfloor L lfloor + 1 - L)$ then, as $a_n$ is cauchy, for any $epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < min(D, epsilon) le D$.
So if $a_n ge L$ we have $rfloor L lfloor < L le a_n$ and $a_n-L < rfloor L lfloor + 1 - L$ so $a_n < rfloor L lfloor + 1$.
And if $a_n < L$ then we have $L - a_n < L - rfloor L lfloor$ so $rfloor L lfloor< a_n <L < rfloor L lfloor + 1$.
Either way $rfloor L lfloor< a_n < rfloor L lfloor + 1$ so $rfloor a_n lfloor = rfloor L lfloor$ and $rfloor a_n lfloorto rfloor L lfloor$
answered Jul 31 at 17:04
fleablood
60.2k22575
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
add a comment |Â
up vote
3
down vote
$lfloor a_nrfloor - lfloor a_mrfloor in mathbb Z$, so for any $0 < epsilon < 1$ the only way to have $|lfloor a_nrfloor - lfloor a_mrfloor| < epsilon$ is to have $|lfloor a_nrfloor - lfloor a_mrfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k in mathbb Z$ so that $k le a_n < k+1$ and $k le a_m < k+1$).
So for $lfloor a_nrfloor$ to be cauchy is for $a_n$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M in mathbb N$ and $k in mathbb Z$ where $n > M implies k le a_n < k+1$).
Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.
A simple example is $a_n=(-1)^n*frac 1n$. Clearly Cauchy and $a_n to 0$. But for even $n$, $a_n > 0$ and $lfloor a_n rfloor = 0$ but for odd $n$, $a_n < 0$ and $lfloor a_n rfloor = -1$.
So the statement is not true.
However if $a_n to L$ and $L not in mathbb Z$ then statement is true.
Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.
Formally:
If $L not in mathbb Z$ then $rfloor L lfloor < L < rfloor L lfloor + 1$ and if $D = min(L - rfloor L lfloor,rfloor L lfloor + 1 - L)$ then, as $a_n$ is cauchy, for any $epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < min(D, epsilon) le D$.
So if $a_n ge L$ we have $rfloor L lfloor < L le a_n$ and $a_n-L < rfloor L lfloor + 1 - L$ so $a_n < rfloor L lfloor + 1$.
And if $a_n < L$ then we have $L - a_n < L - rfloor L lfloor$ so $rfloor L lfloor< a_n <L < rfloor L lfloor + 1$.
Either way $rfloor L lfloor< a_n < rfloor L lfloor + 1$ so $rfloor a_n lfloor = rfloor L lfloor$ and $rfloor a_n lfloorto rfloor L lfloor$
answered Jul 31 at 17:04
fleablood
60.2k22575
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$lfloor a_nrfloor - lfloor a_mrfloor in mathbb Z$, so for any $0 < epsilon < 1$ the only way to have $|lfloor a_nrfloor - lfloor a_mrfloor| < epsilon$ is to have $|lfloor a_nrfloor - lfloor a_mrfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k in mathbb Z$ so that $k le a_n < k+1$ and $k le a_m < k+1$).
So for $lfloor a_nrfloor$ to be cauchy is for $a_n$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M in mathbb N$ and $k in mathbb Z$ where $n > M implies k le a_n < k+1$).
Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.
A simple example is $a_n=(-1)^n*frac 1n$. Clearly Cauchy and $a_n to 0$. But for even $n$, $a_n > 0$ and $lfloor a_n rfloor = 0$ but for odd $n$, $a_n < 0$ and $lfloor a_n rfloor = -1$.
So the statement is not true.
However if $a_n to L$ and $L not in mathbb Z$ then statement is true.
Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.
Formally:
If $L not in mathbb Z$ then $rfloor L lfloor < L < rfloor L lfloor + 1$ and if $D = min(L - rfloor L lfloor,rfloor L lfloor + 1 - L)$ then, as $a_n$ is cauchy, for any $epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < min(D, epsilon) le D$.
So if $a_n ge L$ we have $rfloor L lfloor < L le a_n$ and $a_n-L < rfloor L lfloor + 1 - L$ so $a_n < rfloor L lfloor + 1$.
And if $a_n < L$ then we have $L - a_n < L - rfloor L lfloor$ so $rfloor L lfloor< a_n <L < rfloor L lfloor + 1$.
Either way $rfloor L lfloor< a_n < rfloor L lfloor + 1$ so $rfloor a_n lfloor = rfloor L lfloor$ and $rfloor a_n lfloorto rfloor L lfloor$
answered Jul 31 at 17:04
fleablood
60.2k22575
$lfloor a_nrfloor - lfloor a_mrfloor in mathbb Z$, so for any $0 < epsilon < 1$ the only way to have $|lfloor a_nrfloor - lfloor a_mrfloor| < epsilon$ is to have $|lfloor a_nrfloor - lfloor a_mrfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k in mathbb Z$ so that $k le a_n < k+1$ and $k le a_m < k+1$).
So for $lfloor a_nrfloor$ to be cauchy is for $a_n$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M in mathbb N$ and $k in mathbb Z$ where $n > M implies k le a_n < k+1$).
Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.
A simple example is $a_n=(-1)^n*frac 1n$. Clearly Cauchy and $a_n to 0$. But for even $n$, $a_n > 0$ and $lfloor a_n rfloor = 0$ but for odd $n$, $a_n < 0$ and $lfloor a_n rfloor = -1$.
So the statement is not true.
However if $a_n to L$ and $L not in mathbb Z$ then statement is true.
Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.
Formally:
If $L not in mathbb Z$ then $rfloor L lfloor < L < rfloor L lfloor + 1$ and if $D = min(L - rfloor L lfloor,rfloor L lfloor + 1 - L)$ then, as $a_n$ is cauchy, for any $epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < min(D, epsilon) le D$.
So if $a_n ge L$ we have $rfloor L lfloor < L le a_n$ and $a_n-L < rfloor L lfloor + 1 - L$ so $a_n < rfloor L lfloor + 1$.
And if $a_n < L$ then we have $L - a_n < L - rfloor L lfloor$ so $rfloor L lfloor< a_n <L < rfloor L lfloor + 1$.
Either way $rfloor L lfloor< a_n < rfloor L lfloor + 1$ so $rfloor a_n lfloor = rfloor L lfloor$ and $rfloor a_n lfloorto rfloor L lfloor$
answered Jul 31 at 17:04
fleablood
60.2k22575
answered Jul 31 at 17:04
fleablood
60.2k22575
answered Jul 31 at 17:04
fleablood
60.2k22575
answered Jul 31 at 17:04
fleablood
60.2k22575
60.2k22575
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
add a comment |Â
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
Well done, (+1) from me!
– asdf
Jul 31 at 17:11
add a comment |Â
5
I don't think you are required to show that.
– Kenny Lau
Jul 31 at 16:17
5
$frac(-1)^nn$ is a Cauchy sequence, $lfloor frac(-1)^nn rfloor$ is not
– Rumpelstiltskin
Jul 31 at 16:20
3
Were you actually required to show that, or did the problem say "prove or give a counterexample"? If the former you need to speak to whoever assigned the problem. If the latter, as seems more likely: In the future don't modify the problem before asking about it!
– David C. Ullrich
Jul 31 at 16:34