Flirting Sequences (Real Analysis)
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I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:
In a metric space, a sequence ${P_n}_n$ $ $ flirts with $p$ iff for each $epsilon > 0$, there is a $n in mathbb{N}$ and $m > n$ such that $0 < d(p_n,p) < epsilon$ and $d(p_n,p) < d(p_m,p)$. The sequence ${P_n}_n$ $ $ is a flirting sequence if there is a point $p$ such that ${P_n}$ flirts with $p$.
Give an example of a sequence that flirts with $1$.
I am trying to $(1)$ understand the concept of "flirting" and $(2)$ figure out an example of a sequence that flirts with $1$. Actually, I would be happy to see an example of a any sequence that flirts with something.
I translated the definition like this:
In a metric space, the sequence ${P_n}$ flirts with $p$ iff
$$
forall epsilon > 0 quad exists(m,n in mathbb{N}, m >n): 0 < d(p_n,p) < epsilon quad text{and} quad d(p_n,p) < d(p_m,p).
$$
I have concluded that in $mathbb{R}$ with the usual metric, the sequences {$2-frac1n$} and {$1-frac1n$} do not flirt with $1$.
Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?
real-analysis sequences-and-series metric-spaces real-numbers
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
$endgroup$
|
show 3 more comments
$begingroup$
I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:
In a metric space, a sequence ${P_n}_n$ $ $ flirts with $p$ iff for each $epsilon > 0$, there is a $n in mathbb{N}$ and $m > n$ such that $0 < d(p_n,p) < epsilon$ and $d(p_n,p) < d(p_m,p)$. The sequence ${P_n}_n$ $ $ is a flirting sequence if there is a point $p$ such that ${P_n}$ flirts with $p$.
Give an example of a sequence that flirts with $1$.
I am trying to $(1)$ understand the concept of "flirting" and $(2)$ figure out an example of a sequence that flirts with $1$. Actually, I would be happy to see an example of a any sequence that flirts with something.
I translated the definition like this:
In a metric space, the sequence ${P_n}$ flirts with $p$ iff
$$
forall epsilon > 0 quad exists(m,n in mathbb{N}, m >n): 0 < d(p_n,p) < epsilon quad text{and} quad d(p_n,p) < d(p_m,p).
$$
I have concluded that in $mathbb{R}$ with the usual metric, the sequences {$2-frac1n$} and {$1-frac1n$} do not flirt with $1$.
Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?
real-analysis sequences-and-series metric-spaces real-numbers
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
$endgroup$
1
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
$endgroup$
– lulu
Dec 13 '18 at 21:50
$begingroup$
Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
$endgroup$
– mcmath27
Dec 13 '18 at 21:56
1
$begingroup$
@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
$endgroup$
– lulu
Dec 13 '18 at 21:58
$begingroup$
Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
$endgroup$
– lulu
Dec 13 '18 at 22:00
$begingroup$
@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
$endgroup$
– mcmath27
Dec 13 '18 at 22:04
|
show 3 more comments
$begingroup$
I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:
In a metric space, a sequence ${P_n}_n$ $ $ flirts with $p$ iff for each $epsilon > 0$, there is a $n in mathbb{N}$ and $m > n$ such that $0 < d(p_n,p) < epsilon$ and $d(p_n,p) < d(p_m,p)$. The sequence ${P_n}_n$ $ $ is a flirting sequence if there is a point $p$ such that ${P_n}$ flirts with $p$.
Give an example of a sequence that flirts with $1$.
I am trying to $(1)$ understand the concept of "flirting" and $(2)$ figure out an example of a sequence that flirts with $1$. Actually, I would be happy to see an example of a any sequence that flirts with something.
I translated the definition like this:
In a metric space, the sequence ${P_n}$ flirts with $p$ iff
$$
forall epsilon > 0 quad exists(m,n in mathbb{N}, m >n): 0 < d(p_n,p) < epsilon quad text{and} quad d(p_n,p) < d(p_m,p).
$$
I have concluded that in $mathbb{R}$ with the usual metric, the sequences {$2-frac1n$} and {$1-frac1n$} do not flirt with $1$.
Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?
real-analysis sequences-and-series metric-spaces real-numbers
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
$endgroup$
I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:
In a metric space, a sequence ${P_n}_n$ $ $ flirts with $p$ iff for each $epsilon > 0$, there is a $n in mathbb{N}$ and $m > n$ such that $0 < d(p_n,p) < epsilon$ and $d(p_n,p) < d(p_m,p)$. The sequence ${P_n}_n$ $ $ is a flirting sequence if there is a point $p$ such that ${P_n}$ flirts with $p$.
Give an example of a sequence that flirts with $1$.
I am trying to $(1)$ understand the concept of "flirting" and $(2)$ figure out an example of a sequence that flirts with $1$. Actually, I would be happy to see an example of a any sequence that flirts with something.
I translated the definition like this:
In a metric space, the sequence ${P_n}$ flirts with $p$ iff
$$
forall epsilon > 0 quad exists(m,n in mathbb{N}, m >n): 0 < d(p_n,p) < epsilon quad text{and} quad d(p_n,p) < d(p_m,p).
$$
I have concluded that in $mathbb{R}$ with the usual metric, the sequences {$2-frac1n$} and {$1-frac1n$} do not flirt with $1$.
Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?
real-analysis sequences-and-series metric-spaces real-numbers
real-analysis sequences-and-series metric-spaces real-numbers
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
edited Dec 14 '18 at 0:28
mcmath27
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
asked Dec 13 '18 at 21:47
mcmath27mcmath27
84
84
1
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
$endgroup$
– lulu
Dec 13 '18 at 21:50
$begingroup$
Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
$endgroup$
– mcmath27
Dec 13 '18 at 21:56
1
$begingroup$
@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
$endgroup$
– lulu
Dec 13 '18 at 21:58
$begingroup$
Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
$endgroup$
– lulu
Dec 13 '18 at 22:00
$begingroup$
@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
$endgroup$
– mcmath27
Dec 13 '18 at 22:04
|
show 3 more comments
1
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
$endgroup$
– lulu
Dec 13 '18 at 21:50
$begingroup$
Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
$endgroup$
– mcmath27
Dec 13 '18 at 21:56
1
$begingroup$
@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
$endgroup$
– lulu
Dec 13 '18 at 21:58
$begingroup$
Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
$endgroup$
– lulu
Dec 13 '18 at 22:00
$begingroup$
@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
$endgroup$
– mcmath27
Dec 13 '18 at 22:04
1
1
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
$endgroup$
– lulu
Dec 13 '18 at 21:50
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
$endgroup$
– lulu
Dec 13 '18 at 21:50
$begingroup$
Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
$endgroup$
– mcmath27
Dec 13 '18 at 21:56
$begingroup$
Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
$endgroup$
– mcmath27
Dec 13 '18 at 21:56
1
1
$begingroup$
@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
$endgroup$
– lulu
Dec 13 '18 at 21:58
$begingroup$
@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
$endgroup$
– lulu
Dec 13 '18 at 21:58
$begingroup$
Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
$endgroup$
– lulu
Dec 13 '18 at 22:00
$begingroup$
Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
$endgroup$
– lulu
Dec 13 '18 at 22:00
$begingroup$
@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
$endgroup$
– mcmath27
Dec 13 '18 at 22:04
$begingroup$
@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
$endgroup$
– mcmath27
Dec 13 '18 at 22:04
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = begin{cases} 1 & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$ can you see why this is the case?
Note that
a sequence can flirt with $p$ but not converge to $p$: as in the above (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = begin{cases} frac{1}{2^n} & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
add a comment |
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1 Answer
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$begingroup$
Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = begin{cases} 1 & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$ can you see why this is the case?
Note that
a sequence can flirt with $p$ but not converge to $p$: as in the above (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = begin{cases} frac{1}{2^n} & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
add a comment |
$begingroup$
Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = begin{cases} 1 & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$ can you see why this is the case?
Note that
a sequence can flirt with $p$ but not converge to $p$: as in the above (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = begin{cases} frac{1}{2^n} & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
add a comment |
$begingroup$
Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = begin{cases} 1 & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$ can you see why this is the case?
Note that
a sequence can flirt with $p$ but not converge to $p$: as in the above (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = begin{cases} frac{1}{2^n} & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
$endgroup$
Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = begin{cases} 1 & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$ can you see why this is the case?
Note that
a sequence can flirt with $p$ but not converge to $p$: as in the above (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = begin{cases} frac{1}{2^n} & text{ if } n text{ is even}\
frac{1}{n} &text{ if } n text{ is odd}
end{cases}$$
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
answered Dec 13 '18 at 21:51
Clement C.Clement C.
50.4k33890
50.4k33890
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
add a comment |
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise.
$endgroup$
– mcmath27
Dec 13 '18 at 22:15
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
$begingroup$
@mcmath27 Glad this helped!
$endgroup$
– Clement C.
Dec 13 '18 at 22:16
add a comment |
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1
$begingroup$
It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence ${p_n}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements?
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– lulu
Dec 13 '18 at 21:50
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Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $forall epsilon quad d(p_n,1) = |p_n - 1| < epsilon$? Because if we consider $epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence?
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– mcmath27
Dec 13 '18 at 21:56
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@mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work.
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– lulu
Dec 13 '18 at 21:58
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Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well.
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– lulu
Dec 13 '18 at 22:00
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@lulu Ohhh I see what you mean! So for example, with $epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot?
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– mcmath27
Dec 13 '18 at 22:04