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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?

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    $\begingroup$ 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
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    $\begingroup$ @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
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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 $p$: $$p_n = \begin{cases} \frac{1}{2^n} & \text{ if } n \text{ is even}\\ \frac{1}{n} &\text{ if } n \text{ is odd} \end{cases}$$

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  • $\begingroup$ 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
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Remember that a easy example of a flirting sequence is $~-1, ~1/2,~ -1/3,~ 1/4, ~-1/5, ~1/6,~$ etc.

Mainly if it is even it is positive, if it is odd it is negative, just like $~(-1)^n~.~$

I think the formula is $-1/n,~~ +1/(n+1)$

Even better, $-1/(2*(n)),~~ +1/(2*n+1)$.

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