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