Define $A _n = \{x \in F ^c \cap A: d(x,F)\ge 1/n \} $, where $F $ is a closed subset, and $A$ any subset of a metric space $X $.
Then let $B _n =A _{n+1 } \cap (A _n ) ^c$
I have two questions:
1) How can $B _n $ be described? Is $B _n = \{x \in F ^c \cap A :d(x,F)< 1/n\} \cup F \cup A ^c$?
2) Is it true that $ d(B _{n+1 } ,A _n ) \ge \frac {1 } {n (n+1) }$?
Thanks in advance!
You have $$B_n = A_{n+1} \setminus A_n = \{x\in F^C \cap A : \frac1n > d(x, F) \ge \frac1{n+1}\}$$ Now $d(B_{n+1}, A_n) = d(A_{n+2} \setminus A_{n+1}, A_n)$. We know $$d(a, F) < \frac1{n+1}, a\in B_{n+1}$$ and $d(b,F) \ge \frac1n, b\in A_n$ so we can say that $$d(a,b) \le \frac1n + \frac1{n+1} = \frac{2n+1}{n(n+1)}$$ Thus $d(B_{n+1}, A_n) \le \frac{2n+1}{n(n+1)}$. Not sure about the $\ge$ estimate, though. I think in the general setting, we don't get their separation (think of $d(a,b) := 1-\delta_{ab}$ the dirac metric), but in $\mathbb R^n$ we may get that $d(B_{n+1}, A_n) \ge \frac1n-\frac1{n+1} = \frac1{n(n+1)}$.