I am reading John Conway's Function of One complex Varible 1:
In that I encountered following therem that Space with Nested Interval Property Iff It is complete .
I understand that one way form completeness $\to $ Nested Interval Property.
But in other way Around ,
I donot understand why $F_n$ is closed?
BEcause if its closed by assumption then it directly implies convergence of any cauchy sequnce directly.
Where I am missing?
Any Help will be appreciated
Best Answer
As xbh already pointed out in the comments, $F_n$ is not the set $\{x_n,x_{n+1},\ldots\}$, but the closure there of, as indicated by the uppercase dash (or macron?).
I do not understand what you mean by the remark that closed sets implies convergence. If you are implying that a decreasing chain of closed sets always has non-empty intersection, then this is incorrect. Take, as a counterexample, the sets $I_n = [n,\infty)$ for $n=1,2,\ldots$.
As for the proof, the idea is basically as follows. We want to show that $X$ is complete, which is equivalent to showing that every Cauchy sequence in $X$ converges. Try and verify for yourself that a given Cauchy sequence $(x_1,x_2,\ldots)$ converges in $X$ if and only if the intersection of the $F_n$ has a single point (it is precisely this point in the intersection that will be the point to which the Cauchy sequence converges).
Finally, let me note that the comments are referring to multiple limit points for whatever reason. As you already suspected, SRJ, Cauchy sequences always have at most one limit point (elementary exercise in analysis for the reader), so if a Cauchy sequence converges, it will always do so toward a single point.