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I am going through GF Simmons' Intro to Topology and Modern Analysis.

$F_n$ is a decreasing sequence of non-empty closed subsets of the Metric Space.

It seems that the condition $d(F_n)$->$0$ given in the hyothesis, ensures that the $$F=\bigcap_{n=1}^\infty F_n\neq\varnothing $$ rather F contains exactly one point is not very clear.

Why does "the supremum of the distances between any two points in the closed sets" converging to "$0$" implies that the set contains exactly one point?

Krishan
  • 178

1 Answers1

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It is because the $$d(A)=sup\{d(x,y)|x,y\in A \}$$ If this converges to $0$ it means that d(x,y) converges to $0$ for the class of closed sets $F_n$ as n->$\infty$.

We know that $d(x,y)=0$ only for x=y.Therefore there must exist only one point in the set $F=\bigcap_{n=1}^\infty F_n$.

Krishan
  • 178